Here are the key points about properties of matter from the passage: - Properties can be used to identify and classify substances. They distinguish one substance from another. - Physical properties are observed without changing the substance's composition, though its form may change. They include color, hardness, melting/boiling points, etc. Physical changes involve only physical properties. - Chemical properties are related to a substance's composition. They include flammability and reactions like electrolysis that produce new substances. Chemical changes produce new substances through chemical reactions. - Properties can be either intensive (independent of amount) like density, or extensive (dependent on amount) like mass and volume. - The passage gives examples of physical and chemical

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NOT FOR SALE
General Chemistry 1
Quarter 1
Properties of Matter and Its Various Forms
Department of Education • Republic of the Philippines
General Chemistry 1
Quarter 1 – Module 1
Properties of Matter and Its Various Forms
Department of Education • Republic of the Philippines
Senior High School
General Chemistry 1
Properties of Matter and Its Various Forms
Department of Education • Republic of the Philippines
Senior High School

2. General Chemistry 1 - Grade 11
Alternative Delivery Mode
Quarter 1 – Module 1
Properties of Matter and Its Various Forms
First Edition, 2020
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3. General
Chemistry 1
Quarter 1
Properties of Matter and Its Various Forms
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendation
at action@ deped.gov.ph.
We value your feedback and recommendation
General
Chemistry 1
Quarter 1 – Module 1
Properties of Matter and Its Various Forms
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at action@ deped.gov.ph.
We value your feedback and recommendation
Senior High School
Chemistry 1
Properties of Matter and Its Various Forms
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
s to the Department of Education
Senior High School

4. Department of Education ● Republic of the Philippines

5. Table of Contents
What This Module is About ............................................................................................... i
What I Need to Know ........................................................................................................ i
How to Learn from this Module ........................................................................................ ii
Icons of this Module ......................................................................................................... ii
What I Know ................................................................................................................... iii
Lesson 1.1:
Properties of Matter ...................................................................................................... 1
What I Need to Know ................................................................................ 1
What’s New............................................................................................... 1
What Is It................................................................................................... 2
What’s More ............................................................................................. 3
What Is It................................................................................................... 3
What’s More (1)......................................................................................... 4
What’s More (2)......................................................................................... 5
What I Have Learned ................................................................................ 6
Lesson 4.2:
Common Chemical Substances......................................................................... 7
What’s In................................................................................................... 7
What’s New............................................................................................... 7
What Is It................................................................................................. .. 8
What’s More............................................................................................ .. 8
What’s In.................................................................................................... 9
What’s More............................................................................................... 10
What I Have Learned ................................................................................. 11
Summary …………………………………………………………………………………………..12
Assessment: (Post-Test) ………………………………………………………………………….11
Key to Answers.............................................................................................................. …13
Reference …………………………………………………………………………………….........14

6. i
What This Module is About
This module discusses properties of matter and Its various forms, recognizing
common chemical substances, as well as comparing consumer products on the
basis of their components for use, safety, quality, and cost
This module has 2 lessons:
1. Properties of Matter
2. Common Chemical Substances
You are expected to answer and complete the activities given in each lesson.
Strictly follow the instructions in each activity. You may write your answers on the
answer sheets provided.
What I Need to Know
After going through this module, you are expected to;
1. Use properties of matter to identify substances and to separate them
(STEM_GC11MPIa-b-5)
2. Recognize the formulas of common chemical substances (STEM_GC11MPIa-b-9)
3. Compare consumer products on the basis of their components for use, safety,
quality and cost (STEM_GC11MPIa-b-11)
4. Describe various simple separation techniques such as distillation,
chromatography (STEM_GC11MPIa-b-12)

7. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that are in
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what
Learned have learned from the lesson
What I can do These are tasks that are designed to show
case your skills and knowledge gained, and
applied into real-life concerns and situations.
Follow the directions and/or instructions in the activities and exercises diligently.
This part contains learning objectives that
are set for you to learn as you go along the
This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
connects previous lesson with that
An introduction of the new lesson through
various activities, before it will be presented
These are discussions of the activities as a
deepen your discovery and under-
up activities that are in-
tended for you to practice further in order to
Activities designed to process what you
have learned from the lesson
These are tasks that are designed to show-
case your skills and knowledge gained, and
life concerns and situations.

8. iii
What I Know
Pre Assessment
Directions: Choose the best answer among the choices. Encircle the letter of your
answer.
1. Which process is a chemical change?
A. heating to boiling C. slicing into two pieces
B. burning in air D. dissolving in alcohol
2. Which of the following is an example of an element?
A. soil C. water
B. sugar D. oxygen
3. Which is the most suitable separation method can be used to separate a mixture of
different colored ink?
A. chromatography C. evaporation
B. distillation D. filtration
4. What is matter?
A. building blocks to make up organisms C. molecules that makes up all things
B. anything that is measured and seen D. none of the above
5. Which is the most suitable separation method can be used to separate the mixture of
liquids in crude oil?
A. distillation C. chromatography
B. evaporation D. fractional distillation
6. Which of the following is an example of an extensive property?
A. mass C. density
B. color D. boiling point
7. What is the chemical formula of chalk?
A. CaCO3 C. CaCO
B. COOH D. CaCO2
8. Tony Stark, a consumer with light colored hair, would like to buy shampoo but he saw
that it contains high amounts of sulfate. What should Tony do?
A. buy the shampoo C. find a sulfate free shampoo
B. test the shampoo D. find alternatives to shampoo
9. Glass breaking is an example of..?
A. chemical property C. chemical change
B. physical property D. physical change
10.What is one of the importance of knowing the compositions of a product?
A. so we can save money C. so we can use it safely
B. so we can know the quality D. all of the choices

9. 1
Lesson
Properties of Matter
1.1
What I Need to Know
What’s the difference between silver and its alloy? Why does copper conduct
electricity? Why is alcohol volatile? This unit will help you answer these questions
and understand the composition and properties of matter and the changes it
undergoes.
This module discusses about the properties of matter to identify substances
and to separate them (STEM_GC11MPIa-b-5). We will also try to describe various
simple separation techniques such as distillation, chromatography
(STEM_GC11MPIa-b-12).
What’s New
Activity 1: Look and List
Directions: Get only one object that holds a value to you. On the spaces below, write the
different qualitative and quantitative qualities about your chosen object. An example is
shown below:
Object: cellphone
Qualitative Quantitative
Color black, with pink casing, with tempered
glass, yellow wallpaper, made of alloy,
Oppo
3000mAh, 32gb, 4gb ROM, 4mp front
camera, 12mp back camera, 600 pictures,
43 videos
Object: ___________________
Qualitative Quantitative

10. What Is It
Each substance has its own set of characteristics or properties that distinguishes it
from all other substances. Pepper is pungent while sugar is sweet. Sand is grainy. Glass is
breakable. Wood burns. Clothes can be folded. Apple and potato sl
air, turn brown. In these examples, no matter what size or shape these substances have,
their particular property does not change.
By examining materials, you can find similarities and differences in their properties. This will
enable you to organize them into groups.
Properties of Matter
Physical properties
composition of the substance, though their form may change. Certain properties such as
color, viscosity, transparency, melti
and atomic or molecular diameter are usually referred to as physical properties. These
properties are involved in a physical change.
A physical change is observed when a pinch of salt is dissolved
produce a clear salt solution. The salt changes only in form or state from solid to liquid. It can
be recovered from a salt solution by boiling or evaporating it. The salt has not changed in
composition but remains as sodium chloride
Some physical properties cannot be measured (qualitative) while others can be
(quantitative).
Chemical properties
This property is related to the composition of the material. The flammabi
be determined by burning it, producing carbon dioxide. When pure water undergoes
electrolysis (a process by which electric current passes through water), water decomposes
into hydrogen and oxygen. In these examples, a chemical change t
completely different substances. It results in the formation of new chemical substances. A
chemical reaction is usually detected when one of the following is observed: formation of an
insoluble product (precipitate), evolution of gas
Table 1.1 lists some physical and chemical properties of a substance
the physical property compare with its chemical property?
Physical Properties
Crystals are colorless needles
74°C
Insoluble in water but soluble in ethyl alcohol
Density, 0.883g/Ml
Table 1.1 Properties of Lauric Acid
OTHER PROPERTIES
Properties of matter may be used to describe them. An
property that changes when the amount of material changes. Examples are mass, length,
and volume. An intensive property
2
Each substance has its own set of characteristics or properties that distinguishes it
from all other substances. Pepper is pungent while sugar is sweet. Sand is grainy. Glass is
breakable. Wood burns. Clothes can be folded. Apple and potato slices, when exposed to
air, turn brown. In these examples, no matter what size or shape these substances have,
their particular property does not change.
By examining materials, you can find similarities and differences in their properties. This will
you to organize them into groups.
Physical properties are properties that are observed without changing the
composition of the substance, though their form may change. Certain properties such as
color, viscosity, transparency, melting point, boiling point, hardness, density, specific heat,
and atomic or molecular diameter are usually referred to as physical properties. These
properties are involved in a physical change.
A physical change is observed when a pinch of salt is dissolved in a cup of water to
produce a clear salt solution. The salt changes only in form or state from solid to liquid. It can
solution by boiling or evaporating it. The salt has not changed in
composition but remains as sodium chloride (NaCl).
Some physical properties cannot be measured (qualitative) while others can be
Chemical properties ae observed when matter is involved in a chemical change.
This property is related to the composition of the material. The flammabi
be determined by burning it, producing carbon dioxide. When pure water undergoes
electrolysis (a process by which electric current passes through water), water decomposes
into hydrogen and oxygen. In these examples, a chemical change takes place to form two
completely different substances. It results in the formation of new chemical substances. A
chemical reaction is usually detected when one of the following is observed: formation of an
insoluble product (precipitate), evolution of gas (bubbles), or change in color.
Table 1.1 lists some physical and chemical properties of a substance-lauric acid. How does
the physical property compare with its chemical property?
Physical Properties Chemical Properties
Crystals are colorless needles and melt at Produces soap when combined with sodium
hydroxide
Insoluble in water but soluble in ethyl alcohol Combines with some medicines for better
absorption by the blood
Table 1.1 Properties of Lauric Acid
roperties of matter may be used to describe them. An extensive property
property that changes when the amount of material changes. Examples are mass, length,
intensive property does not depend on the size of the material.
Each substance has its own set of characteristics or properties that distinguishes it
from all other substances. Pepper is pungent while sugar is sweet. Sand is grainy. Glass is
ices, when exposed to
air, turn brown. In these examples, no matter what size or shape these substances have,
By examining materials, you can find similarities and differences in their properties. This will
are properties that are observed without changing the
composition of the substance, though their form may change. Certain properties such as
ng point, boiling point, hardness, density, specific heat,
and atomic or molecular diameter are usually referred to as physical properties. These
in a cup of water to
produce a clear salt solution. The salt changes only in form or state from solid to liquid. It can
solution by boiling or evaporating it. The salt has not changed in
Some physical properties cannot be measured (qualitative) while others can be
ae observed when matter is involved in a chemical change.
This property is related to the composition of the material. The flammability of gasoline can
be determined by burning it, producing carbon dioxide. When pure water undergoes
electrolysis (a process by which electric current passes through water), water decomposes
akes place to form two
completely different substances. It results in the formation of new chemical substances. A
chemical reaction is usually detected when one of the following is observed: formation of an
(bubbles), or change in color.
lauric acid. How does
Chemical Properties
Produces soap when combined with sodium
Combines with some medicines for better
extensive property is a
property that changes when the amount of material changes. Examples are mass, length,
does not depend on the size of the material.

11. Temperature, color, odor, hardness, density, melting and boiling points, and molecular
weight are examples of intensive properties.
The characteristics of a substance, regardless of its shape and size, are cal
intrinsic properties. Color, viscosity, taste, and transparency are examples of
properties that cannot be expressed in numbers. Intrinsic properties assigned with definite
values are boiling point, melting point, density, and refractive index
substance which pertain only to its appearance including is shape, length, mass, and
temperature are called extrinsic properties
What’s More
Activity 2. Physical Property or Chemical Property
Directions: On the spaces provided before each number, write
physical property and write C
____1. Frost forms as the temperature drops on a
____2. A cornstalk grows from a seed that is watered and fertilized.
____3. A match ignites to form ash and a mixture of gases.
____4. Perspiration evaporates when you relax after jogging.
____5. A silver fork tarnishes slowly in
____6. A scab forms over an open cut.
____7. Paper was cut into different sizes
____8. Gasoline fumes are ignited by a spark in a car’s engine cylinder.
____9. Purple iodine vapor appears when solid iodine is warmed.
____10. Electric current decompose
oxygen).
____11. Yellow-green chlorine gas attacks silvery sodium metal to form white crystals of
sodium chloride (table salt.
____12. A magnet separates a mixture of black iron shavings and white sand.
____13. Ice cream melting
____14. An egg turning hard when it is boiled.
____15. Passing an electric current through molten magnesium chloride which yields molten
magnesium and gaseous chlorine.
What Is It
Knowing about a material’s physical and chemical property in order to correctly choose the
most efficient separation technique.
Mixtures are physical combinations of two or more substances. They can be
separated by physical processes. The method of separ
3
Temperature, color, odor, hardness, density, melting and boiling points, and molecular
weight are examples of intensive properties.
The characteristics of a substance, regardless of its shape and size, are cal
intrinsic properties. Color, viscosity, taste, and transparency are examples of
that cannot be expressed in numbers. Intrinsic properties assigned with definite
values are boiling point, melting point, density, and refractive index. The characteristics of a
substance which pertain only to its appearance including is shape, length, mass, and
extrinsic properties.
What’s More
Property or Chemical Property
On the spaces provided before each number, write P if it is a substance’s
if it involves a chemical property. (2 points each number).
____1. Frost forms as the temperature drops on a humid winter night
____2. A cornstalk grows from a seed that is watered and fertilized.
____3. A match ignites to form ash and a mixture of gases.
____4. Perspiration evaporates when you relax after jogging.
____5. A silver fork tarnishes slowly in air.
____6. A scab forms over an open cut.
____7. Paper was cut into different sizes
____8. Gasoline fumes are ignited by a spark in a car’s engine cylinder.
____9. Purple iodine vapor appears when solid iodine is warmed.
____10. Electric current decomposes water into different substances (hydrogen and
green chlorine gas attacks silvery sodium metal to form white crystals of
____12. A magnet separates a mixture of black iron shavings and white sand.
____14. An egg turning hard when it is boiled.
____15. Passing an electric current through molten magnesium chloride which yields molten
magnesium and gaseous chlorine.
Knowing about a material’s physical and chemical property in order to correctly choose the
most efficient separation technique.
Mixtures are physical combinations of two or more substances. They can be
separated by physical processes. The method of separating a mixture into its components
Temperature, color, odor, hardness, density, melting and boiling points, and molecular
The characteristics of a substance, regardless of its shape and size, are called
intrinsic properties. Color, viscosity, taste, and transparency are examples of intrinsic
that cannot be expressed in numbers. Intrinsic properties assigned with definite
. The characteristics of a
substance which pertain only to its appearance including is shape, length, mass, and
if it is a substance’s
if it involves a chemical property. (2 points each number).
s water into different substances (hydrogen and
green chlorine gas attacks silvery sodium metal to form white crystals of
____12. A magnet separates a mixture of black iron shavings and white sand.
____15. Passing an electric current through molten magnesium chloride which yields molten
Knowing about a material’s physical and chemical property in order to correctly choose the
Mixtures are physical combinations of two or more substances. They can be
ating a mixture into its components

12. 4
depends primarily on the properties of each of the components. The following are the more
common methods of separating a mixture:
1. Distillation is used to separate a mixture containing volatile components. This involves
evaporation followed by condensation. Simple distillation is used when the liquid
components in the solution have widely different boiling points or when a dissolved solid
remains in the distilling flask as its solvent distills off. For mixtures of liquids when boiling
points vary within a small range, fractional distillation is recommended. Components of crude
oil are separated into fractions through this process. Petroleum products such as LPG,
kerosene, gasoline, bunker fuel oil, and asphalt are fractions from crude oil.
2. Chromatography is used to separate components from a mixture based on differences in
attraction of these components for a stationary phase (a phase that is immobile) and mobile
phase (a phase where it flows). Other types of chromatography include resin, ion-exchange,
and paper chromatography.
What’s More (1)
Activity 3: Other Separation Methods
Directions: Using different resources such as the internet, books, articles, journals, and
textbooks, research on other separation methods used in chemistry. Write the information
gathered below. (5 points each item)
Separation Methods Purpose Examples where these
methods are applied
Decantation
Filtration
Evaporation

13. 5
What’s More (2)
Activity 4: Chromatography
Directions: you will perform basic chromatography. Prepare all your materials beforehand.
Pass this activity with the deadline indicated by the teacher.
Materials: ordinary filter paper, water-based ink pen (black), Erlenmeyer flask (if available) or
mini coke plastic container (substitute)
Procedure:
1. Cut a small strip of filter paper long enough to reach the bottom of the container or
Erlenmeyer flask while folding about 1cm over the lip. The strip should be about 1” (2.5cm)
wide.
2. Using a water-based ink pen, make a small dot about 0.5” (1.3cm) from one end of the
filter paper.
3. Add enough water to the container to over the bottom of the filter paper but not enough to
reach the ink dot.
4.Place the filter paper in the container with the dotted end facing down.
5. observe and describe what happens to the ink dot.
Questions:
1. What colors rose from the black ink?
_________________________________________________________________________
_________________________________________________________________________
2. If you were to perform the activity again but this time using an ink of different color (e.g.
blue or red), would you still obtain the same results? Explain (You may try this procedure
using a different ink color to find out)
_________________________________________________________________________
_________________________________________________________________________
3. What practical applications does chromatography have?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
Illustration of the setup

14. What I Have Learned
Activity 4: Assessing Your Knowledge
Part A: Which of the following properties are extrinsic and intrinsic? Write E if
extrinsic and I if they are intrinsic.
____1. Temperature
____2. Volume
____3. Boiling temperature
____4. Viscosity
____5. Weight
B. Answer the following questions briefly.
1. Describe how you will separate the following components of the following mixture:
a. palay husk from the grain
_________________________________________________________________________
___________________________________________
_______________________________________________________
b. dissolved dye from water
_________________________________________________________________________
___________________________________________________
_______________________________________________________
6
What I Have Learned
Activity 4: Assessing Your Knowledge
: Which of the following properties are extrinsic and intrinsic? Write E if
extrinsic and I if they are intrinsic.
____6. Density
____7. Mass
____8. Specific gravity
____9. Hardness
____10. Length
owing questions briefly.
1. Describe how you will separate the following components of the following mixture:
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
: Which of the following properties are extrinsic and intrinsic? Write E if they are
1. Describe how you will separate the following components of the following mixture:
_________________________________________________________________________
______________________________
__________________
_________________________________________________________________________
______________________
__________________

15. Lesson
Common Chemical Substances
1.2
What’s In
Previously, we have learned how the
its composition. We have also discussed separation techniques in different mixtures
Now, as we’re already talking about
common chemical substances (
them as well as compare consumer products on the basis of their components for
use, safety, quality, and cost (
What’s New
Activity 2.1 Name That Element!
Directions: Using your periodic table or your prior knowledge, name
in the periodic table.
Element
Symbol
Element Name
1 Cl
2 Sn
3 Ti
4 V
5 Sb
6 Kr
7 Xe
7
Common Chemical Substances
Previously, we have learned how the different properties of matter as well as
its composition. We have also discussed separation techniques in different mixtures
Now, as we’re already talking about substances and mixtures, let’s take a look at
common chemical substances (STEM_GC11MPIa-b-9) and we will try to recog
them as well as compare consumer products on the basis of their components for
use, safety, quality, and cost (STEM_GC11MPIa-b-1)
Activity 2.1 Name That Element!
: Using your periodic table or your prior knowledge, name the following elements
Element Name Element
Symbol
Element Name
8 Ir
9 Mg
10 Be
11 Rn
12 F
13 B
14 Tl
Common Chemical Substances
matter as well as
its composition. We have also discussed separation techniques in different mixtures.
substances and mixtures, let’s take a look at
and we will try to recognize
them as well as compare consumer products on the basis of their components for
the following elements
Element Name

16. 8
What Is It
Matter can be classified into three types based on its composition – elements, compounds,
and mixtures. Elements are two kinds of substances: a substance is matter whose
composition is fixed. Mixtures are not substances because they have a variable composition
1. Elements. An element is the simplest type of matter with unique physical and chemical
properties. It consists of only one kind of atom, and, therefore, cannot be broken down into a
simpler type of matter by any physical or chemical methods. Each element has a name,
such as carbon, fluorine, or oxygen. A sample of fluorine contains only fluorine atoms.
In nature, most elements exist as population of atoms, either separated or in contact
with each other, depending on the physical state. Several elements occur in molecular form:
molecule is an independent structure of two or more atoms bound together. Oxygen, for
example, occurs in air as diatomic (two atom) molecules.
2. Compounds. Compounds are substances formed when two or more elements combine
through a chemical change. Sodium chloride, sugar, and water are examples of compounds.
Another feature of a compound is that its properties are different from the properties of its
compound elements. Example: soft, silvery sodium metal and yellow-green, poisonous
chlorine gas are very different from the compound they form- white, crystalline sodium
chloride, or table salt!
Unlike an element, a compound can be broken down into simpler substances. For
example, am electric current breaks down molten sodium chloride into metallic sodium and
chlorine gas.
3. Mixtures. A mixture consists of two or more substances (elements and/or compounds)
that are mixed together. Because a mixture is NOT a substance, the components of a
mixture can vary in their parts by mass. For example, a mixture of the compounds sodium
chloride and water can have different parts by mass of salt to water. A mixture also retains
many of the properties of its components.
What’s More (1)
Activity 2.2 Research It!
Directions: Using various search engines, books, and the Internet, research the name or
the formula of the different compounds as well as its uses. #1 will serve as an example.
Common
Compounds
(Formula)
Name Uses
1 H2O Water or dihydrogen oxide Most common solvent, we use
it everyday life as we take a
bath, cook our food, and
sustain life
2 SO3

17. 9
3 CH4
4 NH3
5 N2O
6 K2SO4
7 H2O2
8 Hydrofluoric acid
9 Hypobromous acid
10 NO
11
MnSO4
12 Perchlorate
13 Cyanide
14 BaO2
15 NaNO2
16 CaCO3
What’s In
CHEMISTRY IN INDUSTRY
Many science principles are applied in industries. Industrialization not only uplifts the
quality of human existence, but also propels nation to higher levels of economic prosperity.
Many pure substances and mixtures, organic or inorganic in nature, are now
commercially manufactured and used by school and government laboratories, households,
or industries as raw materials for intermediary or final products. They undergo hundreds of
test before they are sold in the market.
Substances may be pure or impure, knowledge of the properties of the components
in the mixture facilitates the manufacture of products to a certain degree of purity. For
example, impure mineral ores are processed into useful metals.

18. Materials exist in gas, solid, or liquid states. Packing them may pose a problem. Materials
used as containers are carefully chosen to ensure that no harmful reactions will take place between
the content and the container. The container must be free
What’s More
Activity 2.3 Becoming a Better Consumer
Directions: Research or read on the different components or ingredients of the following
items. On the third column, decide whether it is safe, has good quality, or for
Product or
Item
Components/Ingredients
1. junk food Salt, corn, monosodium
glutamate (vetsin)
2. soft drink
3 lollipop
4 deodorant
5 detergent
6 shampoo
Guide Questions:
1. Why is it important to know the components or ingredients of the items/products that we
use every day?
_________________________________________________________________________
___________________________________________________
_______________________________________________________
2. What harmful components have you find out from the items above? Explain.
_________________________________________________________________________
_________________________________________________________________________
_______________________________________________________
3. What other options do you have i
_________________________________________________________________________
_________________________________________________________________________
_______________________________________________________
10
Materials exist in gas, solid, or liquid states. Packing them may pose a problem. Materials
used as containers are carefully chosen to ensure that no harmful reactions will take place between
the content and the container. The container must be free from outside contamination.
What’s More (2)
Becoming a Better Consumer
Research or read on the different components or ingredients of the following
items. On the third column, decide whether it is safe, has good quality, or for
Components/Ingredients Safe? Good
Quality?
Salt, corn, monosodium
glutamate (vetsin)
X X
1. Why is it important to know the components or ingredients of the items/products that we
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
2. What harmful components have you find out from the items above? Explain.
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
3. What other options do you have in mind if some of the products have harmful to use?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
Materials exist in gas, solid, or liquid states. Packing them may pose a problem. Materials
used as containers are carefully chosen to ensure that no harmful reactions will take place between
from outside contamination.
Research or read on the different components or ingredients of the following
items. On the third column, decide whether it is safe, has good quality, or for a good price.
Good
price?
Will I
buy?
̸ No
1. Why is it important to know the components or ingredients of the items/products that we
_________________________________________________________________________
______________________
__________________
2. What harmful components have you find out from the items above? Explain.
_________________________________________________________________________
_________________________________________________________________________
__________________
n mind if some of the products have harmful to use?
_________________________________________________________________________
_________________________________________________________________________
_______________

19. What I Have Learned
Activity 2.4 Let’s Sum It Up!
Direction: Make a flash card of the information
Attach a minimum of 10 flash cards in your answer sheet
creative as you can be.
Materials: Short bond paper cut in 1/8 size crosswise.
Procedure: The front portion of the flash card s
in this module and at the back of the card, wr
This activity is essential in helping you retain the knowledge you acquired! Write
remember!
Post Assessment
Directions: Choose the best answer among the choices. Encircle the letter of your
answer.
1. What is matter?
A. building blocks to make up organisms
B. anything that is measured and seen
2. Which is the most suitable separation method can be used to separate the mixture of
liquids in crude oil?
A. distillation
B. evaporation
3. Which is the most suitable separation method can be used to separate a mixture of
different colored ink?
A. chromatography
B. distillation
4. which process is a chemical change?
A. heating to boiling
B. burning in air
5. Which of the following is an example of an element?
A. soil
B. sugar
11
What I Have Learned
Let’s Sum It Up!
Make a flash card of the information in this module that struck you the most.
a minimum of 10 flash cards in your answer sheet when you pass it. Remember, be
Short bond paper cut in 1/8 size crosswise.
The front portion of the flash card should contain a question about
and at the back of the card, write the answer of the question
This activity is essential in helping you retain the knowledge you acquired! Write
Post Assessment
Choose the best answer among the choices. Encircle the letter of your
blocks to make up organisms C. molecules that makes up all things
that is measured and seen D. none of the above
2. Which is the most suitable separation method can be used to separate the mixture of
C. chromatography
D. fractional distillation
3. Which is the most suitable separation method can be used to separate a mixture of
C. evaporation
D. filtration
h process is a chemical change?
C. slicing into two pieces
D. dissolving in alcohol
5. Which of the following is an example of an element?
C. water
D. oxygen
in this module that struck you the most.
when you pass it. Remember, be
hould contain a question about the lessons
ite the answer of the question you wrote.
This activity is essential in helping you retain the knowledge you acquired! Write it down to
Choose the best answer among the choices. Encircle the letter of your
molecules that makes up all things
2. Which is the most suitable separation method can be used to separate the mixture of
D. fractional distillation
3. Which is the most suitable separation method can be used to separate a mixture of
C. slicing into two pieces
D. dissolving in alcohol

20. 12
6. Which of the following is an example of an extensive property?
A. mass C. density
B. color D. boiling point
7. glass breaking is an example of..?
A. chemical property C. chemical change
B. physical property D. physical change
8. Tony Stark, a consumer with light colored hair, would like to buy shampoo but he saw that
it contains high amounts of sulfate. What should Tony do?
A. buy the shampoo C. find a sulfate free shampoo
B. test the shampoo D. find alternatives to shampoo
9. What is the chemical formula of chalk?
A. CaCO3 C. CaCO
B. COOH D. CaCO2
10.What is one of the importance of knowing the compositions of a product?
A. so we can save money C. so we can use it safely
B. so we can know the quality D. all of the choices
SUMMARY:
 Physical properties are properties not responsible for any change in the
composition of a substance, though its form may change.
 Chemical properties involve chemical change which forms two completely
different substances.
 Extensive property is that which changes when the amount of material
undergoes change. Examples of this property are mass, length, and volume.
 Intensive property does not depend on the size or amount of the substance.
Examples of this property are temperature, color, odor, hardness, density,
melting and boiling point, and molecular weight.

21. Key to Answers
LESSON 1
Activity 2
1.P 2. C 3. C 4. P 5. C 6. C 7. P 8. C 9. A 10. C 11. C 12. P 13. P 14. P 15. C
Activity 4
1. E 2. E 3. E 4. I 5. E 6. E 7.E 8. E 9. I 10.
LESSON 2
Activity 1
1. chlorine 2. Tin 3. Titanium 4. Vanadium 5. Antimony 6. Krypton 7. Xenon 8. Iridium 9.
Magnesium 10. Beryllium 11. Radon 12. Fluorine 13. Boron 14. Thallium
Activity 2
Formula
2 SO3 Sulfite
3 CH4 Methane
4 NH3 Ammonium
5 N2O Dinitrogen oxide
6 K2SO4 Potassium sulfate
7 H2O2 Hydrogen peroxide
8 HF Hydrofluoric acid
9 HBrO Hypobromous acid
10 NO Nitrogen oxide
11 MnSO4 Manganese sulfate
12 ClO4 Perchlorate
13 BaO2 Barium oxide
14 NaNO2 Sodium nitrate
15 CaCO3 Calcium carbonate
Pretest
1 B
2 D
3 A
4 C
5 D
6 D
7 A
13
Key to Answers
1.P 2. C 3. C 4. P 5. C 6. C 7. P 8. C 9. A 10. C 11. C 12. P 13. P 14. P 15. C
1. E 2. E 3. E 4. I 5. E 6. E 7.E 8. E 9. I 10. E
1. chlorine 2. Tin 3. Titanium 4. Vanadium 5. Antimony 6. Krypton 7. Xenon 8. Iridium 9.
Magnesium 10. Beryllium 11. Radon 12. Fluorine 13. Boron 14. Thallium
Name
Ammonium
Dinitrogen oxide
Potassium sulfate
Hydrogen peroxide
Hydrofluoric acid
Hypobromous acid
Nitrogen oxide
Manganese sulfate
Perchlorate
Barium oxide
Sodium nitrate
Calcium carbonate
Post Test
1C
2D
3A
4B
5D
6D
7D
1.P 2. C 3. C 4. P 5. C 6. C 7. P 8. C 9. A 10. C 11. C 12. P 13. P 14. P 15. C
1. chlorine 2. Tin 3. Titanium 4. Vanadium 5. Antimony 6. Krypton 7. Xenon 8. Iridium 9.

22. 14
8 C 8C
9 D 9A
10 D 10D
REFERENCES:
1. Echija, Elena, Cecilia Bayquen, Rafeal Alfonso, and Elmarita De Vera.
2020. Frontliners in Science and Technology. Makati City: Diwa
Scholastic Press
2. Silberberg, Martin. 2016. General Chemistry 1&2. McGrawHill Education.

23. 15
For inquiries and feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)

24. 16

25. Government Property
NOT FOR SALE
General Chemistry 1
Quarter 1
Isotopes, Naming Chemical Compounds,
and Calculating Empirical Formula
Department of Education • Republic of the Philippines
General Chemistry 1
Quarter 1 – Module 2
Isotopes, Naming Chemical Compounds,
and Calculating Empirical Formula
Department of Education • Republic of the Philippines
Senior High School
Photo credit: https://bit.ly/2Zl1XfR
General Chemistry 1
Isotopes, Naming Chemical Compounds,
and Calculating Empirical Formula
Department of Education • Republic of the Philippines
Senior High School

26. General Chemistry 1 - Grade 11
Alternative Delivery Mode
Quarter 1 - Module 2
Isotopes, Naming Chemical Compounds, and Calculating Empirical Formula
First Edition, 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any
work of the Government of the Philippines. However, prior approval of the
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exploitation of such work for profit. Such agency or office may, among other things,
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copyright holders. Every effort has been exerted to locate and seek permission to
use these materials from their respective copyright owners. The publisher and
authors do not represent nor claim ownership over them.
Published by the Department of Education – Division of Cagayan de Oro
Schools Division Superintendent: Dr. Cherry Mae L. Limbaco, CESO V
Development Team of the Module
Author: Marian Grace C. Esmade
Reviewers: Jean S. Macasero, EPS - Science
Management Team
Chairperson: Cherry Mae L. Limbaco, PhD, CESO V
Schools Division Superintendent
Co-Chairpersons: Alicia E. Anghay, PhD, CESE
Asst. Schools Division Superintendent
Members Lorebina C. Carrasco, OIC-CID Chief
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Printed in the Philippines by
Department of Education – Bureau of Learning Resources (DepEd-BLR)
Office Address: Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
Telefax: (08822)855-0048
E-mail Address: cagayandeoro.city@deped.gov.ph

27. General
Chemistry 1
Quarter 1
Isotopes, Naming Chemical Compounds, and
Calculating Empirical Formula
This instructional material was collaboratively developed and reviewed
by educators from public and private
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at action@ deped.gov.ph.
We value your feedback and re
Department of Education
General
Chemistry 1
Quarter 1 – Module 2
Isotopes, Naming Chemical Compounds, and
Calculating Empirical Formula
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at action@ deped.gov.ph.
We value your feedback and recommendation
Department of Education ● Republic of the Philippines
Senior High School
Chemistry 1
Isotopes, Naming Chemical Compounds, and
Calculating Empirical Formula
This instructional material was collaboratively developed and reviewed
schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
● Republic of the Philippines
Senior High School

28. Table of Contents
What This Module is About ............................................................................................... i
What I Need to Know ........................................................................................................ i
How to Learn from this Module ........................................................................................ ii
Icons of this Module ......................................................................................................... ii
What I Know ................................................................................................................... iii
Lesson 1:
Isotopes ................................................................................................................................... 1
What I Need to Know ................................................................................ 1
What’s In................................................................................................... 1
What’s New............................................................................................... 2
What Is It................................................................................................... 2
What’s More 1........................................................................................... 3
What’s More 2........................................................................................... 3
Lesson 2:
Naming Compounds..................................................................................................... 5
What’s In................................................................................................... 5
What Is It................................................................................................... 5
What’s More............................................................................................ .. 6
What Is It................................................................................................. .. 7
What’s More 2............................................................................................ 7
What’s More 3............................................................................................ 8
What I Can Do............................................................................................ 8
Lesson 3:
Calculating Empirical Formula............................................................................ 9
What I Need to Know ................................................................................ 9
What’s In................................................................................................... 9
What’s New............................................................................................. .. 10
What Is It................................................................................................. .. 10
What’s More .............................................................................................. 11
What I Have Learned ................................................................................. 12
Summary …………………………………………………………………………………………..13
Assessment: (Post-Test) ………………………………………………………………………….12
Key to Answers.............................................................................................................. …14
Reference …………………………………………………………………………………….........14

29. i
What This Module is About
This module
This module has 3 lessons:
1. Isotopes
2. Naming Compounds
3. Calculating the empirical formula
You are expected to answer and complete the activities given in each lesson.
Strictly follow the instructions in each activity. You may write your answers on the
answer sheets provided.
What I Need to Know
After going through this module, you are expected to;
1. Recognize common isotopes and their uses STEM_GC11AMIc-e-19
2. Represent compounds using chemical formulas, structural formulas and models
STEM_GC11AMIc-e-21
3. Name compounds given their formula and write formula given the name of the compound
STEM_GC11AMIc-e-23
4. Calculate the empirical formula from the percent composition of a compound
STEM_GC11PCIf-32

30. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that are in
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show
case your skills and knowledge gained, and
applied into real-life concerns and situations.
Follow the directions and/or instructions in the activities and exercises diligently.
This part contains learning objectives that
to learn as you go along the
This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
This part connects previous lesson with that
An introduction of the new lesson through
various activities, before it will be presented
These are discussions of the activities as a
way to deepen your discovery and under-
up activities that are in-
tended for you to practice further in order to
Activities designed to process what you
lesson
These are tasks that are designed to show-
case your skills and knowledge gained, and
life concerns and situations.

31. iii
Pre Assessment
Direction: Match the term in column A to the phrase that describes it in Column B.
Write the letter of your answer on the space provided before each term.
COLUMN A COLUMN B
_______ 1. Compound A. number of protons and neutrons in the
nucleus
_______ 2. Nucleus B. positively charged particle
_______ 3. Neutron C. small but dense core of the atom
_______ 4. Proton D. negatively charged particle
_______ 5. Shells E. energy states in which electrons can exist
______ 6. Molecules F. combination of two or more elements
______ 7. Ionic bond G. used to represent a compound
______ 8. Metallic bond H. particle with no charge
______ 9. Covalent bond I. number of protons in the nucleus
______ 10. Chemical formula J. element having the same atomic number but
different atomic mass

32. 1
Lesson
Isotopes
1
What I Need to Know
In this first lesson, you will recognize common isotopes and their uses (STEM_GC11AMIc-e-
19) by reading the short description and answering the activities given. Enjoy learning!
What’s In
Review on Protons, Neutrons, and Electrons
Atoms of an element have a constant or fixed number of protons. The atomic number gives
the number of protons in the nucleus of the atom. For the neutral atom, the number of
protons is equal to the number of its electrons. The atomic number is often represented by
the symbol Z.
Z = nuclear charge = number of protons = number of electrons in neutral form
The mass number is represented by the symbol A. The sum of the number of protons and
neutrons is called the mass number. An atom may be represented by the nuclear symbol A
ZE
where:
E = symbol of the element
A = mass number
Z = atomic number
A = number of protons (Z) + number of neutrons
The protons and neutrons are collectively called nucleons.
Let us look at some examples of nuclear symbols of elements
1.24
12Mg (Magnesium)
Atomic numver (Z)=12
# of protons: 12
# of electrons: 12
Number of neutrons: (A-Z)
(24-12) = 12

33. What’s New
Activity 1: Subatomic Particles
Directions: Complete the table below by supplying the correct quantity or number of
particles. The example before may be used as a guide.
Element Atomic
mass
Phosphorus 31
Barium
Chromium
Bismuth 209
Magnesium
What Is It
John Dalton’s atomic theory states that all atoms of an element have the same mass.
However, with modern science and techniques, isotopes have been discovered. This
discovery revised Dalton’s statement that atoms of the same element have the same mass
numbers. In other words, isotopes are atoms of the same element but has different numbers
of neutrons.
For example, all carbon atoms (Z=6) have 6 protons as well as electrons, bu
naturally occurring carbon atoms have 6 neutrons (A=12).
Sample Problem 1. Determining the number of Subatomic Particles in the Isotopes of an
Element
Problem: Silicon (Si) is a major component in semiconductor chips. It has three natur
occurring isotopes, 28
Si, 29
Si, and
electrons in each silicon isotope.
Plan: The mass number (A) of each of the three isotopes is given, which is the sum of
protons and neutrons. Using the periodic
equals the number of electrons. We obtain the number of neutrons by subtracting (A
Solution: According to the periodic table, the atomic number of Silicon is 14. Therefore,
28
Si has 14 protons, 14 electr
29
Si has 14 protons, 14 electrons, and 15 neutrons (29
30
Si has 14 protons, 14 electrons, and 16 neutrons (30
2
What’s New
Activity 1: Subatomic Particles
Complete the table below by supplying the correct quantity or number of
particles. The example before may be used as a guide.
Atomic
Number
No. of
Protons
No. of
Neutrons
15
56 81
28
83
12 12
John Dalton’s atomic theory states that all atoms of an element have the same mass.
However, with modern science and techniques, isotopes have been discovered. This
revised Dalton’s statement that atoms of the same element have the same mass
numbers. In other words, isotopes are atoms of the same element but has different numbers
For example, all carbon atoms (Z=6) have 6 protons as well as electrons, bu
naturally occurring carbon atoms have 6 neutrons (A=12).
Sample Problem 1. Determining the number of Subatomic Particles in the Isotopes of an
Problem: Silicon (Si) is a major component in semiconductor chips. It has three natur
Si, and 30
Si. Determine the number of protons, neutrons, and
electrons in each silicon isotope.
Plan: The mass number (A) of each of the three isotopes is given, which is the sum of
protons and neutrons. Using the periodic table, we can find the atomic number (Z) which
equals the number of electrons. We obtain the number of neutrons by subtracting (A
Solution: According to the periodic table, the atomic number of Silicon is 14. Therefore,
Si has 14 protons, 14 electrons, and 14 neutrons (28-14)
Si has 14 protons, 14 electrons, and 15 neutrons (29-14)
Si has 14 protons, 14 electrons, and 16 neutrons (30-14)
Complete the table below by supplying the correct quantity or number of
No. of
Electrons
24
John Dalton’s atomic theory states that all atoms of an element have the same mass.
However, with modern science and techniques, isotopes have been discovered. This
revised Dalton’s statement that atoms of the same element have the same mass
numbers. In other words, isotopes are atoms of the same element but has different numbers
For example, all carbon atoms (Z=6) have 6 protons as well as electrons, but only 98.89% of
Sample Problem 1. Determining the number of Subatomic Particles in the Isotopes of an
Problem: Silicon (Si) is a major component in semiconductor chips. It has three naturally
Si. Determine the number of protons, neutrons, and
Plan: The mass number (A) of each of the three isotopes is given, which is the sum of
table, we can find the atomic number (Z) which
equals the number of electrons. We obtain the number of neutrons by subtracting (A-Z).
Solution: According to the periodic table, the atomic number of Silicon is 14. Therefore,

34. 3
What’s More (1)
Activity 2: Do it Yourself
Directions: Using the sample before, determine the number of protons, neutrons, and
electrons in these problems. Encircle and put a label on each subatomic particle.
A. 11
5Q
Solution and Answer:
B. 41
20R
Solution and Answer:
C. 131
53X
Solution and Answer:
D. What elements do Q, R, and X represent?
What’s More (2)
Activity 3: Uses of Isotopes
Directions: Research and know more about the isotopes given below. On the spaces given,
draw the uses of the isotopes. Write a short caption below each drawing.

35. 4
Sulfur Isotopes Krypton Isotopes
Uranium Isotopes Silver Isotopes

36. Lesson Compounds:
Names
2
What’s In
In a chemical formula, element symbols, and, often, numerical subscripts show the type and
number of each atom in the smallest unit of the substance. In this lesson, you will
to write the names and formulas of ionic and simple covalent compounds
visualize molecules.
What Is It
Sometimes the atoms of an element are found alone. At other times they are found joined
together. When two or more atoms combine, whether these are the same or different, they
form a molecule. If these two or
formed.
There are three types of chemical bonds: ionic bond,
An ionic bond is formed when one atom shifts or transfers an electron to another atom. This
happens commonly when atoms with one valence electron, the alkali metals, elements in
Group IA are combined with seven valence electrons, the halogens or elements belonging to
Group VIIA. A good example is table salt. When sodium (Na+) reacts with chlorine (Cl
form the molecule sodium chloride (table salt), which is written as NaCl. Elements in Group
IIA may combine with elements in Group VIA. In general, atoms will form c
the bonding will cause all atoms involved to have a stable outer electron shell or eight
electrons. This rule is called the
when the outermost electron shell has eight electrons.
Sometimes atoms form bonds in which they share electrons. This is called
Water (H2O) is an example of covalent bond. Two electrons, one from each atom of
hydrogen, is shared with one atom of oxygen, since oxygen needs two more electrons for i
to become stable. Another example is carbon dioxide,
four valence electrons. It can complete its outer shell by sharing two pairs of electrons with
oxygen atom and two pairs with another one.
The last type is the metallic bond
a non-metal, in metallic bond, a metal shares electrons with another metal.
6
Compounds: Formulas and
Names
In a chemical formula, element symbols, and, often, numerical subscripts show the type and
number of each atom in the smallest unit of the substance. In this lesson, you will
to write the names and formulas of ionic and simple covalent compounds and how to
Sometimes the atoms of an element are found alone. At other times they are found joined
together. When two or more atoms combine, whether these are the same or different, they
. If these two or more molecules are fitted to combine, a chemical bond is
There are three types of chemical bonds: ionic bond, covalent bond and metallic bond.
is formed when one atom shifts or transfers an electron to another atom. This
happens commonly when atoms with one valence electron, the alkali metals, elements in
Group IA are combined with seven valence electrons, the halogens or elements belonging to
oup VIIA. A good example is table salt. When sodium (Na+) reacts with chlorine (Cl
form the molecule sodium chloride (table salt), which is written as NaCl. Elements in Group
IIA may combine with elements in Group VIA. In general, atoms will form c
the bonding will cause all atoms involved to have a stable outer electron shell or eight
electrons. This rule is called the OCTET RULE. It states that atoms are in stable condition
when the outermost electron shell has eight electrons.
metimes atoms form bonds in which they share electrons. This is called
is an example of covalent bond. Two electrons, one from each atom of
hydrogen, is shared with one atom of oxygen, since oxygen needs two more electrons for i
to become stable. Another example is carbon dioxide,(CO2). Carbon from Group IVA has
four valence electrons. It can complete its outer shell by sharing two pairs of electrons with
oxygen atom and two pairs with another one.
bond. While in ionic and covalent bonds, a metal combines with
metal, in metallic bond, a metal shares electrons with another metal.
Formulas and
In a chemical formula, element symbols, and, often, numerical subscripts show the type and
number of each atom in the smallest unit of the substance. In this lesson, you will learn how
and how to
Sometimes the atoms of an element are found alone. At other times they are found joined
together. When two or more atoms combine, whether these are the same or different, they
more molecules are fitted to combine, a chemical bond is
covalent bond and metallic bond.
is formed when one atom shifts or transfers an electron to another atom. This
happens commonly when atoms with one valence electron, the alkali metals, elements in
Group IA are combined with seven valence electrons, the halogens or elements belonging to
oup VIIA. A good example is table salt. When sodium (Na+) reacts with chlorine (Cl-), they
form the molecule sodium chloride (table salt), which is written as NaCl. Elements in Group
IIA may combine with elements in Group VIA. In general, atoms will form chemical bonds if
the bonding will cause all atoms involved to have a stable outer electron shell or eight
. It states that atoms are in stable condition
metimes atoms form bonds in which they share electrons. This is called covalent bond.
is an example of covalent bond. Two electrons, one from each atom of
hydrogen, is shared with one atom of oxygen, since oxygen needs two more electrons for it
Carbon from Group IVA has
four valence electrons. It can complete its outer shell by sharing two pairs of electrons with
. While in ionic and covalent bonds, a metal combines with
metal, in metallic bond, a metal shares electrons with another metal.

37. 7
What’s More
Activity #3: Naming Chemical Compounds
Directions: Complete the table below. Refer to Table of Compounds and their
Molecular formula for your answer: The first two (2) numbers were done for you.
Given: Table of compounds with their molecular formula
MOLECULE/COMPOUND MOLECULAR FOMRULA
Table salt NaCl
Vinegar CH3COOH
Table sugar C12H22O11
Muriatic acid HCl
Salitre KNO3
Agua Oxigenada H2O2
Rust Fe2O2
Sand SiO
Naphthalene ball C10H6
White wash Ca(OH)2
Washing soda CaCO3
Molecule/Compound Chemical
Formula
Elements
Present
No. of Atoms
in each
element
Total No. of
Atoms
Present
Table salt NaCl Na, Cl 1 Na, 1 Cl
Vinegar CH3COOH C, H, O 2 C, 4 H, 2 O
Table sugar C12H22O11
Muriatic acid HCl
Salitre KNO3
Agua Oxigenada H2O2
Rust Fe2O2
Sand SiO
Naphthalene ball C10H6
White wash Ca(OH)2
Washing soda CaCO3
What’s In
Two or more elements may combine by means of a chemical bond to form a compound. By
combining the symbols of the participating atoms, a chemical formula is formed. A chemical
formula is a group of symbols used to represent a compound. This is also called a molecular
formula. More than one atom is indicated by a numerical subscript. For instance, H2O means
that the water molecule consists of two atoms of hydrogen and one of oxygen.

38. 8
What Is It
Certain combinations of atoms form stable groups called radicals or polyatomic ion, which
form chemical bonds as an intact unit. The valence number of these radicals is taken as one.
If a molecule contains more than one of a given radical, its written formula emphasizes this
by using parentheses. Calcium phosphate, a major constituent of bones and teeth, is written
Ca3(PO4)2.
Some Polyatomic Ions
MONOVALENT 1-
BIVALENT 2-
TRIVALENT 3-
Ammonium NH4 (1+) Carbonate CO3 Phosphate PO4
Acetate C2H3O2 Chromate CrO4 Borate BO3
Chlorate ClO3 Oxalate C2O4
Chlorite ClO2 Sulfate SO4
Bicarbonate HCO3 Sulfite SO3
Bisulfate HSO4 Peroxide O2
Hydroxide OH
Nitrate NO3
Nitrite NO2
In writing a chemical formula, follow these rules:
1. Write the correct symbols of the elements and the polyatomic ions.
2. Determine the charge or valence number of the elements and the ions.
3. Indicate the charge by writing it on the right superscript
4. Exchange their valence numbers using the CRISS-CROSS METHOD.
Example: Write the chemical formula of the following compounds:
What’s More (2)
Activity #4: Writing Chemical Formulas

39. 9
Directions: Write the chemical formula of the following compounds.
COMPOND FORMULA COMPOND FORMULA
Zinc oxide Calcium carbonate
Lithium hydride Potassium chloride
Magnesium chloride Hydrogen fluoride
Aluminum nitrite Zinc nitrate
Sodium hydroxide Magnesium sulfate
What’s More (3)
Activity #5: Writing Chemical Formulas: Practicing More
Directions: Write the chemical formula of the following compounds.
1. Sodium bromide _______________
2. Barium chloride _______________
3. Aluminum hydroxide _______________
4. Sodium oxalate _______________
5. Potassium oxide _______________
6. Nitrogen phosphate _______________
7. Hydrogen sulfide _______________
8. Zinc chloride _______________
9. Silicon oxide _______________
10. Ammonium sulfate _______________
What I Can Do
Activity #6: Model Making (Portfolio Assessment)
Directions: You will represent compounds by using two/three dimensional models. First,
research or choose which compound you want to make a model of. Follow the procedure
and prepare the materials needed for the model making. You are HIGHLY ENCOURAGED
to use recyclable materials. Deductions will be made if your output is not made out of
recyclable materials.
Materials:
Recycled sticks (balloon sticks, bamboo stick, etc)
Recycled plastic caps of softdrinks
Glue / Glue stick
Procedure:
1. Have a copy of the compound of your choice.
2. Imitate the compound structure by using the caps representing the atoms and the sticks
representing the bonds.
3. Place your output in your portfolio folder or clear book.
Lesson

40. 10
What I Need to Know
In this third lesson of your Module 2, you will calculate the empirical formula from the percent
composition of a compound (STEM_GC11PCIf-32). Empirical formulas show the simplest
ratio among atoms in a compound.
What’s In
Percentage Composition
When a chemist has discovered a new compound, the first question to answer is,
what is the formula? To answer, he begins with analyzing the compound to determine
amounts of the elements for a given amount of the compound. This is expressed as
percentage composition. He then determines the empirical formula from this percentage
composition. If the compound is a molecular substance, he must also find the molecular
weight of the compound in order to determine the molecular formula.
Empirical formula
When a chemist analyzes an unknown compound, the first step is usually the
determination of the compound’s empirical formula. The empirical formula is therefore the
simplest formula of any compound; it is always written so that the subscripts in the molecular
formula are converted to the smallest possible whole numbers. It is derived from smallest
ratio of moles of all the atoms present in the molecule of a compound.
Calculating Empirical Formula
3

41. What’s New
Activity 1: Check the Label
Directions: Get any food product and check
composition of your food below:
Guide Questions:
1. Given the composition of your food, which of the following is good for your health? Which
of them is bad?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
2. Why is it always important to ch
_________________________________________________________________________
_________________________________________________________________________
_____________________________________________________________________
What Is It
The empirical formula is derived from mass analysis. It shows the lowest whole
numbers of moles, and thus the relative number of atoms, of each element in the compound.
For example, in hydrogen peroxide, there is 1 part by mass of hydrogen for every 16 par
by mass of oxygen. Because the atomic mass of hydrogen is 1.008 amu and that of oxygen
is 16.00 amu, there is one H atom for every O atom. Thus, the empirical formula is HO.
A chemist studying an unknown compound goes through a three step process to fi
the empirical formula:
1. determine the mass (g) of each component element
2. convert each mass (g) to amount (mol), and write a preliminary formula
11
: Get any food product and check the nutrition facts table. List down the
composition of your food below:
1. Given the composition of your food, which of the following is good for your health? Which
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
hy is it always important to check the label when we buy food products?
_________________________________________________________________________
_________________________________________________________________________
_____________________________________________________________________
The empirical formula is derived from mass analysis. It shows the lowest whole
numbers of moles, and thus the relative number of atoms, of each element in the compound.
For example, in hydrogen peroxide, there is 1 part by mass of hydrogen for every 16 par
. Because the atomic mass of hydrogen is 1.008 amu and that of oxygen
is 16.00 amu, there is one H atom for every O atom. Thus, the empirical formula is HO.
A chemist studying an unknown compound goes through a three step process to fi
1. determine the mass (g) of each component element
2. convert each mass (g) to amount (mol), and write a preliminary formula
the nutrition facts table. List down the
1. Given the composition of your food, which of the following is good for your health? Which
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
eck the label when we buy food products?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
The empirical formula is derived from mass analysis. It shows the lowest whole
numbers of moles, and thus the relative number of atoms, of each element in the compound.
For example, in hydrogen peroxide, there is 1 part by mass of hydrogen for every 16 parts
. Because the atomic mass of hydrogen is 1.008 amu and that of oxygen
is 16.00 amu, there is one H atom for every O atom. Thus, the empirical formula is HO.
A chemist studying an unknown compound goes through a three step process to find
2. convert each mass (g) to amount (mol), and write a preliminary formula

42. 12
3. convert the amounts (mol) mathematically to whole number (integer) subscripts.
To accomplish the math conversion,
 Divide each subscript by the smallest subscript, and
 If necessary, multiply through by the smallest integer that turns all subscripts
into integers.
Example:
A sample of an unknown compound contains 0.21 mol of zinc, 0.14 mol of phosphorus, and
0.56 mol of oxygen. What is the empirical formula?
Step 1: We are given the amount (mol) of each element as fractions. We use these fractional
amounts directly in a preliminary formula as subscripts of the element symbols. Then, we
convert the fractions to whole numbers.
Step 2: SOLUTION
Using the fractions to write the preliminary formula, with the symbols Zn for zinc, P
for phosphorus, and O for oxygen:
Zn0.21P0.14O0.56
Converting the fraction to whole numbers:
1. Divide each subscript by the smallest one, which in this case is 0.14:
.
.
.
.
.
.
Zn1.5P1.0O4.0
2. Multiply through by the smallest integer that turns all subscripts into
integers. We multiply by 2 to make the 1.5 (subscript of Zn) into an integer.
Zn(1.5x2)P(1.0x2)O(4.0x2) Zn3P2O8
What’s More
Activity #2: Solving
Directions: Find the empirical formula of the following compounds.
(a) 0.063mol of chlorine atoms combined with 0.22 mol of oxygen atoms
(b) 3.33 mol of C, 6.6 mol of H, and 3.33 mol of O

43. What I Have Learned
Activity #3: Infographic (Portfolio Output)
Directions: You will make an infographic on the things you have learned in this
module. It may be a digital art
materials. Prepare them and follow the procedures.
What is an infographic?
It is a visual representation of data. It is also a visual aid created to better understand
concepts. Infographics are useful
than words to explain complex concepts. Use your creativity in making one!
Materials:
 Coloring materials (crayons, oil pastel, watercolor, etc)
 Art materials (art paper, cartolina, scissors, glue,
 For digital art/s, print your output
Procedure:
 Prepare your materials. Use recycled materials as much as possible.
 Research or design your infographic based on your own creativity.
 Feature the concepts you have learned in this module: iso
formula, and naming compounds in a creative manner.
Post Assessment
Word Search Puzzle
Directions: Search and shade the word or words in the puzzle that complete the
sentences below:
1. ____________ are used to represent an element
2. ____________ is a small dense core of an atom
3. Atoms combine to form ________________
4. The simplest form of matter is/are _______________
5. the _________________ is derived from mass analysis.
6. The chemical name of Li2
7. The chemical name of ZnO is _______________
8. the _______________ has the same atomic number but has a differ
physical properties
9. The chemical name of NaCl is __________________
10. The common name of C
13
What I Have Learned
: Infographic (Portfolio Output)
Directions: You will make an infographic on the things you have learned in this
a digital art or handcrafted. You are encouraged to use recycled
materials. Prepare them and follow the procedures.
It is a visual representation of data. It is also a visual aid created to better understand
Infographics are useful for critical thinking and literacy models. Use more images
than words to explain complex concepts. Use your creativity in making one!
(crayons, oil pastel, watercolor, etc)
(art paper, cartolina, scissors, glue, stickers, etc)
For digital art/s, print your output
Prepare your materials. Use recycled materials as much as possible.
Research or design your infographic based on your own creativity.
Feature the concepts you have learned in this module: isotopes, calculating empirical
formula, and naming compounds in a creative manner.
Post Assessment
Search and shade the word or words in the puzzle that complete the
1. ____________ are used to represent an element
2. ____________ is a small dense core of an atom
3. Atoms combine to form ________________
4. The simplest form of matter is/are _______________
5. the _________________ is derived from mass analysis.
2O is _______________
The chemical name of ZnO is _______________
8. the _______________ has the same atomic number but has a differ
of NaCl is __________________
12H22O11 _________________________
Directions: You will make an infographic on the things you have learned in this
or handcrafted. You are encouraged to use recycled
It is a visual representation of data. It is also a visual aid created to better understand
for critical thinking and literacy models. Use more images
than words to explain complex concepts. Use your creativity in making one!
Prepare your materials. Use recycled materials as much as possible.
Research or design your infographic based on your own creativity.
topes, calculating empirical
Search and shade the word or words in the puzzle that complete the
8. the _______________ has the same atomic number but has a different mass &

44. 14
SUMMARY:
 Atoms of an element have a constant or fixed number of protons. The atomic number
gives the number of protons in the nucleus of the atom. For the neutral atom, the
number of protons is equal to the number of its electrons.
 isotopes are atoms of the same element but has different numbers of neutrons.
 There are three types of chemical bonds: ionic bond, covalent bond and metallic
bond.
 An ionic bond is formed when one atom shifts or transfers an electron to another
atom.
 The OCTET RULE states that atoms are in stable condition when the outermost
electron shell has eight electrons.
 Sometimes atoms form bonds in which they share electrons. This is called covalent
bond.
 While in ionic and covalent bonds, a metal combines with a non-metal, in metallic
bond, a metal shares electrons with another metal.
 The empirical formula is derived from mass analysis. It shows the lowest whole
numbers of moles, and thus the relative number of atoms, of each element in the
compound.

45. Key to Answers
Pre Assesment
Lesson 1
Activity #1
Element Atomic
mass
Phosphorus 31
Barium 137
Chromium 52
Bismuth 209
Magnesium 24
Activity 2: Do it Yourself
A. 11
5Q
5p+, 6n0, 5e-; Q = B
B. 41
20R
20p+, 21n0, 20e-; R= Ca
C. 131
53X
53p+, 78n0, 53e-; X= I
Lesson 2
Activity #3
Molecule/Compound Chemical
Formula
Table salt NaCl
Vinegar CH
Table sugar C12
Muriatic acid HCl
Salitre KNO
Agua Oxigenada H2O
Rust Fe2
Sand SiO
Naphthalene ball C10
15
Key to Answers
Atomic
Number
No. of
Protons
No. of
Neutrons
15 15 16
56 56 81
24 24 28
83 83 126
12 12 12
Chemical
Formula
Elements
Present
No. of Atoms
in each
element
NaCl Na, Cl 1 Na, 1 Cl
CH3COOH C, H, O 2 C, 4 H, 2 O
12H22O11 C, H, O 12 C; 22 H;
11 O
HCl H, C 1 H; 1 Cl
KNO3 K, N, O 1 K, 1 N, 3 O
O2 H, O 2 H, 2 O
2O2 Fe, O 2 Fe, 3 O
SiO Si, O 1 Si, 1 O
10H6 C, H 10 C, 5 H
No. of
Electrons
15
56
24
83
12
No. of Atoms Total No. of
Atoms
Present
2
8
45
2
4
4
5
2
15

46. 16
5White wash Ca(OH)2 1 Ca, 2 O, 2
H
5
W5ashing soda CaCO3 1 Ca, 1 C, 3
O
5
Activity #4
COMPOND FORMULA COMPOND FORMULA
Zinc oxide ZnO Calcium carbonate CaCO3
Lithium hydride LiH Potassium chloride KCl
Magnesium chloride MgCl2 Hydrogen fluoride HF
Aluminum nitrite Al(NO)3 Zinc nitrate Zn(NO3)2
Sodium hydroxide NaOH Magnesium sulfate MgSO4
Activity #5
Post Assessment
1 symbols
2 nucleus
3 elements
4 atom
5 Empirical formula
6. lithium oxide
7. zinc oxide
8. isotope
9. Sodim chloride
10. sugar

47. 17
REFERENCES:
1. Echija, Elena, Cecilia Bayquen, Rafeal Alfonso, and Elmarita De Vera.
2020. Frontliners in Science and Technology. Makati City: Diwa
Scholastic Press
2. Silberberg, Martin. 2016. General Chemistry 1&2. McGrawHill Education.

48. 18
For inquiries and feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
DepEd Division of Cagayan de Oro City
Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
Telefax: ((08822)855-0048
E-mail Address: cagayandeoro.city@deped.gov.ph

49. Government Property
NOT FOR SALE
NO
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General Chemistry 1
Quarter 1
Determining Molar Mass
Chemical Reactions and Equations
Department of Education
General Chemistry 1
Quarter 1 – Module 3
Determining Molar Mass
Chemical Reactions and Equations
Department of Education ● Republic of the Philippines
Senior High School
General Chemistry 1
Determining Molar Mass
Chemical Reactions and Equations
● Republic of the Philippines
Senior High School

50. General Chemistry 1- Grade 12
Alternative Delivery Mode
Quarter 1 – Module 3
Determining Molar Mass
Chemical Reactions and Equations
First Edition, 2020
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Published by the Department of Education – Division of Cagayan de Oro
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Development Team of the Module
Author: Marian Grace C Esmade
Reviewers: Jean S. Macasero,Ph.DEPS,Shirley A. Merida, Ph.D.EPS,
Duque Caguindangan, Eleanor Rollan, Rosemarie Dullante,
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51. General Chemistry 1
Quarter 1
Determining Molar Mass
Chemical Reactions and Equations
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at action@ deped.gov.ph.
We value your feedback and recommendations.
Department of
General Chemistry 1
Quarter 1 – Module 3
Determining Molar Mass
Chemical Reactions and Equations
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at action@ deped.gov.ph.
We value your feedback and recommendations.
Department of Education ● Republic of the Philippines
General Chemistry 1
Module 3
Determining Molar Mass
Chemical Reactions and Equations
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
of the Philippines

52. This page is intentionally blank

53. Table of Contents
What This Module is About............................................................................................................ i
What I Need to Know...................................................................................................................... ii
How to Learn from this Module .................................................................................................... .ii
Icons of this Module....................................................................................................................... .iii
What I Know................................................................................................................................... ..iii
Lesson 1:
Index Fossils........................................................................................................................
What I Need to Know........................................................................................... 1
What I know .......................................................................................................... 2
What Is It................................................................................................................ 3
What’s New: …..................................................................................................... 5
What’s In: ….......................................................................................................... 6
What’s More .......................................................................................................... 7
What I Have Learned: ......................................................................................... 8
Lesson 2:
History of the Earth Through Geological Time ...................................
What I need to know ............................................................................................ 10
What I know .......................................................................................................... 10
What is it .............................................................................................................. ..13
What’s New ......................................................................................................... ..16
What is it .................................................................................................................17
What’s More ........................................................................................................ ..18
What is it .................................................................................................................19
What’s New ......................................................................................................... ..21
What’s I can Do................................................................................................... ..22
What I have learned..............................................................................................23

54. Summary ......................................................................................................................................
Assessment: (Post-Test) ............................................................................................................ 23
Key to Answers............................................................................................................................. 26
References.................................................................................................................................... 29

55. This page is intentionally blank

56. Module 2
What This Module is About
Chemical transformations are happening in everyday life. It is not enough to
describe matter and the changes that it undergoes qualitatively. They often need
more accurate, quantitative observations. In chemical reactions, a huge number of
atoms and molecules are involved. Instead of dealing with particles that cannot be
seen, chemists work with bulk of matter. These observations use measurable
quantities and calculations that are essential in understanding chemical reactions.
This module contains discussions and activities that will explain and illustrate this
topic.
What I Need to Know
At the end of this module, you should be able to:
1. Calculate molecular formula given molar mass; and
2. Write and balanced chemical equations

57. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests a
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that are in
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show
case your skills and knowledge gained, and
applied into real-life concerns and situations.
II
Follow the directions and/or instructions in the activities and exercises diligently.
This part contains learning objectives that
are set for you to learn as you go along the
This is an assessment as to your level of
knowledge to the subject matter at hand,
ifically to gauge prior related
This part connects previous lesson with that
An introduction of the new lesson through
various activities, before it will be presented
These are discussions of the activities as a
way to deepen your discovery and under-
up activities that are in-
tended for you to practice further in order to
Activities designed to process what you
have learned from the lesson
These are tasks that are designed to show-
case your skills and knowledge gained, and
life concerns and situations.

58. What I Know
Pretest: MULTIPLE CHOICE:
Directions: Read and understand each item and choose the letter of the correct answer.
Use separate answer sheet of paper.
1. Which is not a type of chemical reaction?
A. Combination
B. Substitution
2. What scientific law is explained using balanced chemical equations?
A. Law of chemical reactions
B. Law of conservation of energy
3. The electrolysis of water into hydrogen and oxygen is an example of ______
reaction.
A. Combination
B. Decomposition
4. Which of the following symb
A. (aq)
B. (s)
5. In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
A. Equal sign
B. Coefficient
6. The symbol(s) written after
A. Soluble
B. Solution
7. Which of the following is TRUE of a balanced equation?
A. The total number of atoms remains the same.
B. The kinds of atoms remain the same.
C. The total number of molecules remains the same.
D. The number of atoms of each element remains the same.
8. Which of the following is true of a chemical reaction?
A. Only physical changes occur.
B. New substances must form.
same.
9. When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
the following word equations best represents this reaction?
A. sulfur + oxygen → sulfur dioxide
B. sulfur dioxide + oxygen
10. Why can’t you change the formula in order to balance a reaction?
A.The number of atoms in the left should be equal to the atoms on the right of the
arrow.
B.. The elements have definite mass ratios.
C. The molecule is indestructible.
D. The elements that combine have definite composition and fixed proportion of
elements by mass.
What I Know – Pre Assessment
Pretest: MULTIPLE CHOICE:
Read and understand each item and choose the letter of the correct answer.
sheet of paper.
Which is not a type of chemical reaction?
C. decomposition
D. evaporation
What scientific law is explained using balanced chemical equations?
Law of chemical reactions C. law of conservation of mass
Law of conservation of energy D. law of definite proportions
The electrolysis of water into hydrogen and oxygen is an example of ______
C. double replacement
D. single replacement
Which of the following symbols means a substance is in water solution?
C. (l)
D. (w)
In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
C. plus sign
D. arrow
6. The symbol(s) written after a formula in a chemical equation stands for
C. solid
D. synthesis
7. Which of the following is TRUE of a balanced equation?
A. The total number of atoms remains the same.
B. The kinds of atoms remain the same.
tal number of molecules remains the same.
D. The number of atoms of each element remains the same.
8. Which of the following is true of a chemical reaction?
A. Only physical changes occur. C. Only changes of state occur.
B. New substances must form. D. Chemical properties remain the
9. When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
s best represents this reaction?
→ sulfur dioxide C. sulfur dioxide → sulfur + oxygen
B. sulfur dioxide + oxygen → sulfur D. sulfur → sulfur dioxide + oxygen
10. Why can’t you change the formula in order to balance a reaction?
A.The number of atoms in the left should be equal to the atoms on the right of the
B.. The elements have definite mass ratios.
C. The molecule is indestructible.
D. The elements that combine have definite composition and fixed proportion of
Read and understand each item and choose the letter of the correct answer.
C. decomposition
. evaporation
What scientific law is explained using balanced chemical equations?
C. law of conservation of mass
law of definite proportions
The electrolysis of water into hydrogen and oxygen is an example of ______
C. double replacement
D. single replacement
ols means a substance is in water solution?
In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
a formula in a chemical equation stands for
C. Only changes of state occur.
D. Chemical properties remain the
9. When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
sulfur + oxygen
→ sulfur dioxide + oxygen
A.The number of atoms in the left should be equal to the atoms on the right of the
D. The elements that combine have definite composition and fixed proportion of

59. This page is intentionally blank

60. Lesson Writing and Balancing
Chemical Equations
2
What I Need to Know
We will learn how to write sample chemical reactions and translate them into
chemical equations. This module will lead you to the beauty of the chemical
language to the extent that you will be able to write it, read it and interpret its
meaning.
What’s New
Activity #1: Ice Cube Experiment
Directions: Prepare all the materials needed, follow the procedures, and answer the guide
questions.
Materials:
Ice cube
Tap water
Clear plastic cup
Weighing scale (any)
Procedure:
1. Pour water into a clear plastic cup so that it is about 1/3-full.
2. Weigh using the weighing scale. Take note of the mass.
3. Add 1 piece of ice.
4. Weigh the cup, water, and ice. Record the combined mass.
Question:
1. What is the initial mass? ___________
2. What is the final mass? ___________
3. Does the mass change or stay the same as the ice cube melts? ____________
4. Why does this happen?
___________________________________________________________________
5. If you weighed a stick of butter and then let it melt, do you think it would weigh more,
less, or the same afterwards? Why?
___________________________________________________________________

61. What Is It
Law of Conservation of Mass
In the late eighteenth century, Antoine Lavoisier, a French chemist, recognized the
importance of accurate measurements. He extensively studies and explained the nature of
combustion. He found out that combustion involved reaction with oxygen. His experime
in which he carefully weighed the reactants and products of various reactions, suggested
that mass is neither created nor destroyed. Lavoisier’s discovery of this
of mass was the basis for the development in chemistry in the ninete
A chemical change involves reorganization of the atoms in one or more substances.
The law of conservation of mass requires that there must be exactly as many atoms among
the combined products of a chemical reaction as in its combined reacta
this better, let us define words that will be used in this lesson.
material in a chemical reaction. Products are the substance formed as a result of a chemical
reaction. In a chemical equation, reactants are fo
the right side. A chemical reaction can therefore be summarized as
For example, when the methane (CH4) in natural gas combines with oxygen (O2) in the air
and burns, carbon dioxide (CO2)
The balanced chemical equation for this reaction is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
The reactants in this reaction are methane and oxygen gas. The products are carbon dioxide
and water.
A balanced equation conforms to the
number of atoms of each kind on the left side of the reaction is equal to the number of atoms
of each kind on the right side of the equation.
There is 1 atom of carbon on the left side and 1 atom of carbon
There are 4 atoms of hydrogen on the left side. On the right side, there are 2 atoms of
hydrogen that is multiplied by the coefficient 2 found on the left side of H
On the left side of the equation, there are 2 atoms of oxygen multi
coefficient 2 found on the left side of O
CO2 and 1 atom of oxygen in H
H2O.
To get a clearer view of the number of atoms of
Kind of Atom
C
H
O
Thus, the balanced equation above conforms to the law of conservation of mass.
Let us have another example.
When aluminum and barium oxide are heated together, a vigorous reaction begins,
and elemental barium and aluminum oxide, Al2O3, are formed. The equation is
Law of Conservation of Mass
In the late eighteenth century, Antoine Lavoisier, a French chemist, recognized the
importance of accurate measurements. He extensively studies and explained the nature of
combustion. He found out that combustion involved reaction with oxygen. His experime
in which he carefully weighed the reactants and products of various reactions, suggested
that mass is neither created nor destroyed. Lavoisier’s discovery of this law of conservation
was the basis for the development in chemistry in the nineteenth century.
A chemical change involves reorganization of the atoms in one or more substances.
The law of conservation of mass requires that there must be exactly as many atoms among
the combined products of a chemical reaction as in its combined reactants. To understand
this better, let us define words that will be used in this lesson. Reactants
material in a chemical reaction. Products are the substance formed as a result of a chemical
reaction. In a chemical equation, reactants are found on the left side and the products are on
the right side. A chemical reaction can therefore be summarized as
Reactants → Products
For example, when the methane (CH4) in natural gas combines with oxygen (O2) in the air
and burns, carbon dioxide (CO2) and water (H2O) are formed.
The balanced chemical equation for this reaction is:
→ CO2(g) + 2H2O(g)
The reactants in this reaction are methane and oxygen gas. The products are carbon dioxide
A balanced equation conforms to the law of conservation of mass. Let us check if the
number of atoms of each kind on the left side of the reaction is equal to the number of atoms
of each kind on the right side of the equation.
There is 1 atom of carbon on the left side and 1 atom of carbon
There are 4 atoms of hydrogen on the left side. On the right side, there are 2 atoms of
hydrogen that is multiplied by the coefficient 2 found on the left side of H2O.
On the left side of the equation, there are 2 atoms of oxygen multi
coefficient 2 found on the left side of O2. On the right side, there are two atoms of oxygen in
and 1 atom of oxygen in H2O multiplied by the coefficient 2 found on the left side of
To get a clearer view of the number of atoms of each kind, look at the table below.
No. on the Left side No. on the Right side
1
4
2 x 2 = 4 2 + 1(2) = 4
Thus, the balanced equation above conforms to the law of conservation of mass.
Let us have another example.
aluminum and barium oxide are heated together, a vigorous reaction begins,
and elemental barium and aluminum oxide, Al2O3, are formed. The equation is
In the late eighteenth century, Antoine Lavoisier, a French chemist, recognized the
importance of accurate measurements. He extensively studies and explained the nature of
combustion. He found out that combustion involved reaction with oxygen. His experiments,
in which he carefully weighed the reactants and products of various reactions, suggested
law of conservation
enth century.
A chemical change involves reorganization of the atoms in one or more substances.
The law of conservation of mass requires that there must be exactly as many atoms among
nts. To understand
Reactants are the starting
material in a chemical reaction. Products are the substance formed as a result of a chemical
und on the left side and the products are on
For example, when the methane (CH4) in natural gas combines with oxygen (O2) in the air
The reactants in this reaction are methane and oxygen gas. The products are carbon dioxide
law of conservation of mass. Let us check if the
number of atoms of each kind on the left side of the reaction is equal to the number of atoms
There is 1 atom of carbon on the left side and 1 atom of carbon on the right side.
There are 4 atoms of hydrogen on the left side. On the right side, there are 2 atoms of
O.
On the left side of the equation, there are 2 atoms of oxygen multiplied by the
. On the right side, there are two atoms of oxygen in
O multiplied by the coefficient 2 found on the left side of
each kind, look at the table below.
No. on the Right side
1
2 x 2 = 4
2 + 1(2) = 4
Thus, the balanced equation above conforms to the law of conservation of mass.
aluminum and barium oxide are heated together, a vigorous reaction begins,
and elemental barium and aluminum oxide, Al2O3, are formed. The equation is

62. 2Al(l) + 3BaO(s) → Al2O3(s) + 3Ba(l)
a. Identify the reactants and products.
b. Check the equation if it conforms with the Law of Conservation of Mass.
Answer:
a. The reactants are Al and BaO. The products are Al2O3 and Ba.
b.
Kind of
Atom
No. on the Left side No. on the Right
side
Al 1 x 2 = 2 2
Ba 1 x 3 = 3 1 x 3 = 3
O 1 x 3 = 3 3
It conforms with the Law of Conservation of Mass.
What’s More
Activity #2: Let’s Practice!
Directions: Identify the reactants and products of the reaction on (a). Check the
equation if it conforms with the Law of Conservation of Mass by filling in the table.
1. Cl2(g) + 2 KBr(aq) → 2KCl(aq) + Br2(l)
(a) _____________________________________________________
(b)
2. 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g)
(a) _______________________________________________
Kind of
Atom
No. on the Left side No. on the Right
side
(b)
3. K2CrO4(aq) + 2AgNO3(aq) → Ag2CrO4(s) + KNO3(aq)
(a) _____________________________________________________
(b)
Kind of
Atom
No. on the Left side No. on the Right
side
Kind of
Atom
No. on the Left side No. on the Right
side

63. What’s In
You may ask, “Why must chemical equations be balanced?” Simply because it’s the
law! Matter cannot be created or destroyed in chemical reactions as stated in the
conservation of mass. In every
the products as started in the reactants. Balanced chemical equations show that mass is
conserved in chemical reactions. In our next topic, we will learn
equations and balance them out.
What Is It
Writing Chemical Reactions and Balancing Chemical Equations
The three fundamental chemical laws are the basis for writing chemical reactions. The
conversion of substances to other subst
represented by a chemical equation. The chemical equation is very important because it
provides two types of information: the nature of reactants and products, and the relative
number of each. This lesson is brok
A. Writing Chemical Reactions
Writing chemical reactions are important for chemists. It is an important language that
translates the reactions into easy, readable and understandable sentence, which we call the
chemical equation. Word reactions are translated into symbols for easy reading. By looking
at the chemical equations, the reader can easily interpret what transpired in the reaction.
For example: The reaction of hydrogen and oxygen to give water is represe
follows:
There are conventions and simple rules to follow in writing chemical equations. They are
as follows:
1. As mentioned in previous texts, the starting material or substances called reactants
are written on the left side and
on the right side.
2. An arrow (→) is used to represent the conversion of the reactants to products. This
may literally mean “to yield” or “to form”. The plus sign (+) means “to react with” or “to
combine with”.
3. It is recommended that the states of the substances be indicated by placing the
following symbols after the formula of the substance (e.g. s, l, g, aq)
4. In a chemical reaction, the law of conservation of mass holds. A balanced equation
conforms to this law. As mentioned in the previous lesson, the number of atoms of
each kind on the left and right sides of the arrow must be equal.
In this example, we will write the chemical equation for this reaction:
Two molecules of acetylene gas will react with 5
4 molecules of carbon dioxide gas and two molecules of water vapor.
Strategy:
1. Identify the reactants and products: C
You may ask, “Why must chemical equations be balanced?” Simply because it’s the
law! Matter cannot be created or destroyed in chemical reactions as stated in the
. In every chemical reaction, the same mass of matter must end up in
the products as started in the reactants. Balanced chemical equations show that mass is
conserved in chemical reactions. In our next topic, we will learn how to write chemical
equations and balance them out.
Writing Chemical Reactions and Balancing Chemical Equations
The three fundamental chemical laws are the basis for writing chemical reactions. The
conversion of substances to other substances during a chemical reaction is usually
represented by a chemical equation. The chemical equation is very important because it
provides two types of information: the nature of reactants and products, and the relative
number of each. This lesson is broken in two parts to make your learning easier.
Writing Chemical Reactions
Writing chemical reactions are important for chemists. It is an important language that
translates the reactions into easy, readable and understandable sentence, which we call the
chemical equation. Word reactions are translated into symbols for easy reading. By looking
at the chemical equations, the reader can easily interpret what transpired in the reaction.
For example: The reaction of hydrogen and oxygen to give water is represe
2H2 + O2 = 2H2O
There are conventions and simple rules to follow in writing chemical equations. They are
As mentioned in previous texts, the starting material or substances called reactants
are written on the left side and the resulting substances called products are written
→) is used to represent the conversion of the reactants to products. This
may literally mean “to yield” or “to form”. The plus sign (+) means “to react with” or “to
It is recommended that the states of the substances be indicated by placing the
following symbols after the formula of the substance (e.g. s, l, g, aq)
In a chemical reaction, the law of conservation of mass holds. A balanced equation
is law. As mentioned in the previous lesson, the number of atoms of
each kind on the left and right sides of the arrow must be equal.
In this example, we will write the chemical equation for this reaction:
Two molecules of acetylene gas will react with 5 molecules of oxygen gas to produce
4 molecules of carbon dioxide gas and two molecules of water vapor.
Identify the reactants and products: C2H2 + O2 → CO2 + H2O
You may ask, “Why must chemical equations be balanced?” Simply because it’s the
law! Matter cannot be created or destroyed in chemical reactions as stated in the law of
, the same mass of matter must end up in
the products as started in the reactants. Balanced chemical equations show that mass is
how to write chemical
The three fundamental chemical laws are the basis for writing chemical reactions. The
ances during a chemical reaction is usually
represented by a chemical equation. The chemical equation is very important because it
provides two types of information: the nature of reactants and products, and the relative
en in two parts to make your learning easier.
Writing chemical reactions are important for chemists. It is an important language that
translates the reactions into easy, readable and understandable sentence, which we call the
chemical equation. Word reactions are translated into symbols for easy reading. By looking
at the chemical equations, the reader can easily interpret what transpired in the reaction.
For example: The reaction of hydrogen and oxygen to give water is represented as
There are conventions and simple rules to follow in writing chemical equations. They are
As mentioned in previous texts, the starting material or substances called reactants
the resulting substances called products are written
→) is used to represent the conversion of the reactants to products. This
may literally mean “to yield” or “to form”. The plus sign (+) means “to react with” or “to
It is recommended that the states of the substances be indicated by placing the
following symbols after the formula of the substance (e.g. s, l, g, aq)
In a chemical reaction, the law of conservation of mass holds. A balanced equation
is law. As mentioned in the previous lesson, the number of atoms of
molecules of oxygen gas to produce

64. 2. Indicate the states of the substances by placing their symbols on the right side o
the substances
3. Affix the number of molecules as coefficients at the left side of the substances
4. Check the equation if it conforms with the Law of Conservation of Mass
Kind of
Atom
C
H
O
B. Balancing Equations
In general, a balanced chemical equation is written in two steps:
1. Write the formula and state or phase of the
2. Balance the number of atoms of each kind by using coefficients. Write the
coefficients on the left side of the substances.
Note: The chemical formulas must not be changed. The subscripts must not be
changed. Only the coefficients
Example 1. Write the balanced equation for the reaction of solid magnesium with oxygen in
air to produce magnesium oxide.
Step 1. Write the chemical equation
Step 2. Balance the number of atoms
Upon inspection, there are 2 atoms of oxygen in the left side, and only 1 atom of
oxygen on the right side. Thus, we shall focus on balancing the oxygen atom on the right
side by putting a coefficient 2 at the left side of MgO.
The oxygen atom is now balanced, but there is only 1 atom of Mg on the left side and
two atoms on the right side. The next move is to put a coefficient 2 on the left side of Mg to
balance the Mg atoms.
Step 3. Check the equatio
Mass.
What’s More
Activity #3: Self Test
Directions: Part A. Write the chemical equations of the following reactions:
1. Aqueous aluminum nitrate reacts with aqueous sodium hydroxide to form aqueous
aluminum hydroxide and aqueous sodium nitrate.
____________________________________________________________
2. Iron reacts with sulfuric acid to produce iron (III) sulfate and hyd
____________________________________________________________
3. Oxygen gas reacts with carbon sulfide to produce carbon dioxide and sulfur dioxide.
Indicate the states of the substances by placing their symbols on the right side o
C2H2(g) + O2(g) → CO2(g) + H2O(g)
Affix the number of molecules as coefficients at the left side of the substances
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
Check the equation if it conforms with the Law of Conservation of Mass
No. on the Left side No. on the Right
side
2 x 2 = 4 1 x 4 = 4
2 x 2 = 4 2 x 2 = 4
2 x 5 = 10 4(2) + 1(2) = 10
Balancing Equations
In general, a balanced chemical equation is written in two steps:
Write the formula and state or phase of the reactants and products.
Balance the number of atoms of each kind by using coefficients. Write the
coefficients on the left side of the substances.
Note: The chemical formulas must not be changed. The subscripts must not be
changed. Only the coefficients are to be adjusted.
Example 1. Write the balanced equation for the reaction of solid magnesium with oxygen in
air to produce magnesium oxide.
Step 1. Write the chemical equation
Mg(s) + O2(g) → MgO(s)
Step 2. Balance the number of atoms
Mg(s) + O2(g) → MgO(s)
Upon inspection, there are 2 atoms of oxygen in the left side, and only 1 atom of
oxygen on the right side. Thus, we shall focus on balancing the oxygen atom on the right
side by putting a coefficient 2 at the left side of MgO.
Mg(s) + O2(g) → 2MgO(s)
The oxygen atom is now balanced, but there is only 1 atom of Mg on the left side and
two atoms on the right side. The next move is to put a coefficient 2 on the left side of Mg to
2Mg(s) + O2(g) → 2MgO(s)
Step 3. Check the equation if it conforms with the Law of Conservation of
What’s More (A)
Part A. Write the chemical equations of the following reactions:
Aqueous aluminum nitrate reacts with aqueous sodium hydroxide to form aqueous
aluminum hydroxide and aqueous sodium nitrate.
____________________________________________________________
Iron reacts with sulfuric acid to produce iron (III) sulfate and hydrogen gas.
____________________________________________________________
Oxygen gas reacts with carbon sulfide to produce carbon dioxide and sulfur dioxide.
Indicate the states of the substances by placing their symbols on the right side of
Affix the number of molecules as coefficients at the left side of the substances
Check the equation if it conforms with the Law of Conservation of Mass
No. on the Right
4(2) + 1(2) = 10
reactants and products.
Balance the number of atoms of each kind by using coefficients. Write the
Note: The chemical formulas must not be changed. The subscripts must not be
Example 1. Write the balanced equation for the reaction of solid magnesium with oxygen in
Upon inspection, there are 2 atoms of oxygen in the left side, and only 1 atom of
oxygen on the right side. Thus, we shall focus on balancing the oxygen atom on the right
The oxygen atom is now balanced, but there is only 1 atom of Mg on the left side and
two atoms on the right side. The next move is to put a coefficient 2 on the left side of Mg to
n if it conforms with the Law of Conservation of
Part A. Write the chemical equations of the following reactions:
Aqueous aluminum nitrate reacts with aqueous sodium hydroxide to form aqueous
____________________________________________________________
rogen gas.
____________________________________________________________
Oxygen gas reacts with carbon sulfide to produce carbon dioxide and sulfur dioxide.

65. ____________________________________________________________
Part B. Balance the following
Mg + N2 → Mg3N2
Cl2(g) + KBr(aq) → KCl(aq) + Br
C2H6 + O2 → CO2 + H2O
What’s More
Activity 4. Reflection
Direction: Answer the guide question in 5
punctuation.
We balance equations for a lot of reasons
represent what happens when we observe the chemical reaction in the real world. It
also obeys an important guiding principle which is the law of conservation of mas
As a student, how do you balance the different aspects of your life?
What is your guiding principle in life?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_____________________________________________________
____________________________________________________________
Part B. Balance the following chemical reactions
+ Br2(l)
What’s More (B)
Answer the guide question in 5-10 sentences. Observe proper grammar
We balance equations for a lot of reasons. Frist, we want the equations to
represent what happens when we observe the chemical reaction in the real world. It
also obeys an important guiding principle which is the law of conservation of mas
As a student, how do you balance the different aspects of your life?
What is your guiding principle in life?
___________________________________________________________________
___________________________________________________________________
_________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
____________________________________________________________
10 sentences. Observe proper grammar and
. Frist, we want the equations to
represent what happens when we observe the chemical reaction in the real world. It
also obeys an important guiding principle which is the law of conservation of mass.
As a student, how do you balance the different aspects of your life?
___________________________________________________________________
___________________________________________________________________
_________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
______________

66. Lesson
Determining Molar Mass
1
What’s In
All the ideas in this lesson will rely on your understandning a key concept to a uit called the
mole. In daily life, we measure things by counting or by weighing: we weigh rice and flour,
but we count eggs or ballpens. But how exactly can we count or weigh
moleulces, or formula units when all of them are munite objects? As you’ll see, chemists
have devised a unit called the
What Is It
Defining the Mole
The mole (mol) is the SI unit for
substance that contains the same number of atoms in 12 g of carbon
called Avogadrio’s number (in honor of the Italian physicist Amadeo Avogadro), is enormous:
One mole 1(mol) contains 6
Thus,
1 mol of carbon-12 contains
1 mol of H2O contains
1 mol of NaCl contains
Knowing the amount (in moles), the mass (in grams), and the number of
very important as we mix different substances to run a reaction. The central relationship
between masses on the atomic scale and on the macroscopic scale is the same for
elements and compounds,
 Elements – the mass in atomic mass units (amu)
same numerically as the mass in grams (g) of 1 mole of atoms of the element. Each
atom of an element is considered to have
Thus,
1 atom of S has a mass of
mass of 32.07g
Determining Molar Mass
All the ideas in this lesson will rely on your understandning a key concept to a uit called the
mole. In daily life, we measure things by counting or by weighing: we weigh rice and flour,
but we count eggs or ballpens. But how exactly can we count or weigh
moleulces, or formula units when all of them are munite objects? As you’ll see, chemists
have devised a unit called the mole, to count chemical entitles by weighing them.
The mole (mol) is the SI unit for amount of substance. It is defined as the amount of a
substance that contains the same number of atoms in 12 g of carbon-12. This number,
called Avogadrio’s number (in honor of the Italian physicist Amadeo Avogadro), is enormous:
One mole 1(mol) contains 6.022x1023
entities
contains 6.022x1023
carbon-12 atoms
contains 6.022x1023
H2O molecules
contains 6.022x1023
NaCl formula units
Knowing the amount (in moles), the mass (in grams), and the number of
very important as we mix different substances to run a reaction. The central relationship
between masses on the atomic scale and on the macroscopic scale is the same for
the mass in atomic mass units (amu) of one atom of an element is the
same numerically as the mass in grams (g) of 1 mole of atoms of the element. Each
atom of an element is considered to have the atomic mass given in the periodic table.
1 atom of S has a mass of 32.07 amu and 1 mol (6.022x1023
atoms) of S has a
16
S
32.07
Determining Molar Mass
All the ideas in this lesson will rely on your understandning a key concept to a uit called the
mole. In daily life, we measure things by counting or by weighing: we weigh rice and flour,
but we count eggs or ballpens. But how exactly can we count or weigh atoms, ions,
moleulces, or formula units when all of them are munite objects? As you’ll see, chemists
, to count chemical entitles by weighing them.
amount of substance. It is defined as the amount of a
12. This number,
called Avogadrio’s number (in honor of the Italian physicist Amadeo Avogadro), is enormous:
Knowing the amount (in moles), the mass (in grams), and the number of entities beomes
very important as we mix different substances to run a reaction. The central relationship
between masses on the atomic scale and on the macroscopic scale is the same for
of one atom of an element is the
same numerically as the mass in grams (g) of 1 mole of atoms of the element. Each
the atomic mass given in the periodic table.
atoms) of S has a

67. Calculating the Molecular Mass of a Compound
Using the periodic table and the formula of a compound, we calculate the molecular mass
(also called molecular weight) of a formula unit of the compound as the sum of the atomic
masses:
Molecular mass = sum of atomic masses
1. For example, the molecular mass of a water molecule (using atomic masses to four
significant figures from the periodic table) is
Molecular mass of H2O = (2 x atomic mass of H) + (1 x atomic mass of O)
= (2 x 1.008 amu) + 16.00 amu = 18.02 amu
2. Another example would be calculating the molecular mass of tetraphosphorus trisulfide.
First, we write the formula, then multiply the number of atoms of each element by its atomic
mass (which we can find from the periodic table), and find the sum.
Formula: P4S3
Molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S)
= (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu
3. In ionic compounds, it doesn’t consist of molecules so the mass of a formila unit is termed
as the formula mass instead of molecular mass. To calculate the formula mass of a
compound with a polyatomic ion, the number of atoms of each element inside the
parentheses is multiplied by the subscript outside the parentheses. For barium nitrate,
Ba(NO3)2,
Formula mass of Ba(NO3)2
= (1 x atomic mass of Ba) + (2 x atomic mass of N) + (6 x atomic mass of O)
= (137.3 amu + (2 x 14.01 amu) + (6 x 16.00 amu) = 261.3 amu
What’s More
Activity #1: Let’s Practice!
Directions: Familiarizing ourselves with the atomic mass will be a great help in
finding out the molecular mass of a compound. Take out a copy of your periodic
table. Locate and write the atomic mass of the following elements on the space
provided.
Atom Atomic Mass Atom Atomic Mass
1 Cl 6 F
2 N 7 Mg
3 O 8 Al
4 Ca 9 Si
5 P 10 S
What’s More #2

68. Activity #2: Calculating The Molecular Mass
Directions: Now that you are familiar with the atomic mass, let us calculate the
molecular mass of the different compounds. First, write the formula then afterwards
you can write your solutions on the space provided and encircle your final answer. (5
points each number)
1. hydrogen peroxide
Formula: _____________
2. cesium chloride
Formula: _____________
3. sulfuric acid
Formula: _____________
4. potassium sulfate
Formula: _____________

69. 5. sodium oxide
Formula: _____________
6. nitrogen dioxide
Formula: _____________

70. SUMMARY
- A mole of substance is the amount that contains Avogadro’s number
(6.022 x 1023
) of chemical entities (atoms, molecules, or formula units)
- The mass (in grams) of a mole of the entity has the same numerical value
as the mass (in amu) of the individual entity. Thus, the mole allows us to
count entities by weighing them
- Chemical changes have many applications in the modern world. The
creation of new materials like electronic gadgets is an example of this
application.
- The chemical equation shows the formula and the phase of the reactants
and products
- The four basic types of chemical reactions are combination,
decomposition, single displacement, and double displacement

71. What I Have Learned
Post- Test
Directions: Read and understand each item and choose the letter of the correct answer.
Use separate answer sheet of paper.
1. Which of the following is TRUE of a balanced equation?
A. The total number of atoms remains the same.
B. The kinds of atoms remain the same.
C. The total number of molecules remains the same.
D. The number of atoms of each element remains the same.
2. Which of the following is true of a chemical reaction?
A. Only physical changes occur.
B. New substances must form.
same.
3. Which is not a type of chemical reaction?
C. Combination
D. Substitution
4. The electrolysis of water into hydrogen and oxygen is an example of ______ reaction.
C. Combination
D. Decomposition
5. What scientific law is explained using balanced chemi
C. Law of chemical reactions
D. Law of conservation of energy
6. Which of the following symbols means a substance is in water solution?
C. (aq)
D. (s)
7. When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
the following word equations best represents this reaction?
E. A. sulfur + oxygen
oxygen
F. B. sulfur dioxide + oxygen
oxygen
8. In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
A. Equal sign
B. Coefficient
9. The symbol(s) written after a formula in a chemical eq
A. Soluble
B. Solution
What I Have Learned
Read and understand each item and choose the letter of the correct answer.
Use separate answer sheet of paper.
TRUE of a balanced equation?
A. The total number of atoms remains the same.
B. The kinds of atoms remain the same.
C. The total number of molecules remains the same.
D. The number of atoms of each element remains the same.
rue of a chemical reaction?
A. Only physical changes occur. C. Only changes of state occur.
B. New substances must form. D. Chemical properties remain the
Which is not a type of chemical reaction?
C. decomposition
D. evaporation
The electrolysis of water into hydrogen and oxygen is an example of ______ reaction.
C. double replacement
D. single replacement
What scientific law is explained using balanced chemical equations?
Law of chemical reactions C. law of conservation of mass
Law of conservation of energy D. law of definite proportions
Which of the following symbols means a substance is in water solution?
C. (l)
D. (w)
When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
the following word equations best represents this reaction?
A. sulfur + oxygen → sulfur dioxide C. sulfur dioxide
B. sulfur dioxide + oxygen → sulfur D. sulfur → sulfur dioxide +
In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
C. plus sign
D. arrow
. The symbol(s) written after a formula in a chemical equation stands for
C. solid
D. synthesis
Read and understand each item and choose the letter of the correct answer.
C. Only changes of state occur.
D. Chemical properties remain the
C. decomposition
D. evaporation
The electrolysis of water into hydrogen and oxygen is an example of ______ reaction.
C. double replacement
D. single replacement
C. law of conservation of mass
law of definite proportions
Which of the following symbols means a substance is in water solution?
When oxygen is available, sulfur dioxide is produced from the burning of sulfur. Which of
C. sulfur dioxide → sulfur +
→ sulfur dioxide +
In a chemical equation, the symbol that takes the place of the word ‘yield’ is a(n)
uation stands for

72. 10. Why can’t you change the formula in order to balance a reaction?
A. The number of atoms in the left should be equal to the atoms on the right of the
arrow.
B.. The elements have definite mass ratios.
C. The molecule is indestructible.
D. The elements that combine have definite composition and fixed proportion of
elements by mass.

73. REFERENCES:
1. Echija, Elena, Cecilia Bayquen, Rafeal Alfonso, and Elmarita De Vera.
2020. Frontliners in Science and Technology. Makati City: Diwa
Scholastic Press
Silberberg, Martin. 2016. General Chemistry 1&2. McGrawHill Education

74. For inquiries and feedback, please write or call:
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Mass Relationships in Chemical Reactions
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76. General Mathematics- Grade 11
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Quarter 1 - Module 4: Quantifying Chemical Reactions
First Edition, 2020
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77. General Chemistry 1
Quarter 1
Mass Relationships in
Chemical Reactions
This instructional material was collaboratively developed and reviewed
by educators from public and
We encourage teachers and other education stakeholders to email their
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Department of
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General Chemistry 1
Quarter 1 - Module 4
Mass Relationships in
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This instructional material was collaboratively developed and reviewed
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feedback, comments, and recommendations to the Department of Education
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We value your feedback and recommendations.
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Mass Relationships in
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79. Table of Contents
What This Module is About.......................................................................................................................i
What I Need to Know..................................................................................................................................ii
How to Learn from this Module ..............................................................................................................iii
Icons of this Module...................................................................................................................................iii
What I Know.................................................................................................................................................iv
Lesson 1:
Determining the Amount of Reactant and Product in a
Chemical Reaction.....................................................................................................................1
What’s In .................................................................................................. 1
What I Need to Know ................................................................................ 1
What’s New: Fishball anyone? .................................................................. 2
What Is It: Stoichiometry .......................................................................... 2
What’s More: Stoich In Action .................................................................. 4
What I Have Learned: Tell Me What You Know ........................................ 4
What I Can Do: Show Me What You Know ............................................... 5
Additional Activity: Let’s Go Online ........................................................... 5
Lesson 2:
Limiting and Excess Reagent.......................................................................................6
What’s In................................................................................................................ 6
What I Need to Know........................................................................................... 6
What’s New: Excess Baggage ........................................................................ ..7
What Is It: Limiting and Excess Reagent ................................................. 7
What’s More: More Than Enough ............................................................ 9
What I Have Learned: Tell Me What You Know ……………………………. 9
What I Can Do: How Big is the Balloon …............................................... 10
Additional Activity: Let’s Go Online ........................................................... 10
Lesson 3:
Calculating Theoretical Yield and Percent Yield ...................................11
What’s In............................................................................................................................11

80. What I Need to Know.....................................................................................................11
What’s New: Encircle Me ......................................................................................... . 11
What Is It: Yield of the Reaction....................................................................... 12
What’s More: Solve, Solve, Solve..................................................................... 13
What’s More: Let’s Go Online............................................................................ 14
What I Have Learned: Tell Me What You Know …....................................... 14
What I Can Do: Step by Step .................................................................. 14
Summary ................................................................................................................................................... 15
Assessment: (Post-Test) ...................................................................................................................... 16
Key to Answers......................................................................................................................................... 18
References .................................................................................................................... 19

81. This page is intentionally blank

82. What This Module is About
Chemical reactions are apparent in the things around us or in our daily activities. The
quantitative relationship of reactants and products in a chemical reaction is manifested for
example, when we bake a cake or bread, or cook a dish. We make sure that all the
ingredients are present and in correct proportions based from a recipe to make the desired
end-product. Same concept applies to chemical reactions. After learning how to balance a
chemical equation in module 3, this module will help you understand the mass relationships
of reactants and products in a chemical reaction.
The lessons contained in this module are as follows:
 Lesson 1: Determining the Amount of Reactant and Product in a Chemical
Reaction.
 Lesson 2: Limiting and Excess Reagent.
 Lesson 3: Calculating Theoretical Yield and Percent Yield in a Reaction.
i

83. What I Need to Know
At the end of this module, you should be able to:
1. Construct mole or mass ratios for a reaction in order to calculate the amount of
reactant needed or amount of product formed in terms of moles or mass.
(STEM_GC11MRIg-h-38);
2. Determine mass relationship in a chemical reaction (STEM_GC11MRIg-h-42).
3. Explain the concept of limiting reagent in a chemical reaction; identify the excess
reagent(s) (STEM_GC11MRIg-h-40);
4. Calculate percent yield and theoretical yield of the (STEM_GC11MRIg-h-39);
ii

84. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the
• Answer all the given tests and exercises.
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
ow to Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that are in
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to
case your skills and knowledge gained, and
applied into real-life concerns and situations.
iii
activities and exercises diligently.
This part contains learning objectives that
are set for you to learn as you go along the
assessment as to your level of
knowledge to the subject matter at hand,
ifically to gauge prior related
This part connects previous lesson with that
lesson through
various activities, before it will be presented
These are discussions of the activities as a
way to deepen your discovery and under-
activities that are in-
tended for you to practice further in order to
Activities designed to process what you
have learned from the lesson
These are tasks that are designed to show-
case your skills and knowledge gained, and
life concerns and situations.

85. What I Know
Multiple Choice. Read and understand each question and s
answer from among the given choices.
1. Which of the following equations best represent the law of conservation of mass?
A. 1 + 2 → 3
B. 2 + 8 → 4
2. Which of the following statements is true about stoichiometric coefficient?
A. It is written to make the number of each element in a chemical equation the same
in the reactant and product side.
B. the sum of all coefficients in the left must be equal to
the right side of the equation.
C. It is the subscript written after each element.
D. None of the above.
3. In a chemical reaction, stoichiometry refers to:
A. Amount of materials consumed and products formed.
B. The activation energy
C. The rate or reaction
D. Reaction in equilibrium
For items 4 to 7, base your answer from the combustion of butane
the reaction:
2
4. What is the mole-mole factor of
A. 10:2
B. 2:13
5. which pair has the mole-
A. Butane and oxygen
B. Carbon dioxide and water
6. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is
produced?
A. 526.4 moles
B. 106.1 moles
7. How many grams of C4H
A. 453.6 grams
B. 548.1 grams
For items 8 to 11. In the reaction
made to react with 7.2 g of hydrogen gas
8. Which is the limiting reagent?
A. Hydrogen
B. Carbon monoxide
What I Know
Read and understand each question and select the letter of the
answer from among the given choices.
Which of the following equations best represent the law of conservation of mass?
C. 6 → 12 +
D. 2 → 2 + 2
Which of the following statements is true about stoichiometric coefficient?
It is written to make the number of each element in a chemical equation the same
in the reactant and product side.
B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
the right side of the equation.
C. It is the subscript written after each element.
D. None of the above.
In a chemical reaction, stoichiometry refers to:
A. Amount of materials consumed and products formed.
e activation energy
he rate or reaction
eaction in equilibrium
, base your answer from the combustion of butane (C4H
( ) + 13 ( ) → 8 ( ) + 10 ( )
mole factor of butane and water?
C. 8:10
D. 2:10
-mole factor of 8:10?
A. Butane and oxygen C. Oxygen and carbon dioxide
B. Carbon dioxide and water D. Water and butane
6. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is
C. 40.2 moles
D. 8.16 moles
H10 is needed to produce 37.8 moles of CO2?
C. 2.607 grams
D. 8769.6 grams
reaction 2H2 + CO → CH3OH, 6.8 g carbon monoxide gas
7.2 g of hydrogen gas. The reaction produced 5.2 g of methanol
8. Which is the limiting reagent?
C. methanol
D. CO2
iv
elect the letter of the best
Which of the following equations best represent the law of conservation of mass?
Which of the following statements is true about stoichiometric coefficient?
It is written to make the number of each element in a chemical equation the same
the sum of all coefficients in
H10) as shown in
C. Oxygen and carbon dioxide
6. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is
?
6.8 g carbon monoxide gas was
produced 5.2 g of methanol.

86. 9. What is the theoretical yield?
A. 57.6 g C. 7.8 g
B. 50.4 g D. 6.8 g
10. What is the percent yield?
A. 76% C. 68%
B. 78% D. 67%
v

87. This page is intentionally blank

88. Lesson Determining the Amount of
Reactant and Product in a
1
What’s In
You have learned from
before a chemical formula in the reactants or products side in order to balance a chemical
equation. This numerical coefficient in a balanced chemic
determine how much of the reactants are needed to produce a certain amount of product or
how much product is produced given a certain amount of reactant.
you to the concept of mass to mole or mole to mole
reactants and products in a chemical reaction
What I Need to Know
A balanced chemical equation indicates the number of moles required for each
reactant to produce a certain number of moles of the product/s. When you know the ratio of
the product to the reactants, it will be much easier to determine how much of the initia
materials are you going to prepare to achieve a specific amount of product. For instance,
you are selling fishballs in a stick for 10 pesos per stick. Each stick has 6 fish balls. If you
plan to sell 100 sticks of fishballs to
materials like cooking oil and fuel,
bamboo sticks and 500 pieces of raw fish balls from the market.
Similarly, when you have a balanced chemical equation,
the amount of product/s formed from knowing the amount of the reactants. In the same
manner, knowing the amount of product/s formed will help you determine the amount of
initial reactant used up in a single chemical reaction.
In this lesson, you will learn
balanced chemical equations and be able to perform stoichiometric calculations related to
chemical equations.
Determining the Amount of
Reactant and Product in a
Chemical Reaction
from the previous lesson that numerical coefficients are written
before a chemical formula in the reactants or products side in order to balance a chemical
This numerical coefficient in a balanced chemical equation
determine how much of the reactants are needed to produce a certain amount of product or
how much product is produced given a certain amount of reactant. This lesson will introduce
mass to mole or mole to mole ratio to determine the
in a chemical reaction.
What I Need to Know
A balanced chemical equation indicates the number of moles required for each
reactant to produce a certain number of moles of the product/s. When you know the ratio of
the product to the reactants, it will be much easier to determine how much of the initia
materials are you going to prepare to achieve a specific amount of product. For instance,
you are selling fishballs in a stick for 10 pesos per stick. Each stick has 6 fish balls. If you
plan to sell 100 sticks of fishballs to reach a profit of 1000 pesos, excluding all other
materials like cooking oil and fuel, you will know that you needed to buy 100 pieces of
bamboo sticks and 500 pieces of raw fish balls from the market.
Similarly, when you have a balanced chemical equation, you will be able to predict
the amount of product/s formed from knowing the amount of the reactants. In the same
manner, knowing the amount of product/s formed will help you determine the amount of
initial reactant used up in a single chemical reaction.
, you will learn to identify mole ratios of reactants and products from
balanced chemical equations and be able to perform stoichiometric calculations related to
1
Determining the Amount of
Reactant and Product in a
Chemical Reaction
the previous lesson that numerical coefficients are written
before a chemical formula in the reactants or products side in order to balance a chemical
al equation will be used to
determine how much of the reactants are needed to produce a certain amount of product or
This lesson will introduce
to determine the amount of
A balanced chemical equation indicates the number of moles required for each
reactant to produce a certain number of moles of the product/s. When you know the ratio of
the product to the reactants, it will be much easier to determine how much of the initial
materials are you going to prepare to achieve a specific amount of product. For instance,
you are selling fishballs in a stick for 10 pesos per stick. Each stick has 6 fish balls. If you
excluding all other
know that you needed to buy 100 pieces of
you will be able to predict
the amount of product/s formed from knowing the amount of the reactants. In the same
manner, knowing the amount of product/s formed will help you determine the amount of
to identify mole ratios of reactants and products from
balanced chemical equations and be able to perform stoichiometric calculations related to

89. What’s New
Fishball anyone?
Directions: Let us use the given example
determining the amount of reactants and products in a chemical reaction by answering the
questions below. Susan is selling fishballs i
fish balls.
1. Write the equation of fish balls, bamboo stick and fishballs in a stick.
__________________ +
2. If Susan has 100 fish balls, how many bamboo sticks will she need to consume all
the fish balls?
3. If Susan wants to make 50 fishballs in a stick, how many fish balls will she need?
What Is It
Stoichiometry
One of the requirements of a balanced chemical equation is that it follows the
Conservation of Mass, which states that matter is neither created nor destroyed. The
identity and quantity of the elements in the reactants side, though they can change in pairing
or arrangement, must be equal to the identity and quantity of elements in the products side.
To do this, all elements in the left side of the equation must
number in the right side of the equation. A
each element, ion or molecule
the number of the same element
coefficient denoted by a number
substance. The mole method
between reactants and products in a chemical reaction) more understandable.
example the formation of table salt or NaCl
The stoichiometric coefficients in the equation denotes that 2 elemental
react with 2 molecules of hydrochloric acid (
(NaCl) and 2 molecules of hydrogen (
equation can be interpreted and read
moles of NaCl and 1 mole of H
coefficient for the reactants and product
The symbol ≅ means “equivalent to”. Hence, we can make the following mole
factor:
or
Let’s consider a simple example. Ammonia, NH
used in the production of agricultural fertilizers and synthetic fibers. It is produced by the
reaction of nitrogen and hydrogen gases.
What’s New
the given example previously to understand the concept of
determining the amount of reactants and products in a chemical reaction by answering the
questions below. Susan is selling fishballs in a stick for 10 pesos per stick. Each stick has 5
Write the equation of fish balls, bamboo stick and fishballs in a stick.
+ ________________________ → ________________________
If Susan has 100 fish balls, how many bamboo sticks will she need to consume all
If Susan wants to make 50 fishballs in a stick, how many fish balls will she need?
One of the requirements of a balanced chemical equation is that it follows the
, which states that matter is neither created nor destroyed. The
identity and quantity of the elements in the reactants side, though they can change in pairing
or arrangement, must be equal to the identity and quantity of elements in the products side.
do this, all elements in the left side of the equation must be reflected
in the right side of the equation. A stoichiometric coefficient is then added
each element, ion or molecule to make the number of each element in the le
element in the right side of the equation. This stoichiometric
coefficient denoted by a number, can be interpreted as the number of moles of each
mole method approach makes stoichiometry (the quantitat
between reactants and products in a chemical reaction) more understandable.
example the formation of table salt or NaCl:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)
The stoichiometric coefficients in the equation denotes that 2 elemental
hydrochloric acid (HCl) to form 2 molecules o
hydrogen (H2) gas. Following the mole method approach, the
and read as 2 moles of Na react with 2 moles of HCl to form 2
moles of NaCl and 1 mole of H2. Since the equation is balanced, the stoichiometric
coefficient for the reactants and products can be used in a ratio or mole-mole factor
2 ≅ 2
means “equivalent to”. Hence, we can make the following mole
or or or or
Let’s consider a simple example. Ammonia, NH3, is a leading industrial chemical
sed in the production of agricultural fertilizers and synthetic fibers. It is produced by the
reaction of nitrogen and hydrogen gases.
3 ( ) + ( ) → 2 ( )
to understand the concept of
determining the amount of reactants and products in a chemical reaction by answering the
a stick for 10 pesos per stick. Each stick has 5
Write the equation of fish balls, bamboo stick and fishballs in a stick.
_______
If Susan has 100 fish balls, how many bamboo sticks will she need to consume all
If Susan wants to make 50 fishballs in a stick, how many fish balls will she need?
One of the requirements of a balanced chemical equation is that it follows the Law of
, which states that matter is neither created nor destroyed. The
identity and quantity of the elements in the reactants side, though they can change in pairing
or arrangement, must be equal to the identity and quantity of elements in the products side.
be reflected, and of the same
is then added before
in the left side equal to
in the right side of the equation. This stoichiometric
can be interpreted as the number of moles of each
(the quantitative relationship
between reactants and products in a chemical reaction) more understandable. Let’s take for
The stoichiometric coefficients in the equation denotes that 2 elemental sodium (Na)
to form 2 molecules of sodium chloride
. Following the mole method approach, the
as 2 moles of Na react with 2 moles of HCl to form 2
Since the equation is balanced, the stoichiometric
mole factor:
means “equivalent to”. Hence, we can make the following mole-mole
or
, is a leading industrial chemical
sed in the production of agricultural fertilizers and synthetic fibers. It is produced by the

90. 2
As shown in the balanced equation, 3 moles of H2 are stoichiometrically equivalent to
1 mole N2 and to 2 moles NH3. The ratio of moles H2 to moles NH3 in 3:2; the ratio of moles
N2 to moles NH3 is 1:2. Using this ratio, we will be able to calculate the quantities of the
product or reactant like the example below:
a. How many moles of H2 are needed to produce 26.5 moles of NH3?
Solution:
26.5
3
2
= 39.8
b. How many moles of NH3 will be produced if 33.7 moles of N2 reacts completely with H2?
Solution:
33.7
2
1
= 67.4
Note: In using the mole-mole factor or ratio, the arrangement of the numerator and
denominator is done in a way that there is a cancellation of similar units found in the
numerator and denominator. In first example above, moles of NH3 were cancelled while in
the second example, moles of N2 were cancelled.
The stoichiometric coefficients of the reactants and products can be used readily to
determine the mole-mole ratio of the substances involved. However, in laboratory setups,
the mass (in grams) of the substance are given or needed instead of moles for easy
measurement. In this type of problem, there is a need to convert the mass into number of
moles using the molar mass of the substance. Below are the steps used to convert grams to
moles or vice versa:
1. Convert the mass of the substance (A) to number of moles using its molar mass.
( )
1
( )
=
2. Using the number of moles of substance A and the mole-mole ratio of substance A
and B from the balanced equation, compute for the number of moles of B.
=
3. Convert the number of moles of substance B to mass using its molar mass.
( )
1
= ( )
The molar mass of the substance is used as a conversion factor to calculate the
number of moles of the substance given its mass and vice versa. The three steps can be
combined into a single step as follows:
( ) = ( ) ( )
( )
In every conversion, like the steps shown above, similar units found in the numerator
and denominator are cancelled out. Let’s have another example: Solid lithium hydroxide is
used to remove carbon dioxide and is called a CO2 scrubber. The reaction is:
2 ( ) + ( ) → ( ) + ( )

91. How many grams of CO2 can be absorbed by 236.1 g LiOH?
Solution:
( ) = 236.
( ) = 217
Note: Since the given value (236.1g) has 4 significant figures, the answer will also have 4
significant figures.
What’s More
Stoich in Action!
Direction: Use a separate sheet to show your
1. Rust, Fe2O3, form from the reaction of
a. Write the balanced equation of the reaction.
b. What is the mole-mole ratio of Fe to Fe
c. How many moles of Fe
d. What is the molar mass of Fe
e. How many grams of O
2. Given the decomposition reaction:
a. What is the mole-mole ratio of KClO
b. What is the molar mass of KClO
c. How many moles of KClO
d. How many grams of KCl are produced from an initial mass of 568.4 g KClO
What I have learned
Tell me what you Know
Directions: In a separate sheet of paper, write your comprehensive understanding of
this lesson by answering the questions.
1. Make a flowchart for determining the mass of the product from a given mass
of the reactant.
2. How does an imbalanced chemical equation affect your calculation o
amount of reactant or product in a chemical reaction?
3. Describe how you would determine the mole
chemical equation.
4. Explain and give an example of
the conversion of mass to mole
5. Site a situation in your daily life where you can apply the concept of
stoichiometry.
3
can be absorbed by 236.1 g LiOH?
.2
1
23.949
1
2
217.0
Since the given value (236.1g) has 4 significant figures, the answer will also have 4
What’s More
Use a separate sheet to show your computation for the following problems:
form from the reaction of iron and oxygen in the following equation:
( ) + ( ) → ( )
Write the balanced equation of the reaction.
mole ratio of Fe to Fe2O3?
How many moles of Fe2O3 is produced from 39.4 moles of Fe?
What is the molar mass of Fe2O3?
How many grams of O2 are needed to produce 29.8 g of Fe2O3
Given the decomposition reaction: 2 ( ) → 2 ( ) + 3 (
mole ratio of KClO3 to O2?
the molar mass of KClO3?
How many moles of KClO3 are needed to produce 36.6 moles of O
How many grams of KCl are produced from an initial mass of 568.4 g KClO
What I have learned
sheet of paper, write your comprehensive understanding of
this lesson by answering the questions.
for determining the mass of the product from a given mass
How does an imbalanced chemical equation affect your calculation o
amount of reactant or product in a chemical reaction?
Describe how you would determine the mole-mole factor or ratio from a
and give an example of how you would arrange the units and ratio in
the conversion of mass to mole of reactants to products.
Site a situation in your daily life where you can apply the concept of
44.01
1
Since the given value (236.1g) has 4 significant figures, the answer will also have 4
computation for the following problems:
iron and oxygen in the following equation:
3?
( )
are needed to produce 36.6 moles of O2?
How many grams of KCl are produced from an initial mass of 568.4 g KClO3?
sheet of paper, write your comprehensive understanding of
for determining the mass of the product from a given mass
How does an imbalanced chemical equation affect your calculation of the
mole factor or ratio from a
you would arrange the units and ratio in
Site a situation in your daily life where you can apply the concept of

92. What I Can Do
Show Me What You Know
Directions: Solve the following problems. Write your complete solution and answers
on a separate sheet of paper.
1. The fertilizer ammonium sulfate is prepared by the reaction between ammonia
and sulfuric acid:
a. What is the molar mass of NH
b. What is the molar mass of (NH
c. What is the mole-mole ratio of
d. How many moles of ammonium sulfate is produced when 638.4 g of NH
reacts completely with H
e. How many grams of NH
2. In the fermentation process, ethanol is produced from decomposition of glucose.
a. Write a balanced chemical equation of
b. What is the mole-mole ratio of glucose
c. What is the molar mass of glucose?
d. What is the molar mass of ethanol?
e. How much ethanol is produced
glucose?
Additional Activity
Let’s go online!
Directions: Go to the link below and practice what you’ve learned from this lesson:
https://bit.ly/2Ce7nkm
https://bit.ly/2Ce7z34
https://bit.ly/2ZbtkJD
4
What I Can Do
Show Me What You Know
Directions: Solve the following problems. Write your complete solution and answers
sheet of paper.
fertilizer ammonium sulfate is prepared by the reaction between ammonia
2 ( ) + ( ) → ( )
What is the molar mass of NH3?
What is the molar mass of (NH4)2SO4?
mole ratio of NH3 to (NH4)2SO4?
How many moles of ammonium sulfate is produced when 638.4 g of NH
reacts completely with H2SO4?
How many grams of NH3 is needed to produce 50.0 g of (NH4
In the fermentation process, ethanol is produced from decomposition of glucose.
→ +
Glucose Ethanol Carbon dioxide
Write a balanced chemical equation of the decomposition of glucose
mole ratio of glucose ( ) to ethanol (
What is the molar mass of glucose?
What is the molar mass of ethanol?
How much ethanol is produced from the starting material of
Additional Activity
Go to the link below and practice what you’ve learned from this lesson:
Directions: Solve the following problems. Write your complete solution and answers
fertilizer ammonium sulfate is prepared by the reaction between ammonia
How many moles of ammonium sulfate is produced when 638.4 g of NH3
4)2SO4?
In the fermentation process, ethanol is produced from decomposition of glucose.
the decomposition of glucose.
to ethanol ( )
from the starting material of 987.1 g of
Go to the link below and practice what you’ve learned from this lesson:

93. Lesson
Limiting and Excess Reagent
2
What’s In
You have learned from lesson 1 that you can predict the amount of product produced
given the initial amount of reactant, the mole
initial reactant required to produce the desired amount of product can also be co
the same manner. In a real scenario however, the amount of
reaction are not exactly available according to the proportion stoichiometrically identified
from the balanced equation. How to determine which of the reacta
which reactant is used up first in the reaction will be discussed in this lesson.
.
What I Need to Know
Suppose you are preparing cheeseburgers following strictly the proportions: 1 burger
bun, 1 patty and 1 slice of cheese to produce 1 cheeseburger. The number of
cheeseburgers produced depends largely on the availability of materials. There could be
instances where all the ingredients are present in exact
materials. Oftentimes, the number of buns, cheese slices and patties are not the same. You
may have bought more buns than patties or there are more cheese slices than buns
ingredients present in the least amount will determine how many cheeseburgers will be
produced while the ingredient present in excess
cheeseburgers.
In the same analogy, in a chemical reaction, the amount of reacta
in the exact stoichiometrically determined ratio. In this lesson, you will learn how to
determine which reactant is present in excess and which reactant is used up first in a
reaction.
5
Limiting and Excess Reagent
ou have learned from lesson 1 that you can predict the amount of product produced
given the initial amount of reactant, the mole-mole factor and the molar mass. The amount of
initial reactant required to produce the desired amount of product can also be co
the same manner. In a real scenario however, the amount of the reactants
not exactly available according to the proportion stoichiometrically identified
from the balanced equation. How to determine which of the reactants is present in excess or
which reactant is used up first in the reaction will be discussed in this lesson.
What I Need to Know
Suppose you are preparing cheeseburgers following strictly the proportions: 1 burger
bun, 1 patty and 1 slice of cheese to produce 1 cheeseburger. The number of
cheeseburgers produced depends largely on the availability of materials. There could be
es where all the ingredients are present in exact proportion, so there are no excess
materials. Oftentimes, the number of buns, cheese slices and patties are not the same. You
may have bought more buns than patties or there are more cheese slices than buns
ingredients present in the least amount will determine how many cheeseburgers will be
produced while the ingredient present in excess will have leftovers after making
In the same analogy, in a chemical reaction, the amount of reacta
in the exact stoichiometrically determined ratio. In this lesson, you will learn how to
determine which reactant is present in excess and which reactant is used up first in a
Limiting and Excess Reagent
ou have learned from lesson 1 that you can predict the amount of product produced
mole factor and the molar mass. The amount of
initial reactant required to produce the desired amount of product can also be computed in
reactants involved in the
not exactly available according to the proportion stoichiometrically identified
nts is present in excess or
which reactant is used up first in the reaction will be discussed in this lesson.
Suppose you are preparing cheeseburgers following strictly the proportions: 1 burger
bun, 1 patty and 1 slice of cheese to produce 1 cheeseburger. The number of
cheeseburgers produced depends largely on the availability of materials. There could be
, so there are no excess
materials. Oftentimes, the number of buns, cheese slices and patties are not the same. You
may have bought more buns than patties or there are more cheese slices than buns. The
ingredients present in the least amount will determine how many cheeseburgers will be
will have leftovers after making
In the same analogy, in a chemical reaction, the amount of reactants are not present
in the exact stoichiometrically determined ratio. In this lesson, you will learn how to
determine which reactant is present in excess and which reactant is used up first in a

94. 6
What’s New
Excess Baggage
Directions: The table below shows the primary materials needed to produce the items in the
first column. With the given available materials, identify how many items are produced,
which materials are in excess and how many are left of the available materials.
Item Required Available materials Produced
set
Excess
materials
Bicycle 1 bike frame, 2 tires
2 pedals, 1 crank
arm, 1 brake set
68 bike frames, 117
tires, 250 pedals, 72
crank arm, 93 brake
set
Banana
Cue
2 bananas, 1
barbecue stick, ½
cup brown sugar
236 bananas, 150
barbecue stick, 25
cups brown sugar
Cheese
burger
1 burger bun, 1 beef
patty, 1 slice of
cheese
324 burger buns, 12
dozens beef patties,
261 slices of cheese
Table 1 table top, 4 legs, 8
nuts, 8 screws
20 table tops, 50
legs, 50 nuts, 50
screws
Milk tea 1 bag of black tea,
250 ml water, 1/8
cup milk, 2 tbsp
sugar, ¼ cup
tapioca pearls
15 tea bags, 3 liters
water, 2 cups milk,
30 tbsp sugar, 4
cups tapioca pearls
What Is It
Limiting and Excess Reagent
When a chemical reaction is carried out in a flask, the amount of reactants are not
always present in the exact proportion stoichiometrically determined from the balanced
equation. To make sure that all of the more expensive reagent is completely used up and
converted to the desired product, chemists usually add the cheaper reagent in excess
quantity. The reagent that has completely reacted and used up in a reaction is called the
limiting reagent. The excess reagent is the reactant that is present in quantity higher than
what is required to react with the limiting reagent. For instance, you are preparing a ball
dance for your 18th
birthday consisting of pairs of male and female. Upon checking your list
of friends, you found out that you have 18 male friends and 25 female friends. The number
of male friends will limit the number of pairs to 18. All the males will have a partner while
there will be 7 females who will not have a partner for the dance.
Let’s use that context to the balanced chemical equation below:

95. 3 ( ) + ( ) → 2 ( )
7
Ammonia, NH3, is synthesized from the reaction of H2 and N2 gases. Suppose 6
moles of H2 was initially mixed with 4 moles of N2 gas at high pressure. To determine which
of the 2 reactants is the limiting reagent, the amount of NH3 produced must be computed
given the number of moles of H2 and N2 and the mole-mole factor from the balanced
equation.
= #
2
3
= 6
2
3
= 4
=
2
1
= 4
2
1
= 8
From the computed values we determined that if 6 moles of H2 completely reacts
with N2, it can produce 4 moles of NH3 while 4 moles of N2 can produce 8 moles of NH3
when fully used up. Since there is less amount of NH3 produced with 6 moles of H2 than 4
moles of N2, H2 gas is the limiting reagent while the N2 gas is the reagent in excess. To
determine how much of the 4 moles of N2 is in excess, we will use the mole-mole factor of N2
and H2.
= 6
1
3
= 2
The number of moles of N2 required to fully react to 6 moles of H2 is only 2 moles.
Thus, the initial 6 moles of N2 has an excess of 4 moles.
= 6 − 2 = 4
Let’s have another example. The combustion of ethane produces carbon dioxide and
water shown in the reaction below:
2 + 7 → 4 + 6
1. How many moles of CO2 is produced with 56.2 moles C2H6 and 73.4 moles of O2?
Solution:
= 56.2
4
2
= 112.4
= 73.4
4
7
= 41.9
2. Which reagent is the limiting reagent? Which is the excess reagent?
Limiting reagent is O2 while the excess reagent is .
3. How many moles of is in excess?
Moles of required for 73.4 moles on O2:
= 73.4
2
7

96. Excess
Note: If the amount of the initial reactant is expressed in grams (mass) instead of moles, the
number of moles must first be converted into grams using the molar mass of the reactant as
you have learned in the previous lessons. In the same manner, if the problem asks to
determine the mass (in grams) of the product produced by the reaction, the number of moles
of the product must be converted into grams using the molar mass of the product.
What’s More
More Than Enough
Directions: Read carefully the following problems. Find what
Write your complete solution and answer in a separate sheet of paper.
1. In the production of Zinc sulfide
a. How many moles of ZnS is produced when sulfur is co
b. How many grams of ZnS is produced when zinc is completely used up?
c. Which reactant is the limiting reagent?
d. How many grams of the excess reagent is left?
2. An aqueous solution of NaOH can dissolve an aluminum sheet in the reaction:
2
If 126.4 g of NaOH is 97.7 g of Al are made to react,
a. How many moles of NaAlO
b. How many grams of NaAlO
c. Which reactant is the limiting reagent?
d. How many grams of the excess reagent is left?
3. Ferric chloride, FeCl3, reacts with silver nitrate, AgNO
and silver chloride, AgCl.
a. Write the balanced chemical equation of the
b. If 108.5 moles of FeCl
AgCl is produced?
c. How many grams of AgCl is produced?
d. Which reactant is the limiting reagent?
e. Which reactant is the excess reagent?
What I Have Learned
Tell Me What You Know
Directions: Based on what you have learned from this lesson, write a concise explanation of
the following questions. Write your answer on a separate sheet of paper.
1. What is the role of the stoichiometric coefficient in a balanced chemical equation to the
determination of the limiting reactant?
2. Why is it that in most cases, the amounts of reactants and products are indicated in grams
instead of the number for mole
3. How do you determine the limiting reactant in a chemical reaction?
4. If a chemical reaction involves only one reactant, will there be a limiting reagent?
= 20.9
Excess : 56.2 moles – 20.9 moles = 35.3 moles
8
initial reactant is expressed in grams (mass) instead of moles, the
number of moles must first be converted into grams using the molar mass of the reactant as
you have learned in the previous lessons. In the same manner, if the problem asks to
e mass (in grams) of the product produced by the reaction, the number of moles
of the product must be converted into grams using the molar mass of the product.
What’s More
Directions: Read carefully the following problems. Find what is asked for each problem.
Write your complete solution and answer in a separate sheet of paper.
inc sulfide, 36.8 g of zinc is made to react with 19.4 g of sulfur.
+ →
moles of ZnS is produced when sulfur is completely used up?
How many grams of ZnS is produced when zinc is completely used up?
Which reactant is the limiting reagent?
How many grams of the excess reagent is left?
An aqueous solution of NaOH can dissolve an aluminum sheet in the reaction:
+ 2 + 2 → 2 + 3
If 126.4 g of NaOH is 97.7 g of Al are made to react,
How many moles of NaAlO2 is produced if 126.4 g of NaOH reacts completely?
How many grams of NaAlO2 is produced using the same amount of NaOH?
Which reactant is the limiting reagent?
How many grams of the excess reagent is left?
, reacts with silver nitrate, AgNO3 to form ferric nitrate, Fe(NO
Write the balanced chemical equation of the reaction.
If 108.5 moles of FeCl3 is combined with 76.8 moles of AgNO3, how many moles of
How many grams of AgCl is produced?
Which reactant is the limiting reagent?
Which reactant is the excess reagent?
What I Have Learned
: Based on what you have learned from this lesson, write a concise explanation of
the following questions. Write your answer on a separate sheet of paper.
What is the role of the stoichiometric coefficient in a balanced chemical equation to the
determination of the limiting reactant?
2. Why is it that in most cases, the amounts of reactants and products are indicated in grams
instead of the number for moles.
How do you determine the limiting reactant in a chemical reaction?
If a chemical reaction involves only one reactant, will there be a limiting reagent?
20.9 moles = 35.3 moles
initial reactant is expressed in grams (mass) instead of moles, the
number of moles must first be converted into grams using the molar mass of the reactant as
you have learned in the previous lessons. In the same manner, if the problem asks to
e mass (in grams) of the product produced by the reaction, the number of moles
of the product must be converted into grams using the molar mass of the product.
is asked for each problem.
36.8 g of zinc is made to react with 19.4 g of sulfur.
mpletely used up?
How many grams of ZnS is produced when zinc is completely used up?
An aqueous solution of NaOH can dissolve an aluminum sheet in the reaction:
is produced if 126.4 g of NaOH reacts completely?
is produced using the same amount of NaOH?
to form ferric nitrate, Fe(NO3)3
how many moles of
: Based on what you have learned from this lesson, write a concise explanation of
the following questions. Write your answer on a separate sheet of paper.
What is the role of the stoichiometric coefficient in a balanced chemical equation to the
2. Why is it that in most cases, the amounts of reactants and products are indicated in grams
If a chemical reaction involves only one reactant, will there be a limiting reagent?

97. 5. Site one (1) real-life situation where you added more than what is needed to produce
something. Make justifications why you have to do it.
What I Can Do
How big is the balloon!
Directions: This activity is a simple application of the concept of limiting a
reactants that you can do at home.
the balloon after mixing the baking soda and vinegar then answer the guide questions
below. For this simple experiment, y
 3 350mL empty plastic water bottles
 Vinegar
 Baking soda
 3 balloons
 funnel
Procedure:
1. Collect, clean, and dry three (3)1 350 ml water bottles. Label the bottles 1, 2 and 3
2. Pour 5 tablespoons vinegar into bottle 1, 10 tablespoons into bottle 2 and 15
tablespoons into bottle 3.
3. Fit the funnel into the opening of the balloon and ca
soda into each balloon.
4. Carefully press the balloon to remove extra air inside without spilling the baking soda.
5. Fit the balloon snugly into the lip or opening of the water bottle. Make sure that the
baking soda does not
into the bottle, use rubber bands to ensure that air is trapped inside the bottle.
6. Repeat steps 3 to 5 for the other 2 balloons and water bottles.
7. Stretch the balloon over the top of the
vinegar. The baking soda will react with vinegar as shown in the equation
+
Acetic acid
8. Record your observations then answer the questions below.
9. Clean your work area and dispose the materials properly.
After performing the activity. answer
separate sheet of paper.
1. When baking soda was mixed with vinegar, effervescence (fizzing or bubbling) was
observed. What caused the effervescence?
2. How can you conclude that the reaction has gone to completi
3. Which balloon has the least amount of air inside?
4. Which balloon has the most air inside?
5. Will the balloon size grow bigger if more vinegar is in the bottle? Why?
6. Will it take more time to complete the reaction if there were more vinegar in the bottle
with the same amount of baking soda?
7. Which is the limiting reagent?
8. Which is the reagent in excess?
Additional Activity
Let’s go online!
life situation where you added more than what is needed to produce
hing. Make justifications why you have to do it.
9
What I Can Do
Directions: This activity is a simple application of the concept of limiting a
reactants that you can do at home. Follow the instruction properly, observe what happens to
the balloon after mixing the baking soda and vinegar then answer the guide questions
below. For this simple experiment, you will need the following items:
3 350mL empty plastic water bottles
Collect, clean, and dry three (3)1 350 ml water bottles. Label the bottles 1, 2 and 3
Pour 5 tablespoons vinegar into bottle 1, 10 tablespoons into bottle 2 and 15
tablespoons into bottle 3.
Fit the funnel into the opening of the balloon and carefully put 1 tablespoon of baking
soda into each balloon.
Carefully press the balloon to remove extra air inside without spilling the baking soda.
Fit the balloon snugly into the lip or opening of the water bottle. Make sure that the
baking soda does not fall into the bottle while doing this. If the balloon is loosely fit
into the bottle, use rubber bands to ensure that air is trapped inside the bottle.
Repeat steps 3 to 5 for the other 2 balloons and water bottles.
Stretch the balloon over the top of the bottle to slowly pour the baking soda into the
The baking soda will react with vinegar as shown in the equation
+ → + +
Acetic acid Sodium Sodium water
bicarbonate acetate
Record your observations then answer the questions below.
Clean your work area and dispose the materials properly.
After performing the activity. answer the following questions. Write your answer on a
When baking soda was mixed with vinegar, effervescence (fizzing or bubbling) was
observed. What caused the effervescence?
How can you conclude that the reaction has gone to completion?
Which balloon has the least amount of air inside?
Which balloon has the most air inside?
Will the balloon size grow bigger if more vinegar is in the bottle? Why?
Will it take more time to complete the reaction if there were more vinegar in the bottle
ith the same amount of baking soda?
Which is the limiting reagent?
Which is the reagent in excess?
Additional Activity
life situation where you added more than what is needed to produce
Directions: This activity is a simple application of the concept of limiting and excess
serve what happens to
the balloon after mixing the baking soda and vinegar then answer the guide questions
Collect, clean, and dry three (3)1 350 ml water bottles. Label the bottles 1, 2 and 3
Pour 5 tablespoons vinegar into bottle 1, 10 tablespoons into bottle 2 and 15
refully put 1 tablespoon of baking
Carefully press the balloon to remove extra air inside without spilling the baking soda.
Fit the balloon snugly into the lip or opening of the water bottle. Make sure that the
fall into the bottle while doing this. If the balloon is loosely fit
into the bottle, use rubber bands to ensure that air is trapped inside the bottle.
bottle to slowly pour the baking soda into the
The baking soda will react with vinegar as shown in the equation:
+
Carbon
dioxide
the following questions. Write your answer on a
When baking soda was mixed with vinegar, effervescence (fizzing or bubbling) was
Will the balloon size grow bigger if more vinegar is in the bottle? Why?
Will it take more time to complete the reaction if there were more vinegar in the bottle

98. Directions: Go to the link below and practice what you’ve learned from this lesson:
https://bit.ly/309SojJ
https://bit.ly/3fGoF8g
https://bit.ly/3fui4ht
Lesson Calculating Theoretical Yield and Percent
Yield in a Reaction
3
What’s In
In your previous lessons, you were taught on how to calculate molecular formula
given molar mass, write and balance chemical equations and construct mole or mass ratios
for a reaction with stoichiometry in order to determine the amount of reactants needed
form the products.
In this lesson, you will learn how to differentiate the theoretical, actual, and percent
yield as well as the process on how to calculate them using stoichiometry.
What I Need to Know
Ideally, in a chemical reaction, it is pr
from the given reactants. But in reality, it is not that easy to achieve. Most of the chemical
reactions obtain are less than the 100% yield of the product/s due to several factors that
could affect the reaction process such as; experimental errors, incomplete reactions,
unexpected side reactions, amount of the reactant/s, undesirable by
external factors, etc. For you to be able to eliminate all of these contributory factors, you
have to obtain an “ideal environment” during the experimentation to achieve 100% yield,
which is close to impossible to attain under normal conditions.
In order to evaluate the success of a chemical reaction, you need to determine the
percent yield by calculating the
What’s New
Encircle Me…
Direction: In this chemical equation, i
You may use another sheet of paper or you may answer directly on this page. That
is why the font is big so that it would be easier for you to encircle.
2NH
Go to the link below and practice what you’ve learned from this lesson:
10
Calculating Theoretical Yield and Percent
Yield in a Reaction
In your previous lessons, you were taught on how to calculate molecular formula
given molar mass, write and balance chemical equations and construct mole or mass ratios
for a reaction with stoichiometry in order to determine the amount of reactants needed
In this lesson, you will learn how to differentiate the theoretical, actual, and percent
s well as the process on how to calculate them using stoichiometry.
What I Need to Know
Ideally, in a chemical reaction, it is predicted to produce a 100% yield of the product
from the given reactants. But in reality, it is not that easy to achieve. Most of the chemical
reactions obtain are less than the 100% yield of the product/s due to several factors that
on process such as; experimental errors, incomplete reactions,
unexpected side reactions, amount of the reactant/s, undesirable by
external factors, etc. For you to be able to eliminate all of these contributory factors, you
an “ideal environment” during the experimentation to achieve 100% yield,
which is close to impossible to attain under normal conditions.
In order to evaluate the success of a chemical reaction, you need to determine the
percent yield by calculating the theoretical yield and the actual yield of the reaction process.
What’s New
In this chemical equation, identify and encircle the following
You may use another sheet of paper or you may answer directly on this page. That
is why the font is big so that it would be easier for you to encircle.
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Go to the link below and practice what you’ve learned from this lesson:
Calculating Theoretical Yield and Percent
In your previous lessons, you were taught on how to calculate molecular formula
given molar mass, write and balance chemical equations and construct mole or mass ratios
for a reaction with stoichiometry in order to determine the amount of reactants needed to
In this lesson, you will learn how to differentiate the theoretical, actual, and percent
s well as the process on how to calculate them using stoichiometry.
edicted to produce a 100% yield of the product
from the given reactants. But in reality, it is not that easy to achieve. Most of the chemical
reactions obtain are less than the 100% yield of the product/s due to several factors that
on process such as; experimental errors, incomplete reactions,
unexpected side reactions, amount of the reactant/s, undesirable by-product/s, other
external factors, etc. For you to be able to eliminate all of these contributory factors, you
an “ideal environment” during the experimentation to achieve 100% yield,
In order to evaluate the success of a chemical reaction, you need to determine the
theoretical yield and the actual yield of the reaction process.
the following parts.
You may use another sheet of paper or you may answer directly on this page. That

99. 11
Direction of the
Reaction
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Product/s
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Reactant/s
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Subscripts
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Coefficients
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
What Is It
Yield of the Reaction
Chemical reaction is a process in which substance/s interact chemically to produce a
new substance/s with different compositions. It is represented by a chemical equation.
Substance/s at the left side of the equation is/are called the reactant/s while the substance/s
located at the right side of the equation is/are called the product/s. Chemical properties of
the element or compound as reactant/s dictate the process in which an element or
compound undergoes changes during the reaction.
Evaluation of the percent yield is important to measure the success of a chemical
reaction. Percent yield is the ratio of the actual yield to the theoretical yield expressed as a
percentage. So, how are you going to compute for the percent yield of a chemical reaction?
You have to know the values of the theoretical yield and the actual yield of the reaction.
Following the formula of:
= %
Theoretical yield is the amount of product that is expected to form based on
stoichiometry. It is the maximum amount of product produced from the given amount/s of
reactant/s. It is calculated based on the stoichiometry of the chemical reaction. While the
actual yield is the amount of product produced during the reaction. It is the amount of
product obtained after the actual reaction and it is normally lesser than the theoretical yield.
It is determined experimentally.
These are the steps in calculating the percent yield of a chemical reaction:

100. 1. Balance the given chemical equation
2. Identify the limiting reactant and the excess reactant
3. Compute for the theoretical yield of the reaction
4. Calculate the percent yield
You can continue solving when asked:
a. Percent error
b. Amount of excess
What’s More
Activity 1: Solve, Solve, Solve…
Direction: Determine the theoretical yield and the actual yield, given the information in each
question. You must show your work, including units, through each step of the calculations.
Use separate papers for your answers for these set of problems. (
adapted and modified from https://bit.ly/2W18laC
1. Cl2(g) +
70.90 g/mol
a. Calculate the theoretical yield of aluminum chloride (in grams) that can be
produced from 10.00 grams of aluminum metal.
b. An experiment was performed and obtained 25.23 grams of aluminum chloride.
Determine the percent yield of aluminum chloride.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
2. V(s) +
50.94 g/mol
a. Calculate the theoretical yield of vanadium (III) oxide, assuming you begin
with 200.00 grams vanadium metal.
b. After the experiment is performed, an experimental yield of 183.2 grams is
produced. Calculate the percent yield for this experiment.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
3. KI(aq) + H2O(l)
166.0 /mol 18.02 g/mol
a. Calculate the mass of manganese (IV) oxide that can be synthesized from
15.00 grams of potassium iodide.
b. Calculate the percent yield of this experiment if a mass of 1.982 grams of
manganese (IV) oxide is produced.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
Balance the given chemical equation
Identify the limiting reactant and the excess reactant
12
Compute for the theoretical yield of the reaction
Calculate the percent yield
You can continue solving when asked:
Percent error
Amount of excess reactant
What’s More
Solve, Solve, Solve…
: Determine the theoretical yield and the actual yield, given the information in each
question. You must show your work, including units, through each step of the calculations.
separate papers for your answers for these set of problems. (These problems are
adapted and modified from https://bit.ly/2W18laC).
Al(s) AlCl3(s)
26.98 g/mol 133.33 g/mol
Calculate the theoretical yield of aluminum chloride (in grams) that can be
produced from 10.00 grams of aluminum metal.
An experiment was performed and obtained 25.23 grams of aluminum chloride.
Determine the percent yield of aluminum chloride.
the percent error.
Compute for the excess amount of the excess reactant.
O2(g) V2O3(s)
32.00 g/mol ∆ 149.88 g/mol
Calculate the theoretical yield of vanadium (III) oxide, assuming you begin
with 200.00 grams vanadium metal.
After the experiment is performed, an experimental yield of 183.2 grams is
produced. Calculate the percent yield for this experiment.
he percent error.
Compute for the excess amount of the excess reactant.
+ KMnO4(aq) I2(s) + MnO2(s)
158.03 g/mol 253.80 g/mol 86.97 g/mol
Calculate the mass of manganese (IV) oxide that can be synthesized from
15.00 grams of potassium iodide.
Calculate the percent yield of this experiment if a mass of 1.982 grams of
manganese (IV) oxide is produced.
Determine the percent error.
Compute for the excess amount of the excess reactant.
: Determine the theoretical yield and the actual yield, given the information in each
question. You must show your work, including units, through each step of the calculations.
These problems are
Calculate the theoretical yield of aluminum chloride (in grams) that can be
An experiment was performed and obtained 25.23 grams of aluminum chloride.
Calculate the theoretical yield of vanadium (III) oxide, assuming you begin
After the experiment is performed, an experimental yield of 183.2 grams is
2(s) + KOH(aq)
86.97 g/mol 56.11 g/mol
Calculate the mass of manganese (IV) oxide that can be synthesized from
Calculate the percent yield of this experiment if a mass of 1.982 grams of

101. Activity 2: Let’s Go Online…
Direction: Refer to the following sites/links for further discussions on how to calculate the
theoretical yield using stoichiometry and percent yield using the
comprehensive summary and personal reflection on what you have learn from the videos.
Write your answers on a separate paper.
https://bit.ly/3gEnN4f
https://bit.ly/2CjAYIS
https://bit.ly/3edzFIM
https://bit.ly/3faDK1W
What I Have
Tell Me What You Know
Based on what you have learned from this lesson, briefly discuss the following questions.
Limit your answers to 3-5 sentences. Use a separate paper for your answers.
1. Why is it important to calculate the percent yield in a
2. In a chemical reaction, explain the difficulty in obtaining the theoretical yield.
3. Discuss the reason why is it, that the actual yield is often lesser than the theoretical
yield?
4. Which is more important to determine in a chemical reacti
the excess reactant? Elaborate your answer.
5. Why do you need to balance first the chemical equation before proceeding to the
calculations of theoretical yield and percent yield?
What I Can Do
Step-by-step…
Direction: Carefully read the following word problems. From the knowledge and skills in
computation that you have acquired in this lesson, do the following:
a. Balance the chemical equation;
b. Identify the limiting reactant and the excess reactant if applicable;
c. Compute for the theoretical yield;
d. Determine the percent yield of the reaction;
e. Calculate the percent error and
f. Compute for the excess amount of the excess reactant if applicable.
13
Let’s Go Online…
Refer to the following sites/links for further discussions on how to calculate the
theoretical yield using stoichiometry and percent yield using the formula. Give your
comprehensive summary and personal reflection on what you have learn from the videos.
Write your answers on a separate paper.
What I Have Learned
Based on what you have learned from this lesson, briefly discuss the following questions.
5 sentences. Use a separate paper for your answers.
Why is it important to calculate the percent yield in a chemical reaction?
In a chemical reaction, explain the difficulty in obtaining the theoretical yield.
Discuss the reason why is it, that the actual yield is often lesser than the theoretical
Which is more important to determine in a chemical reaction, the limiting reactant or
the excess reactant? Elaborate your answer.
Why do you need to balance first the chemical equation before proceeding to the
calculations of theoretical yield and percent yield?
What I Can Do
arefully read the following word problems. From the knowledge and skills in
computation that you have acquired in this lesson, do the following:
Balance the chemical equation;
Identify the limiting reactant and the excess reactant if applicable;
or the theoretical yield;
Determine the percent yield of the reaction;
Calculate the percent error and
Compute for the excess amount of the excess reactant if applicable.
Refer to the following sites/links for further discussions on how to calculate the
formula. Give your
comprehensive summary and personal reflection on what you have learn from the videos.
Based on what you have learned from this lesson, briefly discuss the following questions.
5 sentences. Use a separate paper for your answers.
chemical reaction?
In a chemical reaction, explain the difficulty in obtaining the theoretical yield.
Discuss the reason why is it, that the actual yield is often lesser than the theoretical
on, the limiting reactant or
Why do you need to balance first the chemical equation before proceeding to the
arefully read the following word problems. From the knowledge and skills in
Compute for the excess amount of the excess reactant if applicable.

102. g. Show the complete solutions for your answers in a separate paper.
1. Eighty grams of Silver was obtained from one hundred and forty grams of Silver
nitrate. The Silver metal is prepared by reducing its nitrate. The chemical equation of
the reaction is:
Cu(s) +
2. A hundred grams of Sulfuric acid y
H2SO4(l)
3. Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of
the product.
P4(s) +
4. Aspirin is one of the products when you mix Salicylic acid and
kilogram of Salicylic acid produced one thousand and two hundred grams of Aspirin.
C7H6O3(s) +
Summary
The Law of Conservation of Mass
(product) in a chemical reaction must be equal to the amount of the initial materials
(reactants). The stoichiometric coefficient
left and right side of the equat
factor or ratio between and among substances involved (both product and reactant). The
mole method approach makes stoichiometric calculation easier by expressing the known
and unknown quantities in moles then convert it to grams using the molar mass of the
substance. In some reactions, an excess amount of a less expensive reagent (
excess) is added to make sure that the more expensive reagent is completely converted to
product. The chemical reaction will
the time, the expected amount of product (
amount produced in the reaction (
determined by computing the
theoretical yield multiplied by 100. The closer the value of the percent yield to 100, the more
efficient is the reaction.
Show the complete solutions for your answers in a separate paper.
14
of Silver was obtained from one hundred and forty grams of Silver
nitrate. The Silver metal is prepared by reducing its nitrate. The chemical equation of
14
AgNO3(aq) Cu(NO3)2(aq)
A hundred grams of Sulfuric acid yielded ten grams of water.
H2O(l) + SO3(l)
Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of
Cl2(g) PCl3(l)
Aspirin is one of the products when you mix Salicylic acid and Acetic anhydride. One
kilogram of Salicylic acid produced one thousand and two hundred grams of Aspirin.
C4H6O3(l) C9H8O4(s) +
Law of Conservation of Mass provides that the amount of substances formed
(product) in a chemical reaction must be equal to the amount of the initial materials
stoichiometric coefficient balances the number of elements present in the
left and right side of the equation. This coefficient can be used to determine the
or ratio between and among substances involved (both product and reactant). The
approach makes stoichiometric calculation easier by expressing the known
in moles then convert it to grams using the molar mass of the
In some reactions, an excess amount of a less expensive reagent (
) is added to make sure that the more expensive reagent is completely converted to
cal reaction will stop when all the limiting reagent is used up.
the time, the expected amount of product (theoretical yield) is not the same as the actual
amount produced in the reaction (actual yield). The efficiency of the conducted
determined by computing the percent yield which is the ratio of the actual yield and the
theoretical yield multiplied by 100. The closer the value of the percent yield to 100, the more
Show the complete solutions for your answers in a separate paper.
of Silver was obtained from one hundred and forty grams of Silver
nitrate. The Silver metal is prepared by reducing its nitrate. The chemical equation of
+ Ag(s)
Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of
Acetic anhydride. One
kilogram of Salicylic acid produced one thousand and two hundred grams of Aspirin.
CH3COOH(l)
substances formed
(product) in a chemical reaction must be equal to the amount of the initial materials
balances the number of elements present in the
ion. This coefficient can be used to determine the mole-mole
or ratio between and among substances involved (both product and reactant). The
approach makes stoichiometric calculation easier by expressing the known
in moles then convert it to grams using the molar mass of the
In some reactions, an excess amount of a less expensive reagent (reagent in
) is added to make sure that the more expensive reagent is completely converted to
is used up. Most of
) is not the same as the actual
). The efficiency of the conducted reaction is
which is the ratio of the actual yield and the
theoretical yield multiplied by 100. The closer the value of the percent yield to 100, the more

103. 15

104. Assessment:
Directions. Read and understand each question and select the letter of the best answer
from among the given choices.
1. Which of the following equations best represent the law of conservation of mass?
. + 2 → 3
B. 2 + 8 → 4
2. Which of the following statements is true about stoichiometric coefficient?
A. It is written to make the number of each element in a chemical equation the same
in the reactant and product
B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
the right side of the equation.
C. It is the subscript written after each element.
D. None of the above.
3.. In a chemical reaction, stoichiometry re
A. Amount of materials consumed and products formed.
B. the activation energy
C. the rate or reaction
D. reaction in equilibrium
For items 4 to 6. In the reaction 2H
to react with 7.2 g of hydrogen gas. The reaction produced 5.2 g of methanol.
4. Which is the limiting reagent?
A. Hydrogen
B. Carbon monoxide
5. What is the theoretical yield?
A. 57.6 g
B. 50.4 g
6. What is the percent yield?
A. 76%
B. 78%
For items 7 to 10, base your answer from the combustion of butane (C
reaction:
2
7. What is the mole-mole factor of butane and water?
A. 10:2
B. 2:13
8. which pair has the mole-
A. Butane and oxygen
B. Carbon dioxide and water
Assessment: (Post-Test)
Read and understand each question and select the letter of the best answer
from among the given choices.
Which of the following equations best represent the law of conservation of mass?
C. 6 → 12 +
D. 2 → 2 + 2
Which of the following statements is true about stoichiometric coefficient?
It is written to make the number of each element in a chemical equation the same
in the reactant and product side.
B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
the right side of the equation.
C. It is the subscript written after each element.
D. None of the above.
In a chemical reaction, stoichiometry refers to:
A. Amount of materials consumed and products formed.
B. the activation energy
D. reaction in equilibrium
reaction 2H2 + CO → CH3OH, 6.8 g carbon monoxide gas was made
hydrogen gas. The reaction produced 5.2 g of methanol.
4. Which is the limiting reagent?
C. methanol
D. CO2
What is the theoretical yield?
C. 7.8 g
D. 6.8 g
What is the percent yield?
C. 68%
D. 67%
, base your answer from the combustion of butane (C4H10
( ) + 13 ( ) → 8 ( ) + 10 ( )
mole factor of butane and water?
C. 8:10
D. 2:10
-mole factor of 8:10?
A. Butane and oxygen C. Oxygen and carbon dioxide
B. Carbon dioxide and water D. Water and butane
16
Read and understand each question and select the letter of the best answer
Which of the following equations best represent the law of conservation of mass?
Which of the following statements is true about stoichiometric coefficient?
It is written to make the number of each element in a chemical equation the same
B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
OH, 6.8 g carbon monoxide gas was made
hydrogen gas. The reaction produced 5.2 g of methanol.
10) as shown in the
C. Oxygen and carbon dioxide

105. 9. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is
produced?
A. 526.4 moles C. 40.2 moles
B. 106.1 moles D. 8.16 moles
10. How many grams of C4H10 is needed to produce 37.8 moles of CO2?
A. 453.6 grams C. 2.607 grams
B. 548.1 grams D. 8769.6 grams
17

106. Key to Answers
Pretest Post Test
1. B
2. A
3. A
4. D
5. B
6. C
7. B
8. B
9. C
10. D
Key to Answers
Post Test
18
1. B
2. A
3. A
4. B
5. C
6. D
7. D
8. B
9. C
10. B

107. References
Chang, Raymond. Chemistry, 10th ed. New York, NY: McGraw-Hill
Science/Engineering/Math, 2009.
"Chapter 7.4: Stoichiometry." Chemistry LibreTexts. Last modified June 5, 2019.
https://chem.libretexts.org/Courses/Howard_University/General_Chemistry%3A_An_Atoms_
First_Approach/Unit_3%3A_Stoichiometry/Chapter_7%3A_Stoichiometry/Chapter_7.4%3A_
Stoichiometry.
Commission on Higher Education, General Chemistry 1: Teaching Guide for Senior High,
Manila, 2016.
Department of Education Central Office, Most Essential Learning Competencies (MELCS),
Manila, 2020.
Khan Academy. "Stoichiometry | Chemical reactions and stoichiometry | Chemistry | Khan
Academy." YouTube. August 27, 2009. https://www.youtube.com/watch?v=SjQG3rKSZUQ.
"Limiting Reagents Practice Problems". 2020. Chemistry.Wustl.Edu. Accessed July 12.
http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Plink/limreag/probsetlr.htm.
"Limiting Reagent Stoichiometry (Practice)". 2020. Khan Academy. Accessed July 12.
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/limiting-
reagent-stoichiometry/e/limiting_reagent_stoichiometry.
Maribel, Melissa. "How to Calculate Percent Yield and Theoretical Yield The Best Way -
TUTOR HOTLINE." YouTube. April 30, 2019.
https://www.youtube.com/watch?v=MebTIQNRU5g.
"Quiz #2-4 PRACTICE: Balancing Equations & Mole Ratios | Mr. Carman's Blog".
2020. Kentschools.Net. Accessed July 15. https://www.kentschools.net/ccarman/cp-
chemistry/practice-quizzes/quiz-2-4/.
"Stoichiometry--Molar Mass, Mole Ratios - Quiz". 2020. Quizizz.Com. Accessed July 12.
https://quizizz.com/admin/quiz/58b47527b60c1ba227b6927c/stoichiometry-molar-mass-
mole-ratios.
The Organic Chemistry Tutor. 2015. How To Calculate Theoretical Yield And Percent Yield.
Video. https://www.youtube.com/watch?v=jtAj0s203CI.
The Organic Chemistry Tutor. 2017. Stoichiometry Mole To Mole Conversions - Molar Ratio
Practice Problems. Video. https://www.youtube.com/watch?v=3zmeVamEsWI.
TheChemistrySolution. 2012. Theoretical, Actual And Percent Yield Problems - Chemistry
Tutorial. Video. https://www.youtube.com/watch?v=mmsKDK9WXdE.
19

108. For inquiries and feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
DepEd Division of Cagayan de Oro City
Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
Telefax: ((08822)855-0048
E-mail Address: cagayandeoro.city@deped.gov.ph

109. Government Property
NOT FOR SALE
NOT
General Chemistry 1
Quarter 1
Department of Education
General Chemistry 1
Quarter 1 - Module 5
Gases I
Department of Education ● Republic of the Philippines
Senior High School
=
=
=
Pmixture = P1 + P2
General Chemistry 1
● Republic of the Philippines
Senior High School
2 + P3

110. i
General Mathematics- Grade 12
Alternative Delivery Mode
Quarter 1 - Module 5: Gases I
First Edition, 2020
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111. General
Chemistry 1
Quarter 1
This instructional material was collaboratively developed and reviewed
by educators from public
education stakeholders to email their feedback, comments, and
recommendations to the Department of Education at
We value your feedback and recommendations.
Department of
ii
Senior High School
General
Chemistry 1
Quarter 1 - Module 5
Gases I
This instructional material was collaboratively developed and reviewed
by educators from public schools. We encourage teachers and other
education stakeholders to email their feedback, comments, and
recommendations to the Department of Education at action@deped.gov.ph
We value your feedback and recommendations.
Department of Education ● Republic of the Philippines
Senior High School
Senior High School
Chemistry 1
This instructional material was collaboratively developed and reviewed
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112. iii
Table of Contents
What This Module is About………………………………………………………………………..iv
What I Need to Know………………………………………………………………………………iv
How to Learn from this Module…………………………………………………………………....v
Icons of this Module…………………………………………………………………………………v
What I Know…………………………………………………………………………………………vi
Lesson 1: Pressure and its Units……………………………………………….1
What’s New…………...…………………………………………………………...2
What Is It: Pressure……………………………………………………………….2
What’s More: Reflection on Pressure…………………………………………...2
What Is It: Units of Pressure……………………………………………………..2
What’s More: Conversion of Units………………………………………………4
What I Have Learned……………………………………………………………..5
What I Can Do……………………………………………………………………..6
Lesson 2: Gas Laws……………………………………………………………………..7
What’s In……………………………………………………………………………7
What I Need to Know……………………………………………………………...7
What’s New: Graph Analysis……………………………………………………..8
What Is It: Gas Laws……………………………………………………………...9
What’s More: Gas Laws Calculations…………………………………………..13
What I Have Learned: Picture Analysis………………………………………...15
What Is It: Gas Mixtures………………………………………………………….16
What’s More: Gas Mixtures Calculations……………………………………….17
What I Can Do……………………………………………………………………..19
Summary……………………………………………………………………………………………..20
Assessment: Post-Test………………………………………………………………………….....21
Key to Answers……………………………………………………………………………………...22
References…………………………………………………………………………………………..24

113. iv
Module 5
Gases I
What This Module is About
This module demonstrates your understanding of the mathematical relationship
between the pressure, volume, and temperature of a gas. It also tackles the partial
pressures of a gas and its quantitative relationships of the reactants and products in a
gaseous reaction and behaviour and properties of gases at the molecular level
This module has two (2) lessons:
 Lesson 1: Pressure and its Units
 Lesson 2: Gas Laws
What I Need to Know
After going through this module, you are expected to:
1. Define pressure and give the common units of pressure (STEM_GC11G-Ih-i-43)
2. Use the gas laws to determine pressure, volume, or temperature of a gas under
certain conditions of change (STEM_GC11G-Ih-i-45)
3. Use the Ideal Gas Equation to calculate pressure, volume, temperature, or number of
moles of gas (STEM_GC11G-Ih-i-46)
4. Use Dalton’s Law of Partial Pressure to relate mole fraction and partial pressure of
gases in a mixture. (STEM_GC11DL-Ii-47)
_

114. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and
• Answer all the given tests and exercises.
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
v
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show
case your skills and knowledge gained, and
applied into real-life concerns and situations.
exercises diligently.
This part contains learning objectives that
are set for you to learn as you go along the
to your level of
knowledge to the subject matter at hand,
ifically to gauge prior related
This part connects previous lesson with that
An introduction of the new lesson through
various activities, before it will be presented
These are discussions of the activities as a
way to deepen your discovery and under-
up activities that are in-
tended for you to practice further in order to
Activities designed to process what you
have learned from the lesson
These are tasks that are designed to show-
your skills and knowledge gained, and
life concerns and situations.

115. What I Know
MULTIPLE CHOICE.
Directions: Read and understand each item and choose the letter of the
your answers on
__1. Which of the following descr
a. It is the force exerted by colliding molecules per unit area of container walls.
b. It refers to the force exerted by the wall.
c. The force resulted from the molecules or particles in static.
d. The resulting force from the bodies at rest or in equilibrium.
__2. Which of the following is not a unit of pressure
a. Atm
c. mmHg
__3. Which of the following refers to standard atmospheric pressure
a. Atm
b. Torr
c. mmHg
d. Pa
__4. One (1) atm is equal to what value of mmHg
a. 706 mmHg
b. 273 mmHg
c. 760 mmHg
d. 101.3 kPa
__5. The volume of a given amount of gas is inversely proportional to its pressure at
constant temperature is stated by what law?
a. Avogadro’s Law
b. Charles’s Law
c. Gay-Lussac Law
d. Boyles Law
__6. When volume on a gas goes up what happens to its pressure?
a. stays the same
b. goes down
c. rises
d. rises, then falls
__7. Which of the following refers to the statement of Charles’ Law?
a. The relationship of volume and pressure of a gas is inversely proportional.
b. The relationship of volume and pressure of a gas is directly proportional.
c. The relationship of temperature and volume of a gas is directly proportional.
d. The relationship of temper
__8. When the volume goes down the what happens to its temperature?
a. rises
b. goes up
c. goes down
d. rises, then falls
__9. Which of the following gas laws describes the relationship of volume and moles
a. Boyle’s Law
b. Avogadro’s Law
__10. Which of the following is the ideal gas equation?
a. PV=NRT
b. Pv=nrt
vi
What I Know
Read and understand each item and choose the letter of the
your answers on the space provide before the number.
Which of the following description refers to pressure?
It is the force exerted by colliding molecules per unit area of container walls.
It refers to the force exerted by the wall.
The force resulted from the molecules or particles in static.
The resulting force from the bodies at rest or in equilibrium.
ich of the following is not a unit of pressure?
b. Torr
d. none of the above
refers to standard atmospheric pressure?
One (1) atm is equal to what value of mmHg?
The volume of a given amount of gas is inversely proportional to its pressure at
stated by what law?
hen volume on a gas goes up what happens to its pressure?
Which of the following refers to the statement of Charles’ Law?
The relationship of volume and pressure of a gas is inversely proportional.
The relationship of volume and pressure of a gas is directly proportional.
The relationship of temperature and volume of a gas is directly proportional.
The relationship of temperature and volume of a gas is inversely proportional.
When the volume goes down the what happens to its temperature?
Which of the following gas laws describes the relationship of volume and moles
c. Charles’s Law
d. Ideal Gas Laws
__10. Which of the following is the ideal gas equation?
c. PV=nRT
d. pv=nRT
Read and understand each item and choose the letter of the best answer. Write
It is the force exerted by colliding molecules per unit area of container walls.
The volume of a given amount of gas is inversely proportional to its pressure at
The relationship of volume and pressure of a gas is inversely proportional.
The relationship of volume and pressure of a gas is directly proportional.
The relationship of temperature and volume of a gas is directly proportional.
ature and volume of a gas is inversely proportional.
Which of the following gas laws describes the relationship of volume and moles?
c. Charles’s Law
d. Ideal Gas Laws

116. 1
Lesson Pressure and its Units
1
What I Need to Know
Get a ball or an inflated balloon. Feel the ball. Is it too hard? Too soft? Or
does it feel just right? a basketball player knows the right feel of the ball, they say the
“pressure” is just right. In the same way a jeepney driver can tell right away if the
tire’s pressure is just right, too high, or too low. Everyone knows the important of
pressure, but not all have an idea what is pressure. So, what does pressure mean?
Why does the air inside the tire or a basketball exert pressure? Aside from it what
are other variables that constitutes to the properties of gas?
In this lesson, you are to define pressure and give common units of pressure.
(a) (b)
Look at the picture (a) shown above. What do you think is the role of the pressure of
the gas inside the balloon?
For picture (b) Imagine the tire of a vehicle and the need to pump air into the tire up
to a given pressure.
a. What will happen if the pressure is much lower than what it should be?
b. What will happen if the pressure is much greater than what it should be?
What’s New

117. Matching Type
Direction: Match column A with column B. Write the letter of the correct answer on
the space provided before the number.
Column A
_____1. This is the most easily measured
gas property defined as the force
exerted upon by colliding
molecules per unit area of a
surface.
_____2. It refers to the equivalent unit of
millimeter of mercury (mmHg)
_____3. This instrument is commonly
used to measure the pressure of
a gas.
_____4. An Italian physicist who invented
the barometer
_____5. It refers to a unit commonly used
to express gas pressure.
What Is It
The entire universe is made up of matter including humans,
animals, plants,
or states of matter, namely; solid, liquid, and gas. Moreover, gas behaves differently
from solids and liquids due to differences in their molecular behavior. The movement
of the gas is in random motio
the distance per particles are far greater than the other two states of matter. Under
some conditions of a gas sample, it can be defined in terms of its variables;
temperature, volume, moles, and pres
In the gaseous phase, molecules or the particles collide randomly against
other molecules and against its container or its system. This random collision
resulted changes in momentum which give way to the one of the property of gases
called pressure. Pressure as one of the variables defines as the amount of force
exerted per unit area. It refers to the force exerted by colliding molecules per unit
area of container walls.
2
Match column A with column B. Write the letter of the correct answer on
the space provided before the number.
Column A
This is the most easily measured
gas property defined as the force
by colliding
molecules per unit area of a
a. Torr
b. Atmosphere
It refers to the equivalent unit of
millimeter of mercury (mmHg)
c. Evangelista
Torricelli
This instrument is commonly
used to measure the pressure of d. Pressure
An Italian physicist who invented e. Pascal
It refers to a unit commonly used
to express gas pressure.
f. Barometer
The entire universe is made up of matter including humans,
animals, plants, and even the non-living things. There are three phases
or states of matter, namely; solid, liquid, and gas. Moreover, gas behaves differently
from solids and liquids due to differences in their molecular behavior. The movement
of the gas is in random motion due to the vast empty space in a certain system since
the distance per particles are far greater than the other two states of matter. Under
some conditions of a gas sample, it can be defined in terms of its variables;
temperature, volume, moles, and pressure.
In the gaseous phase, molecules or the particles collide randomly against
other molecules and against its container or its system. This random collision
resulted changes in momentum which give way to the one of the property of gases
Pressure as one of the variables defines as the amount of force
exerted per unit area. It refers to the force exerted by colliding molecules per unit
Match column A with column B. Write the letter of the correct answer on
Column B
a. Torr
b. Atmosphere
c. Evangelista
Torricelli
d. Pressure
e. Pascal
f. Barometer
The entire universe is made up of matter including humans,
living things. There are three phases
or states of matter, namely; solid, liquid, and gas. Moreover, gas behaves differently
from solids and liquids due to differences in their molecular behavior. The movement
n due to the vast empty space in a certain system since
the distance per particles are far greater than the other two states of matter. Under
some conditions of a gas sample, it can be defined in terms of its variables;
In the gaseous phase, molecules or the particles collide randomly against
other molecules and against its container or its system. This random collision
resulted changes in momentum which give way to the one of the property of gases
Pressure as one of the variables defines as the amount of force
exerted per unit area. It refers to the force exerted by colliding molecules per unit

118. What’s More
Time to Ponder!
Direction. Reflect and make an essay
use some facts, based in your daily experiences or relate it in the emotional aspects.
You may use an extra sheet of paper for this activity.
What Is It
What makes pressure quantifiable? How does pressure
measured and expressed? The following details are some of the units that can be
used for pressure:
a. Standard Atmosphere (atm)
and widely used unit for pressure in chemistry. The
1 atm is equal to the pressure that supports a
column of mercury which is exactly 760 mmHg.
b. Torr (or mmHg) 
mercury, where it represents the pressure exerted
by a column of a mercury which exactly equals to
atmosphere. The unit mmHg is called torr, named
after the Italian scientist
also invented Barometer
measure the atmospheric pressure.
c. Pounds per square inch (psi)
gas exerts a pressure of 20 psi, it means the
pressure on the wall of the gas container is 20
3
Reflect and make an essay about “Pressure in Everyday Life”. You may
use some facts, based in your daily experiences or relate it in the emotional aspects.
et of paper for this activity.
What makes pressure quantifiable? How does pressure
measured and expressed? The following details are some of the units that can be
Standard Atmosphere (atm)  It is commonly
and widely used unit for pressure in chemistry. The
1 atm is equal to the pressure that supports a
column of mercury which is exactly 760 mmHg.
mmHg means millimeter of
, where it represents the pressure exerted
lumn of a mercury which exactly equals to
atmosphere. The unit mmHg is called torr, named
after the Italian scientist Evangelista Torricelli, who
Barometer, a device used to
measure the atmospheric pressure.
Pounds per square inch (psi) If we say that
a pressure of 20 psi, it means the
pressure on the wall of the gas container is 20
Mercury Barometer
Pressure
about “Pressure in Everyday Life”. You may
use some facts, based in your daily experiences or relate it in the emotional aspects.
What makes pressure quantifiable? How does pressure
measured and expressed? The following details are some of the units that can be
Mercury Barometer
Pressure Pressure

119. pounds or the force per square inch of the unit area. The atmospheric pressure
at sea level is 14.7 psi.
d. kilopascal(kPa)  A kilopascal is equal to
unit for pressure.
Conversion Factor:
What’s More
Conversion
Direction: Convert the following
question. Show your solution and encircle the final answer.
1. 1 atm to torr
2. 14.7 psi to kPa
3. 760 mmHg to psi
4. 725 torr to atm
5. 35 psi to atm
1 atm = 760 mmHg = 760 torr = 101.3 kPa = 14.7 psi
4
pounds or the force per square inch of the unit area. The atmospheric pressure
at sea level is 14.7 psi.
A kilopascal is equal to 1000 pascals (Pa). It is the standard
What’s More
: Convert the following pressure to its desired unit that is asked in the
question. Show your solution and encircle the final answer.
6. 35 kPa to Pa
7. 450 000 Pa to kPa
8. 5 atm to kPa
9. 3 kPa to psi
10. 530 mmHg to atm
1 atm = 760 mmHg = 760 torr = 101.3 kPa = 14.7 psi
1 kPa = 1000Pa
pounds or the force per square inch of the unit area. The atmospheric pressure
1000 pascals (Pa). It is the standard
pressure to its desired unit that is asked in the
7. 450 000 Pa to kPa
10. 530 mmHg to atm
1 atm = 760 mmHg = 760 torr = 101.3 kPa = 14.7 psi

120. What I Have Learned
Direction: Answer the following questions as directed. For the calculations, show
your solution and encircle the final answer.
1. What is pressure?
2. What are the different units that can be used to measure and express
pressure? Then explain each briefly.
(a.)
(b.)
(c.)
(d.)
3. The pressure of the air on a mountain is 0.978 atm. What will be the pressure
in the units of torr?
4. The pressure inside the tires of a backhoe is 40 psi. What will be the pressure
in the units of atm?
5. The pressure of a certain valley below sea level is 200 kPa, what will be the
height of the mercury column in a barometer?
5
What I Have Learned
Answer the following questions as directed. For the calculations, show
your solution and encircle the final answer.
What are the different units that can be used to measure and express
pressure? Then explain each briefly.
The pressure of the air on a mountain is 0.978 atm. What will be the pressure
The pressure inside the tires of a backhoe is 40 psi. What will be the pressure
The pressure of a certain valley below sea level is 200 kPa, what will be the
height of the mercury column in a barometer?
Answer the following questions as directed. For the calculations, show
What are the different units that can be used to measure and express
The pressure of the air on a mountain is 0.978 atm. What will be the pressure
The pressure inside the tires of a backhoe is 40 psi. What will be the pressure
The pressure of a certain valley below sea level is 200 kPa, what will be the

121. What I Can Do
Performance Task:
Make a short poem about “Pressure in Everyday Life”.
Note:
 You may have you own title as long as within the topic.
 You may add any creativity or even write your poem in calligraphy.
 It is handwritten in an A4 bond paper.
Enrichment Activity:
Watch a video through YouTube link below
https://www.youtube.com/watch?v=NzKAJWTmlwg
6
What I Can Do
Make a short poem about “Pressure in Everyday Life”.
you own title as long as within the topic.
You may add any creativity or even write your poem in calligraphy.
It is handwritten in an A4 bond paper.
through YouTube link below entitled “Pressure Gases
https://www.youtube.com/watch?v=NzKAJWTmlwg
Make a short poem about “Pressure in Everyday Life”.
You may add any creativity or even write your poem in calligraphy.
Pressure Gases”,
https://www.youtube.com/watch?v=NzKAJWTmlwg

122. 7
Lesson Gas Laws
2
What’s In
In lesson 1, you have learned about the definition of pressure, and the
different units that can be used to measure and express pressure which are the
atmospheric pressure (atm), millimeter of mercury (mmHg), pounds per square inch
(psi), Pascal (Pa), and kilopascal (kPa). Moreover, you have also learned one unit to
another unit of pressure, and able to show the solution for correct conversion.
Pressure is one of the variables that describes the properties of gases, so in
continuation, the next topic will help you learn about the other variables which are
the Volume(V), Temperature (T), amount in moles (n) that describes the behavior of
gases under certain condition and the laws that governs how it behaves.
What I Need to Know
Air is all around us, thus gases always involved in our daily activities, from
breathing down to automobile tires or bicycle, balloons, and even lifeboats and vest.
Life won’t be possible without this life-sustaining gas found in the atmosphere.
Scientist have always been curious about how gases behaves. And how it is
different compared to other states of matter. Investigations and experiments on the
behavior of gases leads to the parameters or variables that used to describe the
properties of gases aside from pressure, and these are volume, temperature, and
the amount in moles. The relationship between variables are explained by Gas
Laws.
In addition, you will be able to use gas laws to determine pressure, volume,
temperature of a gas under certain conditions od change. Then, use the ideal gas
equation to calculate pressure, volume, temperature, or number of moles of a gas.

123. What’s New
Graph Analysis
Direction: Based on the given graph, analyze
properties of gases (volume, pressure, temperature, and moles). Write your answer
at the sides of the graph.
8
What’s New
Based on the given graph, analyze and infer the relationship of the
properties of gases (volume, pressure, temperature, and moles). Write your answer
and infer the relationship of the
properties of gases (volume, pressure, temperature, and moles). Write your answer

124. 9
What Is It
Gas Laws governs the behavior of gases and describes the relationship of the
following variables: Pressure, Volume, Temperature, and moles. The relationship
among the variables are led and investigated by Robert Boyle, Jacques Charles, and
Amedeo Avogadro, and the laws where named after them respectively.
Boyle’s Law
The law is named after its proponent, who is a British
chemist, Robert Boyle.
He emphasized the law correctly and stated that “The volume of
a given amount of gas is inversely proportional to its pressure at
constant temperature”. It means, as the volume increases, the
pressure of the gas decreases, and vice versa, provided that the
temperature remains the same. Thus, it is evident that the
relationship between the two variables, volume and pressure is
inversely proportional.
Boyle’s Law is expressed in this mathematical equation:
In terms of proportion: V α 1/P (at constant amount and temperature)
In terms of equation: V = k/P (at constant amount and temperature)
PV=k or
P1V1 = P2 V2
Where;
P1 = initial pressure
V1 = initial volume
P2 = Final pressure
V2 = Final volume
Example: A 2.5 L container has a gas pressure of 4.6 atm. If the volume is
decreased to 1.6 L. What will be the new pressure inside the container?
Given: V1 = 2.5 L V2 = 1.6 L
P1= 4.6 atm P2 = ?
P1V1 = P2V2 , P2 = P1V1 / V2
= 1.6 atm (2.5 L)
1.6 L
= 7.2 atm
Figure 1: Robert Boyle.
Image source
[14]

125. 10
Charles’ Law
It is one of the gas laws and named after the
French scientist Jacques Charles who formulated the law
in 1897. This law states that “The volume of a given
amount of gas is directly proportional to its absolute
temperature and constant pressure.” It means that as the
volume increases, the temperature also increases, and
vice versa, provided that the amount of gas and the
pressure is constant. The temperature should be
expressed in Kelvin (K).
Charles’ Law is expressed in this mathematical equation:
In terms of proportion: V α T (at constant amount and pressure)
In terms of equation: V = kT (at constant amount and pressure)
V / T =k or
V1 / T1= V2 / T2
Where;
V1 = initial volume
T1 = initial Temperature
V2 = Final volume
T2 = Final Temperature
Example: A 3.5 L flexible container holds a gas at 250 K. What will be the new
volume if the temperature is increased to 400K at constant pressure?
Given: V1 = 3.5 L V2 = ?
T1= 250 K T2 = 400K
V1 / T1 = V2 / T2 ; V2 = V1T2 / T1
= 3.5 L (400K)
250 K
= 5.6 L
Figure 2: Jacques Charles.
Image source
[11]

126. Avogadro’s Law
The proponent of this law is named after Amedeo
Avogadro, who is a notable Italian mathematical physicist.
This law state that “The volume of a gas at a given
temperature and pressure is directly proportional to the
number of moles contained in the volume
based on Avogadro’s hypothesis that the same volume of
two gases at constant temperature and pressure contain
the same number of molecules. It means as the volume
increases the amount of substance or the moles also
increases, so the relationship is directly proportional. Keep
in mind that a mole is related to the quantity of molecules
in a substance.
Avogadro’s Law is expressed in this mathematical equation:
In terms of proportion: V α n (at constant temperature and pressure)
In terms of equation: V = k n (at constant temperature and pressure)
V / n =k or
V1 / n1= V
Where;
V1 = initial volume
n1 = initial number of moles of the gas
V2 = Final volume
T2 = Final number of moles of the gas
Example: A 2.4 moles of gas occupies 60.0 L at a certain temperature. What volume
will 3.7 moles of a gas occupy?
Given: n1 = 2.4 moles
V1= 60.0 L
V1 / n1 = V2 / n2 ; V1n
V2
11
The proponent of this law is named after Amedeo
Avogadro, who is a notable Italian mathematical physicist.
The volume of a gas at a given
temperature and pressure is directly proportional to the
number of moles contained in the volume”. This law is
based on Avogadro’s hypothesis that the same volume of
two gases at constant temperature and pressure contain
the same number of molecules. It means as the volume
increases the amount of substance or the moles also
hip is directly proportional. Keep
is related to the quantity of molecules
Avogadro’s Law is expressed in this mathematical equation:
α n (at constant temperature and pressure)
In terms of equation: V = k n (at constant temperature and pressure)
V / n =k or
= V2 / n2
= initial number of moles of the gas
= Final number of moles of the gas
A 2.4 moles of gas occupies 60.0 L at a certain temperature. What volume
will 3.7 moles of a gas occupy?
n2 = 3.7 moles
= 60.0 L V2 = ?
n2=n 1V2
= V1 n2 / n1
= 60.0 L ( 3.7 moles)
2.4 moles
= 92.5 L
Figure
Image source
α n (at constant temperature and pressure)
In terms of equation: V = k n (at constant temperature and pressure)
A 2.4 moles of gas occupies 60.0 L at a certain temperature. What volume
Figure 3: Amedeo Avogadro.
Image source
[15]

127. 12
Ideal gas Equation
It is a single equation that sums up and combines the mathematical
expression of Boyle’s Law, Charles’ Law, and Avogadro’s Law.
Where;
P = Pressure n= moles (refers to the amount of substance)
V= Volume T= Temperature (express in Kelvin (K))
R = The universal gas constant (0.0821 atm.L / mol.K)
The value of the universal gas constant (R) is the same anywhere and anytime. It
can be calculated using the using the standard conditions of mole, pressure, volume,
and temperature. The value can also be derived from the ideal gas equation as
shown below.
PV = nRT
R = PV = 1.00 atm x 22.4 L
nT 1.00 mole x 273 K
Example:
A 3.5 L container holds 0.45 moles of O2 gas at 300K. What is the pressure
inside the container?
Given: V = 3.5 L T = 300 K
N = 0.45 moles P = ?
PV = nRT ; P = nRT
V
= 0.45 moles x 0.0821 x 300 K
1.5 L
PV = nRT
R = 0.0821 atm . L
mole .K
atm . L
mole .K
P = 3.17 atm

128. 13
What’s More
Calculations
A. Direction: Calculate the given problems and use the gas laws to determine
pressure, volume, or temperature of a gas under certain conditions of change.
Show your solutions. Encircle your final answer.
1. The gas inside the tire has a volume of 20.00 L at a pressure of 5.00 atm.
Calculate the pressure of the gas if its volume is reduced to 10.0 at the same
temperature.
2. If 150.00 mL of N2 gas was collected at 760 torr, what is the new volume of
the gas when the pressure is compressed to 740 torr at the same
temperature?
3. At 300 K, the given amount of fluorine gas has a volume of 30.0 L. What will
be the temperature if the gas occupies a volume of 25 L at constant pressure?
4. A certain gas sample has a volume of 40.00 L at 273 K. At constant pressure,
the volume increase to 50.00 L. What will be the final temperature of the gas?
5. At 55.00 L a compressible container contains 5.00 moles of a certain gas. If
3.00 moles of a gas were added to the container, what will be its final volume?

129. 14
B. Direction: Complete the following table and use the Ideal gas equation to
calculate pressure, volume, number of moles, and temperature of a gas. Express
your final answer in two (2) decimal places and show your solution below the
table.
Pressure (P) Volume (V) Temperature (T) Moles (n)
1. 5.00 atm 25.00L 273.15 K ____________
2. ____________ 0.55 L 308 K 0.50 mol
3. 20.00 atm 30.00 L ___________ 25.30 mol
4. 15.00 atm __________ 370.00 K 3.00 mol
5. _________ 10.50 L 280.00 K 10.00 mol

130. What I Have Learned
Picture Analysis
Direction: Analyze the given picture and identify the appropriate gas laws that best
describes the picture.
15
Have Learned
Analyze the given picture and identify the appropriate gas laws that best
1.
2.
3.
4.
5.
Analyze the given picture and identify the appropriate gas laws that best

131. 16

132. 1
2
What Is It
Most of the gases encountered in the surroundings are
are composed of different components. An example of a mixture is air which is a
combination of primarily nitrogen and oxygen and other inert gases. Each individual
component in air exert its own pressure, has its own volume, can have a
temperature that is in thermal
its own molar amount. This just means that the ideal gas equation can also be used
on mixtures of gases.
Say for example, a constant
flue gas, a by-product of combustion, which is composed of nitrogen (1), carbon
dioxide (2), and carbon monoxide (3) at a constant
From the situation above, the number of moles of each component can be
interpreted as n1, n2, and n
respectively, so that the total number of moles can be shown as:
The pressure exerted by the mixture can then be calculated using the ideal
gas equation:
…where V is the volume of the container or in this case, the volume of the
constant-volume piston.
Substituting the two equations, the pressure of the mixture can be expressed
as:
And distributing the ideal gas constant and temperature:
17
1
1
2
1
2
3
3
Most of the gases encountered in the surroundings are mixtures.
are composed of different components. An example of a mixture is air which is a
combination of primarily nitrogen and oxygen and other inert gases. Each individual
ert its own pressure, has its own volume, can have a
thermal equilibrium with the other components, and also has
its own molar amount. This just means that the ideal gas equation can also be used
constant-volume piston shown below contains a sample of
product of combustion, which is composed of nitrogen (1), carbon
dioxide (2), and carbon monoxide (3) at a constant temperature of 30
From the situation above, the number of moles of each component can be
, and n3 for nitrogen, carbon dioxide, and carbon monoxide,
respectively, so that the total number of moles can be shown as:
= + +
The pressure exerted by the mixture can then be calculated using the ideal
=
…where V is the volume of the container or in this case, the volume of the
he two equations, the pressure of the mixture can be expressed
=
( + + )
And distributing the ideal gas constant and temperature:
= + +
mixtures. Mixtures
are composed of different components. An example of a mixture is air which is a
combination of primarily nitrogen and oxygen and other inert gases. Each individual
ert its own pressure, has its own volume, can have a
equilibrium with the other components, and also has
its own molar amount. This just means that the ideal gas equation can also be used
piston shown below contains a sample of
product of combustion, which is composed of nitrogen (1), carbon
temperature of 30°C.
From the situation above, the number of moles of each component can be
carbon monoxide,
The pressure exerted by the mixture can then be calculated using the ideal
…where V is the volume of the container or in this case, the volume of the
he two equations, the pressure of the mixture can be expressed

133. 18
As you may recall, the term nRT/V is equal to P which means that the terms in
the right hand side of the equation equates to the pressure exerted by each
individual component.
= = =
Pressures P1, P2, and P3 are called the partial pressure of each gas.
Combining all the equations, it can be concluded that the pressure exerted by the
mixture is the sum of the pressures exerted by each component. This is known as
the Dalton’s Law of Partial Pressure.
= + +
The application of Dalton’s Law of Partial Pressure can help us learn about
the composition of each component in terms of mole fraction of the component.
= =
Combining these two equations gives the following expression:
= =
…where x1 is the mole fraction of component 1 in the mixture. Rearranging
the above equation leads to:
=
This means that the partial pressure of a component is equal to its mole
fraction multiplied to the pressure exerted by mixture.
What’s More
A. Who’s Greater?
Directions: Try to determine which column exerts greater partial pressure is greater
by putting E on the blank if the column on the left is greater than the column on the
right and putting Z on the blank if the column on the right is greater than the column
on the left.
1.
Ptotal = 0.83
atm
X1 = 0.23
Ptotal = 0.95 atm
X2 = 0.19
2.
Ptotal = 14 psi
n1 = 0.2 mol
ntotal = 0.3 mol
Ptotal = 10 psi
n2 = 12 mol
ntotal = 16 mol

134. 19
3.
Ptotal = 20 kPa
X1 = 0.87
Ptotal = 0.95 atm
X2 = 0.19
4.
Ptotal = 101
kPa
n1 = 0.4 mol
ntotal = 1.2 mol
Ptotal = 110 kPa
n2 = 5 mol
ntotal = 14 mol
5.
Ptotal = 1.2 atm
n1 = 0.4 mol
ntotal = 1.2 mol
Ptotal = 110 kPa
n2 = 5 mol
ntotal = 14 mol
6.
Ptotal = 2.3 atm
X1 = 0.23
Ptotal = 3.4 atm
X2 = 0.19
7.
Ptotal = 0.83
atm
X1 = 0.23
Ptotal = 19 psi
X2 = 0.19
8.
Ptotal = 202
kPa
n1 = 0.4 mol
ntotal = 1.2 mol
Ptotal = 30 psi
n2 = 5 mol
ntotal = 14 mol
9.
Ptotal = 1.2 atm
X1 = 0.5
Ptotal = 3.4 atm
X2 = 0.2
10.
Ptotal = 6 MPa
X1 = 0.6
Ptotal = 5.3 MPa
X2 = 0.5
B. Problem Solving
Directions: In a separate sheet of paper, show your solution and encircle your final
answer.
1. A sample of oxygen gas, which is saturated with water vapour, is kept in a 10-
L vessel at 30°C and has a pressure of 758 Torr. If the pressure of the water
vapour at this temperature is 31.8 Torr, what would be the pressure of the dry
oxygen?
2. If the oxygen gas sample in #1 passed through a drier that decreased the
pressure of the mixture to 750 Torr and the pressure exerted by the water
vapour is only 80% of the saturated vapour pressure at the given temperature,
what would be the pressure of the dry oxygen?
3. In a gas mixture composed of N2, Ne, and He, the partial pressure of N2 is
0.50 atm, that of Ne is 1.1 atm, and the total pressure is 2.4 atm. What is the
partial pressure of He?
4. In a gas mixture composed of N2, Ne, and He, the partial pressure of N2 is
0.50 atm, that of Ne is 1.1 atm, and that of He is 0.80 atm. Calculate the mole
fraction of each gas.
5. A gas mixture contains 2.5 mol N2 and 9.7 mol CO2, and has a pressure of 2.3
atm. What is the partial pressure of each gas?

135. What I Can Do
Performance Task: Poster Making
Draw and illustrate the three (3) laws of gases based on their real
in an A4 size bond paper.
Enrichment Activity:
Watch a video through YouTube link below
https://www.youtube.com/watch?v=BxUS1K7xu30
20
Poster Making
Draw and illustrate the three (3) laws of gases based on their real-
in an A4 size bond paper.
through YouTube link below entitled “The Ideal Gas Law
https://www.youtube.com/watch?v=BxUS1K7xu30
-life application
The Ideal Gas Law”,
https://www.youtube.com/watch?v=BxUS1K7xu30

136. Summary
Gases are everywhere, it behaves differently from other states of matter. The
properties of gases are described by its variables or parameters namely; pressure,
temperature, volume, and the amount of substance or moles under some certain
conditions. The relationship between the variables are govern by gas laws. Gas laws
and categorized into three laws; Boyle’s Law, Charles’s Law, and Avogadro’s Law
which are named after their proponents and scientist who observed such properties
of gases.
Boyle’s law describes
proportional given that temperature is constant, for Charles’s Law, Volume and
Temperature is directly proportional at constant pressure. Avogadro’s law explains
the relationship between volume and
proportional when pressure and temperature is constant. Then, the variables are
combined to form an ideal gas equation.
Dalton’s Law of Partial Pressure tells us that the pressure of each component
in a mixture is equivalent to its
by the mixture.
21
everywhere, it behaves differently from other states of matter. The
properties of gases are described by its variables or parameters namely; pressure,
temperature, volume, and the amount of substance or moles under some certain
p between the variables are govern by gas laws. Gas laws
and categorized into three laws; Boyle’s Law, Charles’s Law, and Avogadro’s Law
which are named after their proponents and scientist who observed such properties
Boyle’s law describes that the relationship of pressure and volume is inversely
proportional given that temperature is constant, for Charles’s Law, Volume and
Temperature is directly proportional at constant pressure. Avogadro’s law explains
the relationship between volume and the amount of substance (moles)
proportional when pressure and temperature is constant. Then, the variables are
combined to form an ideal gas equation.
Dalton’s Law of Partial Pressure tells us that the pressure of each component
is equivalent to its mole fraction multiplied to the total pressure exerted
everywhere, it behaves differently from other states of matter. The
properties of gases are described by its variables or parameters namely; pressure,
temperature, volume, and the amount of substance or moles under some certain
p between the variables are govern by gas laws. Gas laws
and categorized into three laws; Boyle’s Law, Charles’s Law, and Avogadro’s Law
which are named after their proponents and scientist who observed such properties
that the relationship of pressure and volume is inversely
proportional given that temperature is constant, for Charles’s Law, Volume and
Temperature is directly proportional at constant pressure. Avogadro’s law explains
the amount of substance (moles) is directly
proportional when pressure and temperature is constant. Then, the variables are
Dalton’s Law of Partial Pressure tells us that the pressure of each component
mole fraction multiplied to the total pressure exerted

137. Assessment: Post
MULTIPLE CHOICE.
Directions: Read and understand each item and choose the letter of the
your answers on
__1. Which of the following descr
e. It is the force exerted by colliding molecules per unit area of container walls.
f. It refers to the force exerted by the wall.
g. The force resulted from the molecules or particles in static.
h. The resulting force from the bodies at rest or in equilibrium.
__2. Which of the following is not a unit of pressure
b. Atm
d. mmHg
__3. Which of the following refers to standard atmospheric pressure
e. Atm
f. Torr
g. mmHg
h. Pa
__4. One (1) atm is equal to what value of mmHg
e. 706 mmHg
f. 273 mmHg
g. 760 mmHg
h. 101.3 kPa
__5. The volume of a given amount of gas is inversely proportional to its pressure at
constant temperature is stated by what law?
e. Avogadro’s Law
f. Charles’s Law
g. Gay-Lussac Law
h. Boyles Law
__6. When volume on a gas goes up what happens to its pressure?
e. stays the same
f. goes down
g. rises
h. rises, then falls
__7. Which of the following refers to the statement of Charles’ Law?
e. The relationship of volume and pressure of a gas is inversely proportional.
f. The relationship of volume and pressure of a gas is directly proportional.
g. The relationship of temperature and volume of a gas is directly proportional.
h. The relationship of temper
__8. When the volume goes down the what happens to its temperature?
e. rises
f. goes up
g. goes down
h. rises, then falls
__9. Which of the following gas laws describes the relationship of volume and moles
c. Boyle’s Law
d. Avogadro’s Law
__10. Which of the following is the ideal gas equation?
c. PV=NRT
d. Pv=nrt
22
Assessment: Post-Test
Read and understand each item and choose the letter of the
your answers on the space provide before the number.
Which of the following description refers to pressure?
It is the force exerted by colliding molecules per unit area of container walls.
It refers to the force exerted by the wall.
resulted from the molecules or particles in static.
The resulting force from the bodies at rest or in equilibrium.
ich of the following is not a unit of pressure?
b. Torr
d. none of the above
refers to standard atmospheric pressure?
One (1) atm is equal to what value of mmHg?
The volume of a given amount of gas is inversely proportional to its pressure at
stated by what law?
hen volume on a gas goes up what happens to its pressure?
Which of the following refers to the statement of Charles’ Law?
The relationship of volume and pressure of a gas is inversely proportional.
The relationship of volume and pressure of a gas is directly proportional.
The relationship of temperature and volume of a gas is directly proportional.
The relationship of temperature and volume of a gas is inversely proportional.
When the volume goes down the what happens to its temperature?
Which of the following gas laws describes the relationship of volume and moles
c. Charles’s Law
d. Ideal Gas Laws
__10. Which of the following is the ideal gas equation?
c. PV=nRT
d. pv=nRT
Read and understand each item and choose the letter of the best answer. Write
It is the force exerted by colliding molecules per unit area of container walls.
The volume of a given amount of gas is inversely proportional to its pressure at
The relationship of volume and pressure of a gas is inversely proportional.
The relationship of volume and pressure of a gas is directly proportional.
The relationship of temperature and volume of a gas is directly proportional.
ature and volume of a gas is inversely proportional.
Which of the following gas laws describes the relationship of volume and moles?
c. Charles’s Law
d. Ideal Gas Laws

138. Key to Answers
Pretest
Lesson 1
What’s New What’s More
What Have I Learned?
1. Defines as the amount of force exerted per unit area. It refers to the force exerted
by colliding molecules per unit area of container walls.
2. a. Standard Atmosphere (atm)
pressure in chemistry.
b. Torr (or mmHg)  mmHg means millimeter of mercury, where it represents
the pressure exerted by a column of a mercury which exactly equals to
atmosphere.
c. Pounds per square inch (psi)
psi.
d. kilopascal(kPa) 
standard unit for pressure.
3. 743 torr
4. 2.72 atm
5. 1500 kPa
Lesson 2
23
Key to Answers
What’s More
efines as the amount of force exerted per unit area. It refers to the force exerted
by colliding molecules per unit area of container walls.
a. Standard Atmosphere (atm)  It is commonly and widely used unit for
mmHg means millimeter of mercury, where it represents
the pressure exerted by a column of a mercury which exactly equals to
Pounds per square inch (psi) The atmospheric pressure at sea level is 14.7
A kilopascal is equal to 1000 pascals (Pa). It is the
standard unit for pressure.
efines as the amount of force exerted per unit area. It refers to the force exerted
It is commonly and widely used unit for
mmHg means millimeter of mercury, where it represents
the pressure exerted by a column of a mercury which exactly equals to
The atmospheric pressure at sea level is 14.7
A kilopascal is equal to 1000 pascals (Pa). It is the

139. 24
What’s New
Graph 1: A graph showing the relationship between volume and pressure, as
stated by Boyle’s Law. Pressure and Volume is inversely proportional. As the
pressure increases, the volume decreases, and vice versa.
Graph 2: A graph showing the relationship between volume and temperature, as
stated by Charles’ Law. Volume and Temperature is directly proportional. As the
volume increases, the temperature also increases.
Graph 3: A graph showing the relationship between volume and moles (the amount
of the substance), as stated by Avogadro’s Law. Volume and number of moles in a
substance is directly proportional.
What’s More (Gas Laws)
A. B.
What Have I Learned (Gas Laws)
What’s More (Dalton’s Law)
A. B.
Post-test

140. 25

141. References
[1] Commission on Higher Education, General Chemistry 1: Teaching Guide for
Senior High, Manila, 2016.
[2] Department of Education Central Office, Most Essential Learning Competencies
(MELCS), Manila, 2020.
[3] Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
Resource Management Development Team, 2016.
[4] A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
Technology, Quezon City: SD Publications, 2001.
[5] "Clipart Library," [Online]. Available: http://www.clipart
clipart-transparent-background.html . [Accessed 6 July 2020].
[6] [Online]. Available: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
July 2020].
[7] [Online]. Available: https://www.123rf.com/photo_133132698_stock
mercury-barometer-vector
surface-weather-measur.html . [Accessed 6 July 2020].
[8] "Youtube," [Online]. Available:
https://www.youtube.com/watch?v=J
OKdqsMFSB_MyyLAqS&index=69. [Accessed 6 July 2020].
[9] "Youtube," [Online]. Available:
https://www.youtube.com/watch?v=NzKAJWTmlwg. [Accessed 6 July 2020].
[10] [Online]. Available:
http://www.eoht.info/photo/10747477/Boyle%27s+law+%28graph%29.
[Accessed 6 July 2020].
[11] "Google," [Online]. Available:
https://sites.google.com/site/chemistryandfragglerocks/charles
6 July 2020].
[12] "Brilliant," [Online]. Availab
6 July 2020].
[13] "Britannica," [Online]. Available: https://www.britannica.com/biography/Robert
Boyle. [Accessed 7 July 2020].
[14] [Online]. Available: https://www.sciencephoto.com/media/22432
charles-french-balloonist . [Accessed 7 July 2020].
[15] "Britannica," [Online]. Available: https://www.britannica.com/biography/Amedeo
Avogadro. [Accessed 7 July 2020].
[16] [Online]. Available: https://studiousguy.com/examples
[Accessed 7 July 2020].
[17] [Online]. Available: https://chemistrygod.com/boyle
July 2020].
[18] [Online]. Available: https://simple.wikipedia.org/wiki/Balloon .
2020].
26
References
Commission on Higher Education, General Chemistry 1: Teaching Guide for
Senior High, Manila, 2016.
Department of Education Central Office, Most Essential Learning Competencies
(MELCS), Manila, 2020.
Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
Resource Management Development Team, 2016.
A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
Technology, Quezon City: SD Publications, 2001.
ary," [Online]. Available: http://www.clipart-library.com/free/balloon
background.html . [Accessed 6 July 2020].
[Online]. Available: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
: https://www.123rf.com/photo_133132698_stock
vector-illustration-labeled-atmospheric-pressure
measur.html . [Accessed 6 July 2020].
"Youtube," [Online]. Available:
https://www.youtube.com/watch?v=JNOg1OsxMUw&list=PL0o_zxa4K1BWziAv
OKdqsMFSB_MyyLAqS&index=69. [Accessed 6 July 2020].
"Youtube," [Online]. Available:
https://www.youtube.com/watch?v=NzKAJWTmlwg. [Accessed 6 July 2020].
http://www.eoht.info/photo/10747477/Boyle%27s+law+%28graph%29.
[Accessed 6 July 2020].
"Google," [Online]. Available:
https://sites.google.com/site/chemistryandfragglerocks/charles-
"Brilliant," [Online]. Available: https://brilliant.org/wiki/ideal-gas-
"Britannica," [Online]. Available: https://www.britannica.com/biography/Robert
Boyle. [Accessed 7 July 2020].
[Online]. Available: https://www.sciencephoto.com/media/22432
balloonist . [Accessed 7 July 2020].
"Britannica," [Online]. Available: https://www.britannica.com/biography/Amedeo
Avogadro. [Accessed 7 July 2020].
https://studiousguy.com/examples-charles-law
July 2020].
[Online]. Available: https://chemistrygod.com/boyle-law-examples . [Accessed 7
[Online]. Available: https://simple.wikipedia.org/wiki/Balloon . [Accessed 6 July
Commission on Higher Education, General Chemistry 1: Teaching Guide for
Department of Education Central Office, Most Essential Learning Competencies
Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
library.com/free/balloon-
[Online]. Available: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
: https://www.123rf.com/photo_133132698_stock-vector-
pressure-tool-earth-
NOg1OsxMUw&list=PL0o_zxa4K1BWziAv
https://www.youtube.com/watch?v=NzKAJWTmlwg. [Accessed 6 July 2020].
http://www.eoht.info/photo/10747477/Boyle%27s+law+%28graph%29.
-s-law. [Accessed
-law/. [Accessed
"Britannica," [Online]. Available: https://www.britannica.com/biography/Robert-
[Online]. Available: https://www.sciencephoto.com/media/224322/view/jacques-
"Britannica," [Online]. Available: https://www.britannica.com/biography/Amedeo-
law-daily-life/.
examples . [Accessed 7
[Accessed 6 July

142. 27
[19] "Youtube," [Online]. Available:
https://www.youtube.com/watch?v=BxUS1K7xu30 . [Accessed 7 July 2020].

143. 28
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144. Government Property
NOT FOR
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NOT
General Chemistry 1
Quarter 1
Department of Education
General Chemistry 1
Quarter 1 - Module 6
Gases II
Department of Education ● Republic of the Philippines
Senior High School
=
=
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Pmixture = P1 + P2
General Chemistry 1
● Republic of the Philippines
Senior High School
2 + P3

145. i
General Mathematics- Grade 12
Alternative Delivery Mode
Quarter 1 – Module 6: Gases II
First Edition, 2020
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146. General
Chemistry 1
Quarter 1
This instructional material was collaboratively developed and reviewed
by educators from public
education stakeholders to email their feedback, comments, and
recommendations to the Department of Education at
We value your feedback and recommendations.
Department of
ii
Senior High School
General
Chemistry 1
Quarter 1 - Module 6
Gases II
This instructional material was collaboratively developed and reviewed
by educators from public schools. We encourage teachers and other
education stakeholders to email their feedback, comments, and
recommendations to the Department of Education at action@deped.gov.ph
We value your feedback and recommendations.
Department of Education ● Republic of the Philippines
Senior High School
Senior High School
Chemistry 1
This instructional material was collaboratively developed and reviewed
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education stakeholders to email their feedback, comments, and
@deped.gov.ph
of the Philippines
Senior High School

147. iii
Table of Contents
What This Module is About…………………………………………………………………………iv
What I Need to Know………………………………………………………………………………..iv
How to Learn from this Module………………………………………………………………….....v
Icons of this Module………………………………………………………………………………….v
What I Know………………………………………………………………………………………….vi
Lesson 1: Gas Stoichiometry………………………………………………………..1
What I Need to Know…………………………………………………………….....1
What’s New: Balancing Equations………………………………………………...1
What Is It: Gas Stoichiometry………………………………………………………2
What’s More: Gas Stoichiometry Calculations…………………………………...3
What I Have Learned: Reflection…………………………………………………..5
What I Can Do…………………………………………………………………….....5
Lesson 2: Kinetic Molecular Theory of Gases…………………………….6
What’s In……………………………………………………………………………...6
What I Need to Know…………………………………………………………….....6
What’s New: Differentiate………...………………………………………………...6
What Is It: Kinetic Molecular Theory of Gases…………………………………...7
What’s More: Diffusion Calculations………….…………………………………...9
What I Have Learned: Reflection…………………………………………………10
Summary……………………………………………………………………………………………..11
Assessment: (Post-Test)…………………………………………………………………………...12
Key to Answers……………………………………………………………………………………...13
References…………………………………………………………………………………………..15

148. iv
Module 6
Gases II
What This Module is About
This module demonstrates your understanding of the mathematical relationship
between the pressure, volume, and temperature of a gas. It also tackles the partial
pressures of a gas and its quantitative relationships of the reactants and products in a
gaseous reaction and behaviour and properties of gases at the molecular level
This module has two (2) lessons:
 Lesson 1: Gas Stoichiometry
 Lesson 2: Kinetic Molecular Theory of Gases
What I Need to Know
After going through this module, you are expected to:
1. Apply the principles of stoichiometry to determine the amounts (volume, number of
moles, or mass) of gaseous reactants and products (STEM_GC11GS-Ii-j-48)
2. Relate the rate of gas effusion with molar mass (STEM_GC11KMT-Ij-50)
_

149. How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests
Icons of this Module
What I Need to
Know
What I know
What’s In
What’s New
What is It
What’s More
What I Have
Learned
What I can do
v
Learn from this Module
To achieve the objectives cited above, you are to do the following:
Take your time reading the lessons carefully.
Follow the directions and/or instructions in the activities and exercises diligently.
Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under
standing of the concept.
What’s More These are follow-up activities that are in
tended for you to practice further in
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show
case your skills and knowledge gained, and
applied into real-life concerns and situations.
Follow the directions and/or instructions in the activities and exercises diligently.
This part contains learning objectives that
are set for you to learn as you go along the
This is an assessment as to your level of
knowledge to the subject matter at hand,
ifically to gauge prior related
This part connects previous lesson with that
An introduction of the new lesson through
various activities, before it will be presented
These are discussions of the activities as a
way to deepen your discovery and under-
up activities that are in-
tended for you to practice further in order to
Activities designed to process what you
have learned from the lesson
These are tasks that are designed to show-
case your skills and knowledge gained, and
life concerns and situations.

150. What I Know
MULTIPLE CHOICE.
Directions: Read and understand each item and choose the letter of the
your answers on
__1. Balance the gaseous chemical equation: __C
a. 1, 5, 3, 4
b. 2, 3, 4, 1
c. 2, 10, 6, 8
d. 4, 6, 8, 2
__2. Which of the following is NOT a gas mixture?
a. Air
c. Chimney Fumes
__3. In which of the following gas mixtures of N
greatest?
a. 2 moles N2, 3 moles He
b. 3 moles N2, 1 moles He
c. 4 moles N2, 2 moles He
d. 5 moles N2, 5 moles He
__4. Which of the following systems will have the greatest volume at STP
a. 1.00 g N2 (MM = 28 g/mol)
b. 1.00 g NH3 (MM = 17 g/mol)
c. 1.00 g He (MM = 4 g/mol)
d. 1.00 g CO2 (MM = 44 g/mol)
__5. Which of the following will contain the greatest number of molecules at 300K and 1 atm
pressure?
a. 0.01 L
b. 0.10 L
c. 1.00 L
d. 10.0 L
__6. How much carbon dioxide is there in a 3.4 L balloon at SATP?
a. 0.13 g
b. 6.1 g
c. 0.15 mol
d. 6.6 g
__7. In the reaction 2H2 (g) + O
needed to react with 10 L of hydrogen gas at STP
a. 20.0
b. 10.0 L
__8. Which of the following postulates of the Kinetic Molecular Theory for gases explains
why gases exhibit pressure?
a. The molecules are in constant random motion
b. The distance between the molecules is great
c. The molecules collide with the walls of the ves
d. The molecular kinetic energy depends on the temperature.
__9. How will a velocity of a gas molecule vary if its molecular weight is increased
g/mol to 64 g/mol?
a. It will increase.
b. It will decrease.
__10. Which of the following gases is the fastest: He, O
a. O2
b. CO2
vi
What I Know
Read and understand each item and choose the letter of the
your answers on the space provide before the number.
Balance the gaseous chemical equation: __C3H8 + __O2  __CO2 + __H
__2. Which of the following is NOT a gas mixture?
b. Smelly Farts
d. Butane
of the following gas mixtures of N2 and He is the partial pressure of He the
, 3 moles He
, 1 moles He
, 2 moles He
, 5 moles He
Which of the following systems will have the greatest volume at STP?
(MM = 28 g/mol)
(MM = 17 g/mol)
1.00 g He (MM = 4 g/mol)
(MM = 44 g/mol)
Which of the following will contain the greatest number of molecules at 300K and 1 atm
dioxide is there in a 3.4 L balloon at SATP?
+ O2 (g)  H2O(l), how many liters of oxygen gas at STP will be
needed to react with 10 L of hydrogen gas at STP?
c. 5.00 L
d. 100 L
Which of the following postulates of the Kinetic Molecular Theory for gases explains
The molecules are in constant random motion
The distance between the molecules is great
The molecules collide with the walls of the vessel
The molecular kinetic energy depends on the temperature.
How will a velocity of a gas molecule vary if its molecular weight is increased
c. It will double.
d. Nothing happens.
__10. Which of the following gases is the fastest: He, O2, CO2, NH3?
c. He
d. NH3
Read and understand each item and choose the letter of the best answer. Write
+ __H2O
and He is the partial pressure of He the
?
Which of the following will contain the greatest number of molecules at 300K and 1 atm
, how many liters of oxygen gas at STP will be
Which of the following postulates of the Kinetic Molecular Theory for gases explains
How will a velocity of a gas molecule vary if its molecular weight is increased from 32
c. It will double.
d. Nothing happens.

151. 1
Lesson Gas Stoichiometry
1
What I Need to Know
Gases are everywhere. We interact with gases in our day to day life. The air
we breathe, the fire that cooks our food, the formation of our
ozone layer, and even the engines that powers our cars deal
with
gaseous reactions.
In this lesson, you will learn the quantitative relationships of the reactants and
products in a gaseous reaction. We will also be tackling how gases interact to form
products and its impact on its pressure and volume. You will also learn how to
determine the amount of gaseous reactants or products using the principles of
stoichiometry.
What’s New
Balancing Equations. Using your previous knowledge on balancing equations, put
the appropriate coefficients in the following reactions below.
1. ___C3H8 + ___O2  ___CO2 + ___H2O
2. ___CO + ___O2  ___CO2
3. ___H2NCONH2 + ___H2O  ___NH3 + ___CO2
4. ___N2 + ___H2  ___NH3
5. ___H2 + ___O2  ___H2O

152. 2

153. What Is It
In every chemical reaction, there is a certain ratio of the components that
must be available in order for the reaction to proceed. This ratio is called the
stoichiometric ratio. The stoichiometric ratio dictate how much reactants are
needed in order to create the desired products. We can see this as a ‘recipe’ for the
reaction except that the ‘recipe’ come in an equation with coefficients telling us how
much of each component is needed rather than having cups and teaspoons and a
whole bunch of instructions
Similarly, gases that participate in a gaseous reaction follow the same
principle. However, quantifying the exact amount of product that a gaseous reaction
produce is not as easy as reactions that yield solid or liquid products which are
simpler to measure and contain.
definite shape and volume, so how can people tell how much product is formed?
The answer is using the ideal gas equation.
Gaseous reactions are often measured at a certain condition called the
standard temperature and pressure
K) and a pressure of 1 atm.
volume changes which is then measurable.
PV=nRT, we get this expression
Substituting the STP values in the equation, we can infer that the amount of
gaseous products is equal to:
…where VSTP is the volume of the gases in
(L).
A perfect example for this kind of calculation is the reaction in an airbag. An
airbag is a safety device that is used in cars to cushion the passenger in a car crash.
Upon impact, a solid reactant is triggered to form a gas which then inflates the bag
that cushions the passenger.
The expansion of the airbag can then be measured which can be used to
determine the amount of gaseous products for
average volume of 60 litres
negligible. Assuming that the airbag’s volume was measured at STP, we can
determine the amount of nitrogen gas produced by the reaction.
3
In every chemical reaction, there is a certain ratio of the components that
must be available in order for the reaction to proceed. This ratio is called the
. The stoichiometric ratio dictate how much reactants are
eate the desired products. We can see this as a ‘recipe’ for the
reaction except that the ‘recipe’ come in an equation with coefficients telling us how
much of each component is needed rather than having cups and teaspoons and a
whole bunch of instructions in it.
Similarly, gases that participate in a gaseous reaction follow the same
principle. However, quantifying the exact amount of product that a gaseous reaction
produce is not as easy as reactions that yield solid or liquid products which are
o measure and contain. Gases, as defined, is a state of matter with no
definite shape and volume, so how can people tell how much product is formed?
The answer is using the ideal gas equation.
Gaseous reactions are often measured at a certain condition called the
standard temperature and pressure (STP) which is at a temperature of 0
K) and a pressure of 1 atm. In these conditions, the reaction proceeds such that the
s then measurable. Rearranging the ideal gas equation
expression:
=
Substituting the STP values in the equation, we can infer that the amount of
is equal to:
=
1 ×
0.0821 × 273
=
22.4
is the volume of the gases involved measured at STP in litres
A perfect example for this kind of calculation is the reaction in an airbag. An
airbag is a safety device that is used in cars to cushion the passenger in a car crash.
pact, a solid reactant is triggered to form a gas which then inflates the bag
that cushions the passenger.
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
The expansion of the airbag can then be measured which can be used to
unt of gaseous products formed. Let’s say that our airbag has an
average volume of 60 litres and the volume occupied by the solid sodium is
. Assuming that the airbag’s volume was measured at STP, we can
determine the amount of nitrogen gas produced by the reaction.
In every chemical reaction, there is a certain ratio of the components that
must be available in order for the reaction to proceed. This ratio is called the
. The stoichiometric ratio dictate how much reactants are
eate the desired products. We can see this as a ‘recipe’ for the
reaction except that the ‘recipe’ come in an equation with coefficients telling us how
much of each component is needed rather than having cups and teaspoons and a
Similarly, gases that participate in a gaseous reaction follow the same
principle. However, quantifying the exact amount of product that a gaseous reaction
produce is not as easy as reactions that yield solid or liquid products which are
Gases, as defined, is a state of matter with no
definite shape and volume, so how can people tell how much product is formed?
Gaseous reactions are often measured at a certain condition called the
which is at a temperature of 0°C (273
In these conditions, the reaction proceeds such that the
Rearranging the ideal gas equation
Substituting the STP values in the equation, we can infer that the amount of
volved measured at STP in litres
A perfect example for this kind of calculation is the reaction in an airbag. An
airbag is a safety device that is used in cars to cushion the passenger in a car crash.
pact, a solid reactant is triggered to form a gas which then inflates the bag
The expansion of the airbag can then be measured which can be used to
Let’s say that our airbag has an
and the volume occupied by the solid sodium is
. Assuming that the airbag’s volume was measured at STP, we can

154. 4
Substituting that into the equation:
=
22.4
=
60
22.4
= 2.68
This means that for every 60 litres of airbag that is inflated, there are 2.68 mol
of nitrogen gas produced by the reaction.
Gases are also measured at another standard condition called as Standard
Ambient Temperature and Pressure (SATP) which is a more attainable set of
conditions than the STP which is at 25°C (298 K) and 1 atm. Substituting the values
for SATP in the ideal gas equation, this yields:
= =
1 ×
0.0821 × 298
=
24.5
This equation can be used when determining the amount of gaseous products
in moles measured at SATP.
What’s More
Directions: Put a star on the letter that corresponds to the best answer.
(For #1-3) Mr Wangxian installed a pneumatic piston near his door that is
triggered to lock the door in case of an attack and placed 13.0 grams of NaN3 in
the empty piston. It is powered by the reaction below:
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
1. How many moles of NaN3 was available for the reaction? (Molar mass of
NaN3 is 65 g/mol)
A. 0.2 mol C. 0.1 mol
B. 0.3 mol D. 0.4 mol
2. Assuming that all of the NaN3 reacted, how many moles of nitrogen gas was
formed?
A. 0.2 mol C. 0.1 mol
B. 0.3 mol D. 0.4 mol
3. Assuming that all of the NaN3 reacted and the volume occupied by the solid
sodium is negligible, what is the final volume of the piston?
A. 6.27 L C. 6.72 L
B. 7.26 L D. 2.76 L

155. 5
4. 22.4 L of compound A was measured at STP after a gaseous reaction, how
many moles of A was in the sample?
A. 0.1 mol C. 10 mol
B. 0.01 mol D. 1.0 mol
(For #5-7) Acetylene (C2H2) is formed by the reaction of water with calcium
carbide, according to the following equation:
CaC2 (s) + 2 H2O (l)  Ca(OH)2 (aq) + C2H2 (g)
Mr Xiyao was told to prepare 10 L (at STP) of acetylene in the laboratory.
5. How many moles of acetylene should Ms Xiyao produce?
A. 0.644 mol C. 0.464 mol
B. 0.446 mol D. 4.064 mol
6. How many moles of CaC2 is needed to produce 10 L of acetylene?
A. 0.644 mol C. 0.464 mol
B. 0.446 mol D. 4.064 mol
7. How many grams of CaC2 must Ms Xiyao prepare? (Molar mass of CaC2 is
64 g/mol)
A. 68.2 g C. 28.6 g
B. 62.8 g D. 26.8 g
(For #8-10) Mr Sang Cheng burned 50 L of acetylene (C2H2) at STP in a constant
pressure piston. The piston expanded until the reaction was over. The reaction
proceeds:
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2 (g)
8. How many moles of acetylene is available for reaction?
A. 2.23 mol C. 3.23 mol
B. 3.22 mol D. 3.32 mol
9. Assuming that all of the acetylene reacted, how many moles of CO2 was
formed?
A. 6.64 mol C. 6.44 mol
B. 6.46 mol D. 4.46 mol
10. What volume is occupied by the CO2 at STP?
A. 0.10 L C. 10.0 L
B. 1.00 L D. 100 L

156. What I Have Learned
Directions: Look up some cases where gas stoichiometry can be used in our
modern day. In your own words, write about what you discovered and how it is
related to the topic.
What I Can Do
Enrichment Activity:
This activity is optional. Try to do this experiment at home, if you can.
You will need:
 Coke, preferably in a bottle
 Mentos
 Balloon
 Ruler
6
What I Have Learned
Look up some cases where gas stoichiometry can be used in our
modern day. In your own words, write about what you discovered and how it is
What I Can Do
. Try to do this experiment at home, if you can.
Instructions:
, preferably in a bottle 1. Place the mentos in the balloon.
2. Make sure the balloon has no
air and put the balloon over the mouth of
the coke bottle while holding the mentos
mentos inside the balloon
3. Release the mentos into the bottle and
secure the balloon and make sure only gas
can enter the balloon.
4. Once the balloon inflates, take it off, tie it,
and measure its radius in cm.
5. Assuming the balloon is a sphere,
calculate the volume of the balloon
formula V(in mL) = (4/3)πr
moles of CO2 in the balloon.
convert the volume to L first!
Look up some cases where gas stoichiometry can be used in our
modern day. In your own words, write about what you discovered and how it is
. Try to do this experiment at home, if you can.
1. Place the mentos in the balloon.
2. Make sure the balloon has no trapped
air and put the balloon over the mouth of
the coke bottle while holding the mentos
mentos inside the balloon
3. Release the mentos into the bottle and
secure the balloon and make sure only gas
balloon inflates, take it off, tie it,
radius in cm.
5. Assuming the balloon is a sphere,
of the balloon using the
= (4/3)πr3
and solve for the
in the balloon. Take note,
convert the volume to L first!

157. 7

158. Lesson Kinetic Molecular Theory of
Gases
2
What’s In
In the previous lesson, you have learned about how gaseous products are
measured with the use of stoichiometry and
In this lesson, we will be learning how gases move around and how they
interact with the space around it.
What I Need to Know
Gases around us behave differently from the other states of matter
are very flexible. Understanding how
of appreciating our environment.
In this lesson, you will learn about the kinetic molecular theory of gases and
relate the rates of gas effusion with molar mass
What’s New
Differentiate. Describe and differentiate the illustrations in the columns from one
another. You may also guess what state of matter it represents.
8
Kinetic Molecular Theory of
Gases
In the previous lesson, you have learned about how gaseous products are
with the use of stoichiometry and the ideal gas law.
In this lesson, we will be learning how gases move around and how they
interact with the space around it.
What I Need to Know
Gases around us behave differently from the other states of matter
nderstanding how these work around our surroundings is one way
of appreciating our environment.
In this lesson, you will learn about the kinetic molecular theory of gases and
relate the rates of gas effusion with molar mass.
What’s New
Describe and differentiate the illustrations in the columns from one
You may also guess what state of matter it represents.
Kinetic Molecular Theory of
In the previous lesson, you have learned about how gaseous products are
In this lesson, we will be learning how gases move around and how they
Gases around us behave differently from the other states of matter since they
these work around our surroundings is one way
In this lesson, you will learn about the kinetic molecular theory of gases and
Describe and differentiate the illustrations in the columns from one

159. 9
What Is It
Solids, liquids, and gases differ from one another in many ways. One
distinction that is being observed between the three states of matter is how closely
packed their atoms are. Solids are the most rigid among the three with each atom
bundled up almost side to side with no room for movement except for tiny vibrations
which we usually feel as heat. Liquids, on the other hand, are more flexible than
solids with its particles freely sliding on each other. This allows liquids to take the
shape of their containers while keeping its volume constant. Meanwhile, gases are
the most flexible and freely moving with its atoms observing very good social
distancing and can be compressed. This is the reason why gases cannot have a
definite volume and can occupy the shape of its container.
In the previous module, we have tackled the behaviour of gases in certain
conditions such as increasing the temperature and pressure. The logic behind why
we can predict its behaviour is due to its Kinetic Molecular Theory. This theory
states that:
1. Gases are made up of very small molecules,
which are separated by a very expansive space
between them.
2. Because of that, the force of attraction between
two particles is negligible. Therefore, the motion
of each particle is independent of one another.
3. Since they have so much space between them,
the molecules are in constant motion and move
around randomly in all directions.
4. Due to their perpetual state of random motion,
sometimes the particles will collide with each
other and with the walls of the container.
5. The collisions that occur are assumed to be
perfectly elastic, hence, there is no change in momentum in the molecules.
6. The average kinetic energy of the molecules is determined only by the
absolute temperature of the gas.
From what we know of the kinetic energy, it is defined as the energy
possessed by the object in motion. When things are in motion, they have velocity.
Gases in particular are very mobile particles and since these particles move
independently from one another and move randomly, it is hard to pinpoint the
velocity at which each molecule moves. In order to determine its kinetic energy, the
root-mean-square velocity of the gas is used and is defined by the equation:
=
3
…where vrms is the root-mean-square velocity of the gas, R is the ideal gas
constant, T is the absolute temperature in K, and M is the molar mass of the gas in
g/mol.

160. 10
As you can observe in the previous equation, the velocity is directly
dependent on its absolute temperature. This means that the hotter the gas, the more
mobile the gas molecules are.
We can express the previous equation in a proportionality equation by
factoring out the constants together to form this equation:
=
3
= √3 × = ×
…where k is equal to the square root of 3R.
Just like the gas laws, we can make use of this proportionality to compare the
velocities of two different gases of different molar masses at the same absolute
temperature. Let’s take compound 1 and 2 as an example with molar masses M1 and
M2, respectively.
, = × , = ×
Since both are of the same temperature, we can factor out the square root of
the temperature. This will yield this expression:
, = √ ×
1
, = √ ×
1
Rearranging the expression, we get:
, × = √ , × = √
Since both expressions are equal to k multiplied to the square root of T, we
can equate both equations and rearrange them again to get this expression:
, × = , ×
,
,
=
This expression is known as Graham’s Law of Diffusion which states that
relationship between the diffusion rate, or the rate at which the gas moves, is
inversely proportional to the square root of its molar mass.
Let’s take the diffusion of helium gas (He, molar mass = 4 g/mol) and oxygen
gas (O2, molar mass = 16 g/mol). How much faster would helium gas diffuse than
oxygen gas? Let us use the Graham’s Law of Diffusion to determine this.
= =
√16
√4
=
4
2
= 2

161. 11
According to our calculations, the ratio of the diffusion rates of helium to
oxygen gas is 4:2, which means that helium diffuses two times faster than oxygen at
the same temperature.
What’s More
Directions: Put a star on the letter that corresponds to the best answer.
1. Two bottles at the same temperature containing hydrogen sulphide (H2S) and
ammonia (NH3) are placed at equal lengths from you and are opened at the
same time. Which gas would you smell first?
A. Both C. None
B. Hydrogen sulphide D. Ammonia
2. Which of the following gases diffuses the fastest? Oxygen (O2), Hydrogen
(H2), Sulphur dioxide (SO2), or Carbon dioxide (CO2)?
A. Oxygen C. Hydrogen
B. Sulphur dioxide D. Carbon dioxide
3. Which of the following gases diffuses the slowest? Oxygen (O2), Hydrogen
(H2), Sulphur dioxide (SO2), or Carbon dioxide (CO2)?
A. Oxygen C. Hydrogen
B. Sulphur dioxide D. Carbon dioxide
4. Which of the following gases diffuses the fastest? Helium (He), Ammonia
(NH3), Hydrogen sulphide (H2S), or Carbon dioxide (CO2)?
A. Helium C. Ammonia
B. Hydrogen sulphide D. Carbon dioxide
5. Which of the following gases diffuses the slowest? Helium (He), Ammonia
(NH3), Hydrogen sulphide (H2S), or Carbon dioxide (CO2)?
A. Helium C. Ammonia
B. Hydrogen sulphide D. Carbon dioxide
6. Which of the following gases diffuses the fastest? Helium (He), Hydrogen
(H2), Sulphur dioxide (SO2), or Carbon dioxide (CO2)?
A. Helium C. Hydrogen
B. Sulphur dioxide D. Carbon dioxide
7. How much faster does helium gas diffuse than sulphur dioxide?
A. 2x C. 4x
B. 0.5x D. 0.25x
8. How much faster does hydrogen gas diffuse than sulphur dioxide?
A. 5.65x C. 4x
B. 0.18x D. 0.25x

162. 9. How will a gas velocity vary if its molecular weight was decreased
from 144 g/mol to 64 g/mol?
A. It will be faster by 200%
B. It will remain the same.
10.How will a gas velocity vary if its molecular weight was increased
g/mol to 128 g/mol?
A. It will be faster by 200%
B. It will remain the same.
What Have I Learned
Directions: Write an essay about how gas diffusion affects your day to day life. If
possible, cite examples and explain its connection to your day to
Summary
12
How will a gas velocity vary if its molecular weight was decreased
g/mol to 64 g/mol?
A. It will be faster by 200% C. It will be halved.
B. It will remain the same. D. It will be faster by 150%
How will a gas velocity vary if its molecular weight was increased
A. It will be faster by 200% C. It will be halved.
B. It will remain the same. D. It will be faster by 150%
What Have I Learned
Write an essay about how gas diffusion affects your day to day life. If
possible, cite examples and explain its connection to your day to day life.
How will a gas velocity vary if its molecular weight was decreased
C. It will be halved.
D. It will be faster by 150%
How will a gas velocity vary if its molecular weight was increased from 32
D. It will be faster by 150%
Write an essay about how gas diffusion affects your day to day life. If
day life.

163. 13
The amount of gaseous reactants and products are determined by measuring
the volume of the gas at standard conditions. These standard conditions are (1)
standard temperature and pressure (STP) which is at 0°C (273 K) and 1 atm and (2)
standard ambient temperature and pressure (SATP) which is at 25°C (298 K) and 1
atm.
With the integration of the ideal gas law, we can then determine the amount of
gas molecules occupied in a volume. At STP, the amount of gaseous molecules are
determined with the use of the expression:
=
22.4
And at SATP, the amount of gaseous molecules can be determined using the
expression:
=
24.5
Gases behave differently compared to solids and liquids due to its kinetic
molecular theory which states that gases are (1) made up of very small molecules
that are (2) separated by very great distances that (3) causes the force of attraction
between two molecules to be negligible, hence their motion is independent from one
another and since they have so much space between them, (4) the molecules are in
constant motion and move randomly in all directions which (5) eventually causes
them to collide (6) in a perfectly elastic collision which means that no momentum is
lost between two particles and due to their constant state of motion, (6) the average
kinetic energy of the molecules can only be determined by its absolute temperature.
The kinetic energy of the gas can also be determined using the root-mean-
square velocity of the gas which is expressed in this equation:
=
3
This can also be used to compare the diffusion rates between two gases at
the same temperature using the Graham’s Law of Diffusion, as shown by the
expression below:
,
,
=

164. Assessment: (Post
MULTIPLE CHOICE.
Directions: Read and understand each item and choose the letter of the
your answers on
__1. Balance the gaseous chemical equation: __C
e. 1, 5, 3, 4
f. 2, 3, 4, 1
g. 2, 10, 6, 8
h. 4, 6, 8, 2
__2. Which of the following is NOT a gas mixture?
c. Air
d. Chimney Fumes
__3. In which of the following gas mixtures
greatest?
e. 2 moles N2, 3 moles He
f. 3 moles N2, 1 moles He
g. 4 moles N2, 2 moles He
h. 5 moles N2, 5 moles He
__4. Which of the following systems will have the greatest volume at STP
e. 1.00 g N2 (MM = 28 g/mol)
f. 1.00 g NH3 (MM = 17 g/mol)
g. 1.00 g He (MM = 4 g/mol)
h. 1.00 g CO2 (MM = 44 g/mol)
__5. Which of the following will contain the greatest number of molecules at 300K and 1 atm
pressure?
e. 0.01 L
f. 0.10 L
g. 1.00 L
h. 10.0 L
__6. How much carbon dioxide is there in a 3.4 L
e. 0.13 g
f. 6.1 g
g. 0.15 mol
h. 6.6 g
__7. In the reaction 2H2 (g) + O
needed to react with 10 L of hydrogen gas at STP?
b. 20.0
d. 10.0 L
__8. Which of the following postulates of the Kinetic Molecular Theory for gases explains
why gases exhibit pressure?
e. The molecules are in constant random motion
f. The distance between the molecules is great
g. The molecules collide with the walls of the vessel
h. The molecular kinetic energy depends on the temperature.
__9. How will a velocity of a gas molecule vary if its molecular weight is increased from 32
g/mol to 64 g/mol?
c. It will increase.
d. It will decrease.
__10. Which of the following gases is the fastest: He, O
c. O2
d. CO2
14
Assessment: (Post-Test)
Read and understand each item and choose the letter of the
your answers on the space provide before the number.
the gaseous chemical equation: __C3H8 + __O2  __CO2 + __H
__2. Which of the following is NOT a gas mixture?
b. Smelly Farts
d. Butane
_3. In which of the following gas mixtures of N2 and He is the partial pressure of He the
, 3 moles He
, 1 moles He
, 2 moles He
, 5 moles He
Which of the following systems will have the greatest volume at STP?
(MM = 28 g/mol)
(MM = 17 g/mol)
1.00 g He (MM = 4 g/mol)
(MM = 44 g/mol)
Which of the following will contain the greatest number of molecules at 300K and 1 atm
How much carbon dioxide is there in a 3.4 L balloon at SATP?
+ O2 (g)  H2O(l), how many liters of oxygen gas at STP will be
needed to react with 10 L of hydrogen gas at STP?
c. 5.00 L
d. 100 L
Which of the following postulates of the Kinetic Molecular Theory for gases explains
The molecules are in constant random motion
The distance between the molecules is great
The molecules collide with the walls of the vessel
The molecular kinetic energy depends on the temperature.
How will a velocity of a gas molecule vary if its molecular weight is increased from 32
c. It will double.
d. Nothing happens.
__10. Which of the following gases is the fastest: He, O2, CO2, NH3?
c. He
d. NH3
Read and understand each item and choose the letter of the best answer. Write
+ __H2O
and He is the partial pressure of He the
?
Which of the following will contain the greatest number of molecules at 300K and 1 atm
, how many liters of oxygen gas at STP will be
Which of the following postulates of the Kinetic Molecular Theory for gases explains
How will a velocity of a gas molecule vary if its molecular weight is increased from 32
c. It will double.
d. Nothing happens.

165. Key to Answers
Pretest
Lesson 1
What’s New
Lesson 2
What’s New
Column 1: Tightly packed, no room for
Column 2: Can slide past each other, particles still touch each other, liquid
Column 3: Very spacious, particles don’t touch, gas
What’s More
15
Key to Answers
What’s More
Column 1: Tightly packed, no room for movement, solid
Column 2: Can slide past each other, particles still touch each other, liquid
Column 3: Very spacious, particles don’t touch, gas
Column 2: Can slide past each other, particles still touch each other, liquid

166. 16

167. 17
Post-test

168. References
[1] Commission on Higher Education, General Chemistry 1: Teaching Guide for
Senior High, Manila, 2016.
[2] Department of Education Central Office, Most Essential Learning Competencies
(MELCS), Manila, 2020.
[3] Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
Resource Management Development Team, 2016.
[4] A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
Technology, Quezon City: SD Publications, 2001.
[5] [Online]. Available: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
July 2020].
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mercury-barometer-vector
surface-weather-measur.html . [Accessed 6 July 2020].
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[Accessed 6 July 2020].
[10] "Brilliant," [Online]. Available: https://brilliant.org/wiki/ideal
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[12] [Online]. Available: https://www.sciencephoto.com/media/224322/view/jacques
charles-french-balloonist . [Accessed 7 July 2020].
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https://www.youtube.com/watch?v=BxUS1K7xu30 . [Accessed 7 July 2020].
18
References
Commission on Higher Education, General Chemistry 1: Teaching Guide for
Senior High, Manila, 2016.
Department of Education Central Office, Most Essential Learning Competencies
(MELCS), Manila, 2020.
Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
Resource Management Development Team, 2016.
A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
Technology, Quezon City: SD Publications, 2001.
ilable: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
https://www.123rf.com/photo_133132698_stock
vector-illustration-labeled-atmospheric-pressure
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Commission on Higher Education, General Chemistry 1: Teaching Guide for
Department of Education Central Office, Most Essential Learning Competencies
Department of Education, EASE/OHSP II Module 9: Gas Laws, Learning
A. Mapa, T. Fidelino and L. Rabago, Chemistry Textbook in Science and
ilable: https://www.firestonetire.ca/tire/weathergrip. [Accessed 6
https://www.123rf.com/photo_133132698_stock-vector-
pressure-tool-earth-
g1OsxMUw&list=PL0o_zxa4K1BWziAv
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[Online]. Available: https://simple.wikipedia.org/wiki/Balloon . [Accessed 6 July
https://www.youtube.com/watch?v=BxUS1K7xu30 . [Accessed 7 July 2020].

169. 19
For inquiries and feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
DepEd Division of Cagayan de Oro City
Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
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E-mail Address: cagayandeoro.city@deped.gov.ph

170. NOT
General Chemistry 1
Quarter 2 - Module 1
Quantum Mechanical Description and
the Electronic Structure of Atoms
Department of Education ● Republic of the Philippines
Senior High School

171. General Chemistry I- Grade 11
Alternative Delivery Code
Quarter 2 - Module 1: Quantum Mechanical Description and the
Electronic Structure of Atoms
First Edition, 2020
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172. Senior High School
General
Chemistry 1
Quarter 2 - Module 1
Quantum Mechanical Description and the
Electronic Structure of Atoms
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173. Table of Contents
What This Module is About ...................................................................................................................i
What I Need to Know..............................................................................................................................i
How to Learn from this Module ............................................................................................................ii
Icons of this Module ................................................................................................................................ii
What I Know........................................................................................................................................... ..iii
Lesson 1:
Quantum Numbers .......................................................................................................1
What I Need to Know ....................................................................................1
What’s New ...................................................................................................1
What Is It ......................................................................................................2
What’s More: Let’s test your understanding…. ............................................3
What’s More: Identify the Orbital ..................................................................4
What I Have Learned: How much have you learned?..................................4
What I Can Do: I am Electroman…. .............................................................5
Lesson 2:
Electron Configuration and the Magnetic Property of Atoms 6
What’s In........................................................................................................6
What’s New: What is the mystery word?......................................................6
What Is It ......................................................................................................7
What’s More: Electron configuration and Orbital diagrams ….....................9
What’s More: Label it! ...................................................................................9
What I Have Learned: Boarding house analogy...........................................10
What’s I Can Do: What are you eating? .......................................................10
Summary......................................................................................................................................11
Assessment: (Post-Test)............................................................................................................12
Key to Answers............................................................................................................................13
References...................................................................................................................................16

174. i
Module 1
What This Module is About
Early efforts by nineteenth-century physicists to comprehend atoms and
molecules met with only limited success. With the unwavering pursuit of scientists to
come up with different experiments and theories, the flurry of research that ensued
altered our concept of nature forever.
This module comprises activities that will help deepen your understanding on
the properties and characteristics of atoms and how they affect the chemistry present
in our daily lives.
The following are the lessons contained in this module:
• Quantum Numbers
• Electron configuration and the Magnetic Property of Atoms
What I Need to Know
At the end of this module, you should be able to:
1. Use quantum numbers to describe an electron in an atom (STEM_GC11ESIIa-
b-54);
2. Determine the magnetic property of the atom based on its electronic
configuration (STEM_GC11ESIIa-b-57);
3. Draw an orbital diagram to represent the electronic configuration of atoms
(STEM_GC11ESIIa-b-58)

175. ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises
diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge.
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under-
standing of the concept.
What’s More These are follow-up activities that are in-
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show-
case your skills and knowledge gained, and
applied into real-life concerns and situations.

176. iii
What I Know
Pretest: MULTIPLE CHOICE:
Directions: Read and understand each item and choose the letter of the correct
answer. Use separate answer sheet of paper.
1. What do you call the three-dimensional orientation of the orbital in space around
the nucleus?
A. magnetic quantum number C. electron configuration
B. principal quantum number D. geometry
2. What quantum number that indicates the relative size of an orbital?
A. magnetic quantum number C. electron configuration
B. principal quantum number D. geometry
3. How many 4d orbitals can be found in an atom?
A. 9 C. 1
B. 5 D. 7
4. Which of the following combinations is allowed?
A. n=2, ℓ =1, ml= -1, ms= +1/2 C. n=3, ℓ =1, ml=-3, ms= -1/2
B. n=1, ℓ =1, ml= +1, ms= -1/2 D. None of the above
5. Who said that no two electrons can have the same set of four quantum numbers?
A. Heisenberg C. Hund
B. Einstein D. Pauli
6. What do you call the arrangement of electrons within the orbitals of an atom to
know more about an atom’s electronic property?
A. principal quantum number C. electron configuration
B. magnetic quantum number D. geometry
7. What is the other term for the building up principle?
A. Uncertainty Principle C. Roblox Principle
B. Hammer Principle D. Aufbau principle
8. Which element has an electron configuration of 1s2
2s2
2p5
?
A. Carbon (atomic number = 6) C. Fluorine (atomic number = 9)
B. Oxygen (atomic number = 8) D. Hydrogen (atomic number = 1)
9. What do you call the pictorial descriptions of the electrons in an atom?
A. orbital diagrams C. gaussian curve
B. energy diagrams D. cliparts
10. What do you call atoms with unpaired electrons?
A. Paramagnetic C. Single atoms
B. Diamagnetic D. Lone pair

177. 1
I Need to Know
In your previous lesson, you were taught on what are the particles or composition of
an atom? They are the proton and neutron that are located inside, and the electron that is
located outside the nucleus. In this lesson, we will be focusing on the characteristics of an
electron since it has an important role in chemical bonding. Since electron is located outside
the nucleus, it is difficult to determine its exact location. That is why we have to learn about
the behaviors of quantum particles. Of these behaviors, the most we can do is to calculate
probabilities as to the location and behavior of the particles.
According to the Heisenberg’s uncertainty principle, it is impossible that both the
energy and position of an electron can be known at the same time. Thus, as we know more
about the electron’s energy, we know less about its position, and vice versa.
Bohr’s model of the hydrogen atom suggests that the electron orbits the nucleus like
our solar system (e.g. the planets around the sun). However, the quantum mechanical
description of the hydrogen atom has proven that the Bohr’s model of electrons is incorrect. It
states that we don’t know exactly where the electron is, but with high probability, we can
conclude that the electron is most likely to be found in an orbital (Chang, 2010). In this lesson,
you should be able to describe the electrons (e-
) in orbitals using the four quantum numbers.
Figure 1.1. Bohr’s Model (Electron in orbit) Figure 1.2. Quantum Mechanics
(Electron in orbital)
What’s New
Activity 1: True or False
Directions: Carefully read the following statements below and write T if it is TRUE
and F if it is FALSE on the space provided at the left side.
_______1. The quantum mechanical description of electron is more accurate than that
of Bohr’s model.
_______2. No two electrons have the same 4 quantum numbers.
_______3. We can both know the energy and the position of electrons at the same
time.
_______4. Any two electrons in the same orbital must have the same spins.
_______5. The four quantum numbers are used to describe the probable location of
an electron in an atom.
1
Quantum Numbers

178. 2
GUIDE QUESTION:
What is the difference between Bohr’s model and the quantum mechanical
model of an electron? State your answer in 3-5 sentences only.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
__________________________________________
What Is It
To describe the orbitals in which electron can be found, quantum numbers are
required. Quantum numbers are set of values which give us an information about the location
of electrons in the electron cloud of an atom. It can be used to determine the electron
configuration of an atom. According to the Pauli Exclusion Principle, each electron in an atom
has an exclusive set of quantum numbers and no two electrons can have the same
combination of four quantum numbers (Silberberg, 2013).
The Four Quantum Numbers
Table 1 Quantum numbers and their possible values
Quantum Number Symbol Possible Values
Principal Quantum Number n 1,2,3,4… (positive integers)
Angular Momentum Quantum Number ℓ 0,1,2,3… (0 to n-1)
Magnetic Quantum Number ml - ℓ,…-1,0,1…,+ ℓ
Spin Quantum Number ms +1/2, -1/2
1. Principal Quantum Number
The principal quantum number (n), describes the energy of an electron. It refers
to the energy level and the size of the orbital an electron is likely to be found. The value
of n starts from 1 to the shell containing the outermost electron of that atom. The larger
the value of n, the greater is the energy and the larger is the orbital. The group of
orbitals with the same value of n is called an electron shell. All the orbitals that have n
= 2, for example, are said to be in the second shell.
Carbon is in the second period of the periodic table, so, its outermost electron
is in the shell with an energy level 2. Therefore, an electron in Carbon can have an n
value from 1 to 2.
2. Angular/Azimuthal Quantum Number
In chemistry, the angular quantum number (ℓ), defines the shape of an
atomic orbital. It also strongly influences bond angles and chemical bonds. It is defined
in chemistry that if ℓ = 0, it is called an s orbital, ℓ = 1 is a p orbital, ℓ = 2 a d orbital, and
ℓ = 3 an f orbital. The first p orbital (ℓ = 1) is in the second electron shell (n = 2), the first
d orbital (ℓ = 2) is in the third shell (n = 3), and so on. The set of orbitals that have the
same n and l values is called a subshell.
3. Magnetic Quantum Number
The magnetic quantum number (ml), describes the orientation of the orbital
in space and can have integral values between - ℓ and ℓ, including zero.
For example, the p subshell (ℓ = 1) contains three orbitals, so the ml of an
electron in a p subshell will be −1, 0, or 1.

179. 3
The outermost electron of Carbon is in a 2p subshell. This means that for that
electron, n=2 and ℓ = 1. Since ℓ = 1, we can conclude that there are three 2p orbitals
in this subshell because there are three values of ml, given by -1, 0, and 1.
4. Spin Quantum Number
Individual electrons within an orbital has a property represented by the
spin quantum number. Each orbital may hold up to two electrons with opposite spin
directions. Electrons are not really spinning in a physical sense, this is just a
representation of the idea that there are two possible values for the spin quantum
number. When an electron is assigned to spin up, it is represented by an upward arrow
and a value of +1/2. If an electron is spinning down, it is represented by a downward
arrow and a value of -1/2 (Brown, 2015).
Figure 1.3. Representation of the Spin Quantum Number values
What’s More (A)
Activity 2.1: Let’s test your understanding!
Directions: Answer the following questions below as directed on a separate sheet of
paper and submit it to your teacher as soon as you are finished.
1. List the values of n, ℓ, and m/ for orbitals in the 4d subshell.
n value/s ℓ value/s ml value/s
2. What is the total number of orbitals associated with the principal quantum number
n=3? Defend your answer.
For items 3-5, identify if the following set of quantum numbers are correct. If not, indicate
which quantum number is wrong.
3. n=2, ℓ =1, ml= -1, ms= +1/2
4. n=3, ℓ =1, ml=-3, ms= -1/2
5. n=1, ℓ =1, ml= +1, ms= -1/2

180. 4
What’s More (B)
Activity 2.2: Identify the orbital
Directions: Identify which orbital is described by the following sets of quantum
numbers. If the set includes an incorrect value, write “not allowed”.
n ℓ ml Orbital
2 1 -1 2p (example)
1. 1 0 0
2. 3 -3 2
3. 3 2 -2
4. 2 0 -1
5. 0 0 0
6. 4 2 1
What Have I Learned
Activity 3: How much have you learned?
Directions: In your own words, describe the following terms in 2-3 sentences only.
1. Quantum Number
_________________________________________________________________________
_________________________________________________________________________
2. Principal Quantum Number
_________________________________________________________________________
_________________________________________________________________________
3. Angular Quantum Number
________________________________________________________________________
________________________________________________________________________
4. Magnetic Quantum Number
___________________________________________________________________
___________________________________________________________________

181. 5
What I Can Do
Activity 4: I am ELECTRON MAN!
Directions: Imagine yourself as an electron. As an electron, you should keep track of
your location and activity for three days. If quantum numbers give
information about the location of an electron or set of electrons, you could
describe your location in any number of ways (e.g. GPS coordinates,
qualitatively describing your surroundings, google map, etc.). Fill out the
table below with the needed details and answer the questions that follow.
The first row serves as an example.
Electron Name: Special Skill:
Day Time Location Activity
1 9 am Dining Area, Rizal’s House, Cagayan de
Oro City, Philippines
Having breakfast
with family
1 9 am
1 3 pm
1 7 pm
2 9 am
2 3 pm
2 7 pm
3 9 am
3 3 pm
3 7 pm
Follow-up Questions:
1. What is the importance of understanding the role of quantum numbers in
chemistry?
______________________________________________________________
______________________________________________________________
______________________________________________________________
2. How are GPS (Global Positioning System) and quantum numbers related to
each other?
______________________________________________________________
______________________________________________________________
______________________________________________________________
3. Why is it important to be aware of your location and surroundings?
______________________________________________________________
______________________________________________________________
______________________________________________________________

182. 6
What’s In
In lesson 1, we have learned that electrons have four quantum numbers which
describe the location of electrons in an orbital and can be used to determine the electron
configuration of an atom. The electron configuration will be discussed in-depth in this lesson
and the magnetic property of an atom will be determined based on its electron configuration.
Furthermore, the electron configuration of an atom should also be represented through orbital
diagrams.
What’s New
Activity 1: What is the Mystery Word?
Directions: Identify the letter that is described by each item below. Use these letters
to reveal the mystery word in the box.
1. The symbol of the element in the third period with 5 valence electrons and is used in
the manufacture of safety matches.
2. The symbol of the element in the first period with 2 valence electrons and used to
inflate party balloons.
3. The symbols of the two gaseous elements in the second period with valence electrons
equal to 5 and 6, respectively.
4. The first letter of the name of the principle which states that electrons fill atomic orbitals
of the lowest available energy levels first before occupying higher levels in the atom’s
ground state.
5. The symbol of the halogen with a complete electron configuration:1s2
2s2
2p6
3s2
3p6
3d10
4s2
4p6
4d10
5s2
5p5
.
Hint: The “word” is often used as an indicator in acid–base titrations. It turns colorless in acidic
solutions and pink in basic solutions.
Electron Configuration and
the Magnetic Property of
Atoms
2
__ __ __ L P T H __ L E __ N
1 2 3 4 5

183. 7
What Is It
Electron configuration is the arrangement of electrons within the orbitals of an atom
to determine its electronic property. The ground-state electron configuration is the most stable
arrangement of electrons in an atom. All the electrons in an atom reside in the lowest energy
orbitals possible in this arrangement. Since each orbital can accommodate a maximum of two
electrons, using the periodic table, we can predict the electron configuration of all elements.
Valence electrons are the outermost electrons of an atom. They are the highest
energy electrons in an atom and are the most reactive. Valence electrons can be gained, lost,
or shared to form chemical bonds unlike the inner electrons which do not participate in
reactions. The number of valence electrons of each element is equal to its group number on
the Periodic Table (Brown, 2015). Elements with the same number of valence electrons tend
to have similar chemical properties and they are grouped together.
These are the principles and rule to be followed in determining the electron
configuration of an element. They are as follows:
The Aufbau Principle
The electrons in an atom fill up its atomic orbitals according to the Aufbau Principle;
"Aufbau," in German, means "building up." According to this principle, electrons are filled in
the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p…
Figure 2.1. The order in which atomic subshells are filled in a many-electron atom.
Hund’s Rule
Hund’s rule states that before additional electrons with opposite spins can occupy the
same orbitals, single electrons with the same spin must occupy each equal-energy orbital first
(Silberberg, 2013). This is like the seats in a bus. In a bus, you sit alone, rather than with a
stranger, if you have the option. As long as there is an available place on the same level,
electron tends to be alone rather than be beside another electron.

184. 8
Pauli Exclusion Principle
No two electrons can have the same combination of four quantum numbers. A
maximum of two electrons may occupy a single orbital, but only if the electrons have opposite
spins.
Table 2.1 Electron configurations of some lighter elements
The sum of superscripts of the ℓ values in the electron configuration of an element
equals to the total number of electrons of the element. This is one way of checking whether
the electron configuration is correct or incorrect. The superscripts represent the total number
of electrons residing in the said orbital.
There is only one (1) s orbital since the value of ℓ is only equal to zero. There are three
2p orbitals since the values of ml are equal to -1,0 and +1. Orbitals of the same n values have
the same energy and that filling of orbitals should be according to the Aufbau’s principle,
Hund’s rule and Pauli’s exclusion principle. Note that p orbitals have higher energy compared
to s orbitals.
Magnetic Property of an Atom
The behaviour of an atom in relation to magnetic fields is influenced by its electron
configuration. This behaviour is also called as the magnetic property of an atom and is
dependent on the number of electrons an atom has that are spin paired. An atom with
electrons that will be very slightly affected by magnetic fields is called diamagnetic. The
orbitals of this atom are all filled and therefore all its electrons are paired with an electron of
opposite spin. Neon is an example of a diamagnetic atom. Conversely, atoms that do not have
all their electrons spin-paired and are affected by magnetic fields are called paramagnetic.
Lithium and sodium are examples of paramagnetic atoms (Brown, 2015).

185. 9
What’s More (A)
Activity 2.1: Electron Configuration and Orbital Diagrams
Directions: Write the complete electron configuration of the following elements and
draw their orbital diagrams. State whether the element is paramagnetic or
diamagnetic. The first one is done for you.
Element
Number
of
electrons
Electron
Configuration
Orbital Diagram
Magnetic
Property
Lithium 3 Paramagnetic
Oxygen
Fluorine
Bromine
Silicon
Zinc
What’s More (B)
Activity 2.2: Label it!
Directions: Label each part of the following notation of an outermost electron and
state what each part represents.
b
a 3s2
c

186. 10
What I Have Learned
Activity 3: Boarding House Analogy
Directions: Imagine you are the landlord of a very strange boarding house. Your job is
to fill the rooms in the building in the most efficient way possible. The rules you
have to follow are as strange as the building because quantum mechanics is not
like anything you might have expected. State what electron rule is being applied
in the situations stated in the left side of the table. Explain each rule. (This activity
is adapted from birdvilleschoosl.net).
Boarding House Rules Electron Rules
From the Bottom Up: Rooms should be filled
from the ground floor and up. Fill up first the rooms
on the first floor before starting to put new tenants
on the second floor.
Singles First: the owner of the building wants to
have the renters spread out as much as possible.
For that reason, singles are placed first in rooms
before couples. If couples must be placed into a
room then all of the other rooms on that floor must
already have a single in them.
Opposite Gender Only: When two people are to
be situated in a room, they must be of opposite
genders. No men may stay together, and no
women may room together. This is an arbitrary
rule on the part of the owners: in a just world, we
wouldn’t have to follow it, but quantum mechanics
has nothing to do with society’s norm.
What I Can Do
Activity 4: What are you eating?
Directions: Research about the ingredients of “PIATTOS”. Identify at least two elements
present in the food and research about the properties and uses of each element. Fill out the
table with the needed details and answer the questions that follow.
FOOD NAME:
Element Properties Uses
Electron
Configuration
Orbital
Diagram
Paramagnetic/
Diamagnetic?
1.
2.
3.
1.
2.
3.
1.
2.
3.
1.
2.
3.
PIATTOS- CHEESE FLAVOR

187. 11
Follow-up Questions:
1. Why is it important to be aware of the ingredients of the food you eat?
___________________________________________________________________
___________________________________________________________________
____________________________________________________
2. Based on your research, is your favorite food good for your health? Why?
___________________________________________________________________
___________________________________________________________________
____________________________________________________
3. Why is it important to gain knowledge on the properties and characteristics of some
elements?
___________________________________________________________________
___________________________________________________________________
___________________________________________________
SUMMARY
• All electrons have four quantum numbers which describe the location of electrons in
the electron cloud of an atom and can be used to determine the electron configuration
of an atom.
• According to the Pauli Exclusion Principle, each electron in an atom has an exclusive
set of quantum numbers and no two electrons can have the same combination of four
quantum numbers.
• The principle quantum number (n) describes the size of the orbital the electron is
residing. The azimuthal or angular quantum number (l) describes the shape of the
orbital. The magnetic quantum number (ml) defines the orientation of the orbital in
space and the electron spin number (ms) defines the direction that the electron spins
on its own axis.
• Electron configuration is the arrangement of electrons within the orbitals of an atom to
know more about an atom’s electronic property.
• The electrons in an atom fill up its atomic orbitals according to the Aufbau Principle;
"Aufbau," in German, means "building up." According to this principle, electrons are
filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s,
5f, 6d, 7p…
• Hund’s rule states that before additional electrons with opposite spins can occupy the
same orbitals, single electrons with the same spin must occupy each equal-energy
orbital first.
• Atoms with unpaired electrons in their electron configuration and are affected by
magnetic field are paramagnetic. Atoms with no unpaired electrons in their electron
configuration are diamagnetic.

188. 12
Assessment: (Post-Test)
Multiple Choice. Read and understand each item and choose the letter of the correct
answer. Use separate answer sheet of paper.
1. How many orbitals in an atom can have the 4d designation?
A. 9 C. 1
B. 5 D. 7
2. Which of the following combinations is allowed?
A. n=2, ℓ =1, ml= -1, ms= +1/2 C. n=3, ℓ =1, ml=-3, ms= -1/2
B. n=1, ℓ =1, ml= +1, ms= -1/2 D. None of the above
3. Which element has an electron configuration of 1s2
2s2
2p5
?
A. Carbon (atomic number = 6) C. Fluorine (atomic number = 9)
B. Oxygen ((atomic number = 8) D. Hydrogen (atomic number = 1)
4. What principal quantum number can an electron in an f sublevel have?
A. 4 C. 1
B. 5 D. 8
5. Can orbital 3f exist?
A. Yes , because it is the biggest orbital.
B. No, because the third shell, electrons occupy only the s,p and d sublevels.
C. Yes, because the third shell can hold up to more than 18 electrons.
D. None of the above
6. Which element has an electronic configuration of 1s2
2s2
p6
3s2
3p6
4s2
3d3
?
A. Carbon (atomic number = 6) C. Krypton (atomic number = 36)
B. Vanadium (atomic number = 23) D. Copper (atomic number = 29)
7. Which of the orbitals has the highest energy?
A. 1s C. 2p
B. 2s D. 3s
8. Which of the following atom is paramagnetic?
A. Zinc C. Calcium
B. Krypton D. Potassium
9. How many unpaired electrons are there in the ground state of a cobalt atom? What is the
magnetic property of the atom?
A. 3, paramagnetic C. 2, diamagnetic
B. 5, paramagnetic D. 0, diamagnetic
10. Which of the following electrons described by quantum numbers (n, l, ml, ms) has the
highest energy?
A. (3,0,0,+1/2) C. (4,1,0,+1/2)
B. (3,1,‐1, ‐1/2) D. (3,2,0,+1/2)

189. 13
Activity
2.2
1s
Not
allowed
3d
2s
Not
allowed
4d
Activity
1:
T
T
F
F
T
Activity
2.1
1.
n=
4
ℓ
=
2
m
/
=
-2,
-1,
0,
1,
2
2.
9
For
n=3,
the
possible
values
of
/
are
0,
1,
and
2.
One
3s
orbital
(
n=3,
/=
0,
and
m
/
=0);
Three
3p
orbitals
(
n=3,
/=
1,
and
m/=
-1,
0,
1,);
Five
3d
orbitals
(
n=3,
/=2,
and
m/=
-2,-1,
0,
1,
2).
The
total
number
of
orbitals
is
1
+
3
+
5
=
9.
3.
Yes
4.
No;
m
l
=
3
5.
No;
ℓ
=
1
Key to Answers
Pre-test:
1.A
2.
B
3.B
4.A
5.D
6.C
7.D
8.C
9.A
10.A
Lesson
1:

190. 14
Key to Answers

191. 15
Key to Answers
Post-test:
1.B
2.
A
3.C
4.A
5.B
6.B
7.D
8.D
9.A
10.C
Lesson
2
-
Activity
2.1

192. 16
References
“Atomic Structure.” SparkNotes. Accessed August 28, 2020.
https://www.sparknotes.com/chemistry/fundamentals/atomicstructure/
section2/.
Brown, Theodore. Chemistry: The Central Science. New York: Pearson,2015.
Canva. Accessed November 5, 2020. https://www.canva.com/education
Chang, R. and Goldsby, K. Chemistry. New York: McGraw-Hill
Education, 2010.
“Quantum Numbers - Concept.” Brightstorm. Accessed August 28, 2020.
https://www.brightstorm.com/science/chemistry/the-atom/quantum-
numbers/.
“Quantum Numbers for Atoms.” Chemistry LibreTexts. August 15, 2020.
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Che
mistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theor
etical_Chemistry)/Quantum_Mechanics/10:_Multi-
electron_Atoms/Quantum_Numbers_for_Atoms.
Silberberg, Martin. Principles of General Chemistry. Boston: McGraw-
Hill Higher Education, 2013.

193. 17
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194. General Chemistry 1
Quarter 2 - Module 2
Chemical Bonding and the Shapes of
Molecules
Senior High School

195. General Chemistry I- Grade 11
Alternative Delivery Mode
Quarter 2 - Module 2: Chemical Bonding and the Shapes of Molecules
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196. General
Chemistry 1
Quarter 2 - Module 2
Chemical Bonding and the Shapes of
Molecules
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197. Table of Contents
What This Module is About ...................................................................................................................i
What I Need to Know..............................................................................................................................i
How to Learn from this Module ............................................................................................................ii
Icons of this Module ................................................................................................................................ii
What I Know........................................................................................................................................... ..iii
Lesson 1:
Types of Bonding and their Properties .......................................................1
What I Need to Know ....................................................................................1
What Is It ......................................................................................................1
What’s More : …............................................................................................3
What I Have Learned: ...................................................................................4
Lesson 2:
Lewis Structures and Bond Formation ......................................................5
What’s In........................................................................................................5
What’s New ...................................................................................................5
What Is It ......................................................................................................6
What’s More ….............................................................................................10
What I Have Learned ....................................................................................10
Lesson 3:
Molecular Geometry and Polarity ....................................................................11
What’s In........................................................................................................11
What Is It ......................................................................................................11
What’s More ….............................................................................................15
What I Have Learned ....................................................................................19
Summary......................................................................................................................................16
Assessment: (Post-Test)............................................................................................................17
Key to Answers............................................................................................................................18
References...................................................................................................................................20

198. i
Module 2
What This Module is About
When you look at everything around you and what it is made of you, you will
understand that atoms seldom exist on their own just as humans cannot survive alone. More
often, the things around us are made up of different atoms that have been bonded together.
The bonding of atoms or molecules is one of the most essential processes in chemistry
because it permits all sorts of different molecules and combinations of atoms to form, which
then make up matters in the world we live in.
In this module, we examine the relationship between the electronic structure of atoms
and the type of chemical bonds they form. Moreover, the role of chemical bonds and lone
pairs on the geometry and polarity of a molecule is discussed.
What I Need to Know
At the end of this module, you should be able to:
1. Draw the Lewis structure of ions (STEM_GC11CBIId-g-70);
2. Apply the octet rule in the formation of molecular covalent compounds
(STEM_GC11CBIId-g-76);
3. Write the formula of molecular compounds formed by the nonmetallic elements of the
representative block (STEM_GC11CBIId-g-77);
4. Draw Lewis structure of molecular covalent compounds (STEM_GC11CBIId-g-78);
5. Describe the geometry of simple compounds (STEM_GC11CBIId-g-81);
6. Determine the polarity of simple molecules (STEM_GC11CBIId-g-82)

199. ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises
diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge.
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under-
standing of the concept.
What’s More These are follow-up activities that are in-
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show-
case your skills and knowledge gained, and
applied into real-life concerns and situations.

200. iii
What I Know
Pre-test: MULTIPLE CHOICE:
Directions: Read and understand each item and choose the letter of the correct answer.
Use separate answer sheet of paper.
1. Which element will have 5 electrons in its Lewis dot symbol?
A. Argon C. Carbon
B. Boron D. Phosphorus
2. Which element is the least electronegative?
A. Calcium C. Iron
B. Cesium D. Barium
3. The complete transfer of one or more electrons between atoms constitutes in forming
___________.
A. ionic bond C. coordinate covalent bond
B. covalent bond D. dative bond
4. It is important to know the geometry of a molecule because it
A. affects the physical and chemical properties of the substance
B. will give the Lewis structure of the molecule
C. will determine whether the molecule is ionic or covalent
D. B and C
5. If there are four (4) electron pairs around the central atom of a molecule, these
electron pairs are in a _________________arrangement.
A. linear C. tetrahedral
B. trigonal planar D. octahedral
6. A list of non-metals is given below. Which elements can exceed the octet rule?
S, C, Cl, O
A. O C. Cl
B. C D. S
7. Which of the following represents a non-polar covalent bond?
A. H-O C. C-C
B. C-N D. Li-F
8. What type of bond does NO2 forms?
A. ionic bond C. coordinate covalent bond
B. covalent bond D. dative bond
9. Atoms bond to other atoms to obtain a/an _______________ electron configuration.
A. alkali metal C. noble gas
B. halogen D. Chalcogen
10. It is a measure of how equally the electrons in a bond are distributed between the
two atoms involved in a covalent bond.
A. polarity C. ionization energy
B. octet rule D. electron affinity

201. 1
What I Need to Know
Atoms usually interact with other atoms or group of atoms. They might be connected
by strong bonds and formed into molecules or crystals, or they might also form temporary,
weak bonds with other atoms. These bonds hold molecules together and are essential to the
chemistry of our bodies and to the existence of life itself. In this lesson, we examine the models
of chemical bonding and their properties, and how they influence the behavior of the
substances around us.
Source:https://www.google.com.ph/search?q=the+structure+of+flavonoid+isorhamnetin
Figure 1. Three-dimensional structure of the flavonoid isorhamnetin. It shows bonding of
Carbon, Hydrogen and Oxygen atoms to each other.
What Is It
Why do atoms bond at all? The answer is that atoms are trying to reach the most stable
or lowest-energy state that they can. Usually, atoms become more stable when their orbital of
the outermost energy level or valence shell is filled with electrons, satisfying the octet rule. If
atoms do not have this arrangement, they reach it by gaining, losing, or sharing electrons via
chemical bonds. In chemical bonding, only valence electrons, electrons located in valence
shell of the element, are involved.
1
Types of Bonding and their
Properties

202. 2
Source:https://www.google.com.ph/search?q=stable+and+unstable+atoms
Figure 2. Electron Shells and Atom Stability
Types of Bonding
1. Ionic Bond
An ionic bond is formed when metals on the left side (Group 1(A) and Group 2(A) of
the periodic table and non-metals on the right side (except noble gases, group 8A) interact.
Once the electrons have been transferred to the non-metal, both the metal and the non-metal
become ions. The metal becomes positively charged and the nonmetal becomes negatively
charged. An ionic compound is formed when the two oppositely charged ions attract each
other. For instance, positively charged sodium ions and negatively charged chloride ions
attract each other to make sodium chloride, or table salt.
Ionic compounds tend to have higher melting and boiling points. They are hard and
brittle and conduct electricity when dissolved in water. Some common ionic compounds are
magnesium bromide (MgBr2), magnesium oxide (MgO), and potassium bromide (KBr).
Source:https://www.google.com.ph/search?q=Common+elements+that+form+ionic+bonds
Figure 3. Common elements that form ionic bonds

203. 3
2. Covalent Bond
Covalent bonds are formed when atoms reach stability by sharing electrons (rather
than fully gaining or losing them). Covalent bonds are more common than ionic bonds in the
molecules of living organisms. These bonds mostly occur between nonmetals or between two
of the same (or similar) elements. One, two, or three pairs of electrons may be shared between
atoms, resulting in single, double, or triple bonds, respectively. The more electrons that are
shared between two atoms, the shorter and stronger their bond will be.
Source:https://www.google.com.ph/search?q=structure+of+single%2C+double%2C+and+triple+bonds+of+common+covalent+c
ompounds
Figure 4. Structures of Some Common Covalent Compounds
Compounds formed through covalent bonding are brittle solid, have relatively low
melting and boiling points, and are poor conductor of heat and electricity. Several covalent
compounds have high vapor pressure, which makes them volatile and good as fuels. Propane,
methane and gasoline are all covalent compounds that readily undergo combustion, producing
energy as a bi-product.
What’s More
Activity 1: The Name is Bond… Chemical Bond
Directions: Fill out the table below with correct answers. The first one is done for you.
Compound
Type of
Bond
Good conductor of
heat or electricity?
High Boiling
Point?
High Melting
Point?
NaCl ionic Yes Yes Yes
CH4
HCl
N2
O2
H2O
KBr
MgCl2
PCl3
CO
CaF2

204. 4
What Have I Learned
Activity 2: What type of bond do you form?
Directions: Imagine yourself and the people around you as atoms. Just like atoms, you need
to form bonds to be more stable. Your attitude, talents and potentials, love and care, advises
and compliments, and the things you treasure should serve the function of electrons. Based
on your understanding on the concept of ionic and covalent bonds, what type of bond do you
want to form with other people? What do you want to do with your electrons? Support your
answer. Your answer should not be less than 7 sentences and not more than 10 sentences.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________

205. 5
What’s In
In lesson 1, we have learned that atoms transfer or share their electrons to other atoms
to become more stable, thus forming new substances. In this lesson, the bonding models will
be represented through Lewis dot symbols and structures and the bond formation of atoms
will be illustrated. Furthermore, we will learn how to predict bonding and formula of molecular
compounds though Lewis structures and the application of the octet rule.
What’s New
Activity 1: Meet the Nobles
Directions: Study the table below and answer the questions that follow.
Element
Atomic
Number
Electron Configuration
He 2 1s2
Ne 10 1s2
2s2
2p6
Ar 18 1s2
2s2
2p6
3s2
3p6
Kr 36 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
Xe 54 1s2
2s2
2p6
3s2
3p6
3d10
4s2
4p6
4d10
4f14
5s2
5p6
Ra 86 1s2 2s2
2p6
3s2
3p6
3d10
4s2
4p6
4d10
4f14
5s2
5p6
5d10
6s2
6p6
1. To what group in the periodic table do the elements listed belong?
2. Why these elements are called the noble gases? What is their common characteristic?
3. Why are light bulbs filled with argon gas rather than oxygen gas?
4. How many valence electrons do noble gases have? Draw the Lewis dot symbol of the noble
gas Xe.
5. Can we relate the number of valence electrons with the stability of the element?
Lewis Structures and Bond
Formation
2

206. 6
What Is It
Lewis Dot Symbols
Valence electrons of an atom are better represented with Lewis dot symbols. From the
previous module, the number of valence electrons of a main block element is usually
equivalent to its group number. For instance, Carbon is a Group 4(a) element, thus it has 4
valence electrons. The number of valence electrons of Fluorine is 7 because it is in Group7(a).
Magnesium has 2 valence electrons because it belongs to Group2(a).
Lewis dot symbol is very useful when learning about chemical bonding, and chemical
reactions. It consists of the symbol of an element and one dot for each valence electron in an
atom of the element. The dots are placed on the four sides of the symbol—top, bottom, left,
and right—and each side can accommodate up to two electrons. The choice on which sides
to place two electrons rather one electron is arbitrary since all four sides are equivalent. It is
recommended that we spread out the dots as much as possible. In general, we cannot write
simple Lewis dot symbols for the transition metals, lanthanides, and actinides because they
all have incompletely filled inner shells.
Source:https://www.google.com.ph/search?q=Lewis+dot+symbols+of+some+main+block+elements
Figure 1. Lewis Dot Symbols of Some Main Block Elements
The Octet Rule and Bond Formations
The octet rule refers to the tendency of atoms to gain, lose or share electrons to have
eight electrons in the valence shell or attain the same number of electrons as the noble gas
nearest to them in the periodic table. Noble gases are known as stable elements as evidenced
by their general lack of reactivity. All the noble gases except Helium have eight valence
electrons that is why many atoms undergoing reactions end up with eight valence electrons.
Octet rules does not generally apply to the d or f electrons. Only the s and p electrons
are involved in the octet rule, making it useful for the main group elements (elements not in
the transition metal or inner-transition metal blocks). Main group elements have an octet which
corresponds to an electron configuration ending with s2
p6
. However, there are some
exceptions to the octet rule: boron and aluminum readily form compounds in which they have
six valence electrons, rather than the usual eight predicted by the octet rule.
For instance, table salt or NaCl is the result of Na+
ions and Cl-
ions bonding together
in an ionic bond. If sodium metal and chlorine gas react under the right conditions, they will
form salt. Since sodium is a metal, it loses an electron, becomes positively charged and the
chlorine gains that electron and becomes negatively charged. The resulting salt is mostly
unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and

207. 7
chlorine that it is made of. (Note that each line that connects two atoms represents two bonded
electrons.)
Source:https://www.google.com.ph/search?q=Ionic+bond+formation+of+sodium+chloride
Figure 2. Ionic bond formation of Sodium Chloride
In covalently bonded atoms, sharing of electrons is being exhibited by all atoms
involved to satisfy the octet rule. For example, two hydrogen share each of their valence
electron with each other to have two electrons in their valence shells through a single bond.
As a result, each hydrogen atom achieves an electron configuration as helium, the nearest
noble gas which has only 2 valence electrons.
Source:https://www.google.com.ph/search?q=Covalent+bond+formation+of+H2+
Figure 3. Covalent bond formation of H2
Several atoms form multiple bonds to satisfy the octet. If two atoms share two pairs of
electrons, the covalent bond is called a double bond. This type of bond is much stronger and
shorter than single bonds. Carbon dioxide (CO2) is one of the compounds where double bonds
are found.
Source:https://www.google.com.ph/search?q=Electron+sharing+of+CO2+through+double+bonds
Figure 4. Electron sharing of CO2 through double bonds
A triple bond arises when two atoms share three pairs of electrons, as in the nitrogen
molecule (N2). This type of bond is much stronger and shorter than single and double bonds.
Source:https://www.google.com.ph/search?q=Electron+sharing+of+N2+through+a+triple+bond
Figure 5. Electron sharing of N2 through a triple bond

208. 8
Writing Lewis Structures
Lewis structures are diagrams that show the bonding between atoms of a molecule,
and the unbonded electrons that may exist in the molecule. A Lewis structure can be drawn
for any covalently-bonded molecule. Lines are drawn between electrons that are bonded to
one another.
Excess electrons that are not bonded or lone pairs are represented as pair of dots
and are placed next to the atoms on which they reside. According to Brown, these procedures
should be followed when writing Lewis structures:
Note that the central atom is the least electronegative atom in the molecule.
Electronegativity is the ability of an atom to attract toward itself the electrons. It increases
from left to right across a period and decreases down a group. Moreover, check the formal
charge of each atom. It is imperative to remember that atoms are more stable when they have
lesser charge or no charge at all. Negative formal charges should be assigned to more
electronegative atoms.
The overall formal charges of the atoms being bonded should be equal to the overall
charge of the compound.

209. 9
Example: Lewis Structure of CO2
Source: Tapayan, 2020

210. 10
What’s More
Activity 2: Predicting Chemical Products
Directions: The following pairs of atoms form ionic or covalent compounds when bonded.
Complete the table below with the needed details. Two answered rows serve as examples.
What I Have Learned
Activity 3: Spotting Mistakes
Directions: The following Lewis structures are incorrect. Explain what is wrong and give a correct
Lewis structure for the molecule. (Relative positions of atoms are shown correctly.)
Atoms
involved
Type of
Bond
Lewis dot
symbol of
each atom
Charge of each
ion after electron
transfer if ionic
bond is formed
Lewis dot
symbol of each
ion if ionic
bond is formed
Formula of the
Product
Na, Cl ionic Na+
Cl- Na+
NaCl
C, I covalent Not applicable Not applicable CI4
Mg, Cl
Ca, F
Na, O
Ca, N
S, Cl

211. 11
What’s In
The previous lesson helped us understand the compositions of molecules and their
bonds through the Lewis structure theory. However, Lewis structures do not demonstrate one
of the most important features of molecules—their overall shapes. The size and shapes of
molecules are defined by the distances and angles between the nuclei of the atoms involved.
In this lesson, we learn how to portrait a molecule by writing a two-dimensional structure for it
and translating it to a three-dimensional shape, and we examine the effects of molecular shape
on molecular polarity.
What Is It
Valence-Shell Electron-Pair Repulsion (VSEPR) Theory
The approach in predicting molecular geometry is called the Valence Shell Electron
Pair Repulsion Theory (VSEPR). This prediction is anchored from the assumption that all
electron pairs in the valence shell around a central atom repel one another. These valence
shell electron pairs are the ones involved in bonding and they want to stay apart from each
other as possible. The key ideas of the VSEPR theory are:
1. Electron pairs stay as far apart from each other as possible to minimize repulsions.
2. Molecular shape is determined by the number of bond pairs and lone pairs around
the central atom.
3. Treat multiple bonds as if they were single bonds (In making the prediction).
4. Lone pairs occupy more volume than bond pairs. Lone pair-lone pair repulsions
are greater than lone-pair-bond pair repulsions which in turn are greater than bond
pair-bond pair repulsions.
VSEPR focuses not only on electron pairs, but it also focusses on electron groups. An
electron group can be an electron pair, a lone pair, a single unpaired electron, a double bond
or a triple bond on the central atom. The actual determinants of molecular shape are classified
into two groups: the electron-group and the molecular geometry. Electron-group geometry
is determined by the number of electron groups or the number of atoms bonded to the central
atom. Molecular geometry, on the other hand, depends on not only on the number of electron
groups or the number of atoms bonded to the central atom, but also on the number of lone
pairs or unbonded pair of electrons on the central atom.
Molecular Geometry and
Polarity
Lesson
3

212. 12
Table 1. Summary of Molecular Geometry
Geometry Type # of Electron Pairs Ideal Bond Angle Examples
linear AB2 2 180° BeCl2
trigonal planar AB3 3 120° BF3
tetrahedral AB4 4 109.5° CH4
trigonal bipyramidal AB5 5 90°, 120° PCl5
octohedral AB6 6 90° SF6
bent AB2E 3 120° (119°) SO2
trigonal pyramidal AB3E 4 109.5° (107.5°) NH3
bent AB2E2 4 109.5° (104.5°) H2O
seesaw AB4E 5 180°,120° (173.1°,101.6°) SF4
T-shape AB3E2 5 90°,180° (87.5°,<180°) ClF3
linear AB2E3 5 180° XeF2
square pyramidal AB5E 6 90° (84.8°) BrF5
square planar AB4E2 6 90° XeF4
Source: https://www.thoughtco.com/introduction-to-molecular-geometry-603800
Figure 1. Molecular Geometry Examples
Source: https://courses.lumenlearning.com/boundless-chemistry/chapter/molecular-geometry/
Figure 2. VSEPR Geometries
Seesaw
Tetrahedral
Trigonal
Pyramidal
Bent
Trigonal
bipyramidal
T-shape
Linear
Octahedral
Square
Pyramid
Square
Planar
Seesaw

213. 13
Predicting Molecular Geometry
1.
2.
..
Molecular Polarity
Polarity is a physical property of compounds which relates other physical properties
such as solubility, boiling and melting points and intermolecular interactions between
molecules. It is a measure of how equally the electrons in a bond are distributed between the
two atoms involved in a covalent bond. Bond polarity increases with the increasing
electronegativity difference between the atoms in a molecule since there is a shift in electron
density towards the more electronegative atom. In a few cases, a symmetrical arrangement
gives rise to a non-polar molecule even though a molecule may have polar bonds.
The shift in electron density is symbolized by a crossed arrow ( ) with the arrow
pointing toward the direction of the shift. For instance, the shift in electron density points toward
fluorine in the polar HF molecule since it is more electronegative than hydrogen.

214. 14
Molecules whose atoms have equal or nearly equal electronegativities are nonpolar. A
molecule with polar bonds, but the molecular geometry is symmetrical allowing the bond
dipoles to cancel each other out like in the molecule C02 is also nonpolar.
Source:http://oer2go.org/mods/en-boundless-static/www.boundless.com/chemistry/textbooks/boundless-chemistry-
textbook/basic-concepts-of-chemical-bonding-9/electronegativity-74/bond-polarity-344-7948/images/molecular-dipole-
moment/index.html
Figure 3. Example of Non-polar molecule (CO2)
Polar molecules exhibit dipole moments while nonpolar molecules do not. In the
presence of an electric field, the positive end of the molecules positions itself towards the
negative plate. The molecular geometry determines whether the molecule is polar or not.
Source:https://int.search.tb.ask.com/search/AJimage.jhtml
Figure 4. Some Examples of Common Polar Compounds
15

215. 15
What’s More
Activity 1: Synthesizing concepts
Directions: Complete the table below with correct answers. The first one is done for you.
Covalent
Compound
Lewis Structure with
correct geometry
Geometry Name Polar or Nonpolar?
CO2 Linear Nonpolar
CH4
NH3
SF4
What I Have Learned
Activity 2: Reasoning Challenge
Directions: Answer the following questions in 2-3 sentences only.
1. VSEPR theory specifies “valence shell” electrons. Explain why these are the most critical
electrons for determining molecular shape?
2. Draw the Lewis structure of ozone, O3. Describe why ozone has a bent shape instead of a
linear shape.

216. 16
SUMMARY
• Atoms form bonds because they are trying to reach the most stable or lowest-energy
state that they can. In chemical bonding, only valence electrons, electrons located in
valence shell of the element, are involved.
• An ionic bond is formed when metals on the left side (Group 1(A) and Group 2(A) of
the periodic table and nonmetals on the right side (except noble gases, group 8A)
interact.
• Covalent bonds are formed when atoms reach stability by sharing electrons (rather
than fully gaining or losing them).
• These bonds mostly occur between nonmetals or between two of the same (or similar)
elements. One, two, or three pairs of electrons may be shared between atoms,
resulting in single, double, or triple bonds, respectively. The more electrons that are
shared between two atoms, the shorter and stronger their bond will be.
• Lewis dot symbol is very useful when learning about chemical bonding, and chemical
reactions. It consists of the symbol of an element and one dot for each valence electron
in an atom of the element.
• Valence electrons of an atom are better represented with Lewis dot symbols. From the
previous module, the number of valence electrons of a main block element is usually
equivalent to its group number.
• The octet rule refers to the tendency of atoms to gain, lose or share electrons to have
eight electrons in the valence shell or attain the same number of electrons as the noble
gas nearest to them in the periodic table.
• Lewis structures are diagrams that show the bonding between atoms of a molecule,
and the unbonded electrons that may exist in the molecule.
• Electronegativity is the ability of an atom to attract toward itself the electrons. It
increases from left to right across a period and decreases down a group.
• Molecular geometry relates to the three-dimensional arrangement of atoms in a
molecule. The size and shape of a molecule of a substance, together with the strength
and polarity of its bonds, affect the physical and chemical properties of molecules and
play an important role in their interactions especially in the life-sustaining processes in
cells.
• The approach in predicting molecular geometry is called the Valence Shell Electron
Pair Repulsion Theory (VSEPR).
• Electron-group geometry is determined by the number of electron groups or the
number of atoms bonded to the central atom. Molecular geometry, on the other hand,
depends on not only on the number of electron groups or the number of atoms bonded
to the central atom, but also on the number of lone pairs or unbonded pair of electrons
on the central atom.
• Polarity is a physical property of compounds which relates other physical properties
such as solubility, boiling and melting points and intermolecular interactions between
molecules.
• Molecules whose atoms have equal or nearly equal electronegativities are nonpolar. A
molecule with polar bonds, but the molecular geometry is symmetrical allowing the
bond dipoles to cancel each other out like in the molecule C02 is also nonpolar.

217. 17
Assessment: (Post-Test)
Multiple Choice. Answer the questions that follow. Choose the best answer
among the given choices for each item.
1. The number of lone pairs in CO2 molecule are ______________.
A. 1 C. 3
B. 2 D. 4
2. A molecule with trigonal planar geometry:
A. H2O C. BF3
B. CO2 D. CH4
3. In ammonia, the tetrahedral shape gets distorted because of the lone pair and becomes
__________.
A. trigonal pyramidal C. linear
B. T-shaped D. bent
4. The number of lone pairs of electrons around the central oxygen atom in In the Lewis
structure of the OF2 molecule is:
A. 1 C. 3
B. 2 D. 4
5. Which one of the formulas for ionic compounds below is incorrect?
A. SrCl2 C. Al3P2
B. AlCl3 D. Cs2S
6. Write the singly bonded Lewis dot structure for BF3. Which of the following statements
best describes this structure?
A. It obeys the octet rule on all atoms.
B. It has less than an octet on at least one atom.
C. It has a lone pair of electrons on the boron atom.
D. It has less than an octet of electrons on all atoms.
7. Which of the following elements can only form one bond in a Lewis structure?
A. N C. C
B. O D. H
8. Which of the following is a nonpolar covalent molecule?
A. NH3 C. HCl
B. H2O D. CCl4
9. Which of the following has the highest boiling point?
A. NaCl C. CH3CH2CH3
B. HBr D. CCl4
10. Which of the following molecules has a dipole moment?
A. CCl4 C. NH3
B. O2 D. CO2

218. 18
Key to Answers
Lesson 1
Activity 1
Lesson 2
Activity 2

219. 19
Lesson 3
Activity 1
Lesson 3
Activity 2

220. 20
References
Books
Brown, Theodore. Chemistry: The Central Science. New York: Pearson,2015.
Chang, R. and Goldsby, K. Chemistry. New York: McGraw-Hill
Education, 2010.
Silberberg, Martin. Principles of General Chemistry. Boston: McGraw- Hill Higher
Education, 2013.
Internet Sources
Common elements that form ionic bonds. Accessed October 31, 2020
https://www.google.com.ph/search?q=Common+elements+that+form+ionic+b
onds+&tbm=isch&ved=2ahUKEwidxam
Common Polar Compounds. Accessed November 2, 2020
https://int.search.tb.ask.com/search/AJimage.jhtml
Covalent bond formation of H2. Accessed November 1, 2020
https://www.google.com.ph/search?q=Covalent+bond+formation+of+H2
Electron Shells and Atom Stability. Accessed October 31, 2020.
https://www.google.com.ph/search?q=stable+and+unstable+atoms&tbm=isch
&ved=2ahUKEwiMr-
Electron sharing of CO2 through double bonds. Accessed November 1, 2020
https://www.google.com.ph/search?q=Electron+sharing+of+CO2+through+do
uble+bonds
Electron sharing of N2 through a triple bond. Accessed November 2, 2020
https://www.google.com.ph/search?q=Electron+sharing+of+N2+through+a+tri
ple+bond
“How to Draw a Lewis Structure”. Thoughtco. Accessed September 1, 2020.
https://www.thoughtco.com/how-to-draw-a-lewis-structure-603983
“Ionic and Covalent Bonds.”Chemistry LibreTexts. Accessed August 29, 2020.
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_M
odules (Organic_Chemistry)/Fundamentals/Ionic_and_Covalent_Bonds
Ionic bond formation of Sodium Chloride. Accessed October 31, 2020
https://www.google.com.ph/search?q=Ionic+bond+formation+of+sodium+chlo
ride
“Lewis Dot Symbols of Some Main Block Elements”. Accessed October 31, 2020
https://www.google.com.ph/search?q=Lewis+dot+symbols+of+some+main+bl
ock+elements

221. 21
“Molecular Geometry.” Boundless Chemistry. Accessed September 3, 2020.
https://courses.lumenlearning.com/boundless-chemistry/chapter/molecular-
geometry/
Molview. Accessed September 2020. molview.org
Molecular Geometry Examples. Accessed November 2, 2020
https://www.thoughtco.com/introduction-to-molecular-geometry-603800
“Molecular Polarity.” Chemistry LibreTexts. Accessed September 6, 2020.
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry
_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemi
stry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Molec
ular_Polarity
Non-polar molecule (CO2). Accessed November 2, 2020
http://oer2go.org/mods/en-boundless
static/www.boundless.com/chemistry/textbooks/boundless-chemistry-
Structures of Some Common Covalent Compounds. Accessed October 31, 2020
https://www.google.com.ph/search?q=structure+of+single%2C+double%2C+
and+triple+bonds+of+common+covalent+compounds
“The Octet Rule.” Chemistry LibreTexts. Accessed August 30, 2020.
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry
_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemi
stry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configuration
s/The_Octet_Rule
“Three-dimensional structure of the flavonoid isorhamnetin. Accessed October 31,
2020
https://www.google.com.ph/search?q=the+structure+of+flavonoid+isorhamnetin
VSEPR Geometries. Accessed November 2, 2020
https://courses.lumenlearning.com/boundless-chemistry/chapter/molecular-
geometry/
20

222. 22
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223. General Chemistry 1
Quarter 2 - Module 3 – Organic Compounds
Department of Education ● Republic of the Philippines
Senior High School

224. General Chemistry 1 – Grade 11
Alternative Delivery Mode
Quarter 2 - Module 3: Organic Compounds
First Edition, 2020
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225. General
Chemistry 1
Quarter 2 - Module 3 - Organic Compounds
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226. Table of Contents
What This Module is About...........................................................................................................i
What I Need to Know.....................................................................................................................ii
How to Learn from this Module ...................................................................................................ii
Icons of this Module......................................................................................................................iii
What I Know...................................................................................................................................iii
Second Quarter – Module 3
Lesson 1: Hydrocarbons and Functional Groups
What I Need to Know .................................................................................... 1
What’s New: ................................................................................................ 1
What Is It ....................................................................................................... 2
What’s More .................................................................................................. 6
What I Have Learned .................................................................................... 6
Lesson 2: Structural Isomerism and Organic Reactions
What I Need to Know .................................................................................... 7
What’s New ................................................................................................... 7
What Is It: ................................................................................................... 8
What’s More A............................................................................................. 13
What’s More B............................................................................................. 13
What I Have Learned .................................................................................. 14
What I Can Do (A)......................................................................................... 14
What I Can Do (B)......................................................................................... 15
Summary............................................................................................................................. 16
Assessment: (Post-Test) .................................................................................................... 17
Key to Answers................................................................................................................... 18
References.......................................................................................................................... 19

227. i
What This Module is About
Organic compounds are considered as carbon-containing chemical compounds
of living things because of their association with organisms and life processes. Their
structure, properties, reactions, compositions and preparation are the core topics of
organic chemistry.
About 200 years ago, it is believed that organic compounds needed a ‘life force’
to be produced. Other compounds like rock that were from nonliving things were
referred to as inorganic. The synthesis of urea (an organic compound) from amonium
cyanate (an inorganic compound) as an experiment of Friedrich Wöhler in 1828
dispelled the belief that organic compounds could only be formed by nature. The range
of application of organic compounds includes, but is not limited to petrochemicals,
food, pharmaceuticals, explosives, fragrances, paints and cosmetics.
In this module, we describe the structures, properties and reactions of
hydrocarbons and organic functional groups. Moreover, the preparation of selected
organic compounds is introduced for you to have a gist of the complexity of organic
synthesis.
What I Need to Know
At the end of this module, you should be able to:
1. Describe the different functional groups (STEM_GC11OCIIg-j-87);
2. Describe structural isomerism and give examples (STEM_GC11CBIId-g-89);
3. Describe some simple reactions of organic compounds: combustion of organic
fuels, addition, condensation, and saponification of fats (STEM_GC11CBIId-g-
90);
4. Describe the preparation of selected organic compounds (STEM_GC11CBIId-
g-97)

228. ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge.
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under-
standing of the concept.
What’s More These are follow-up activities that are in-
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show-
case your skills and knowledge gained, and
applied into real-life concerns and situations.

229. iii
What I Know
Pretest: MULTIPLE CHOICE:
Directions: Read and understand each item and choose the letter of the correct
answer.
1. Which element is present in all organic compounds?
A. Helium C. Carbon
B. Nitrogen D. Oxygen
2. What products are obtained when CH4(g) burns completely in an excess of oxygen?
A. CO2 C. Hydrogen gas
B. CO2 and H2O D. Explosion
3. A hydrocarbon molecule is saturated if the molecule contains ___________
A. Single covalent bonds, only C. A double covalent bond, only
B. A triple covalent bond D. Single and double covalent bonds
4. A functional group with a carbonyl group functionality is _________________.
A. Alcohol C. Ester
B. Amine D. Ketone
5. Choose the incorrect option regarding Isomerism:
A. They differ in both physical and chemical properties.
B. They have the different molecular formula.
C. Chain isomers differ in the arrangement of their skeleton.
D. They have the same molecular formula.
6. Which of the following is formed when an alcohol is dehydrated?
A. Aldehyde C. Alkene
B. Ketone D. Amine
7. The reaction of water with alkene to produce an alcohol is a/an ______ reaction.
A. Addition C. Combustion
B. Condensation D. Saponification
8. The organic starting materials for the preparation of an ester could be_________
A. a ketone and alcohol C. an acid and alcohol
B. water and oxygen D. alkane and aldehyde
9. Hydrolysis (saponification) of a fat would yield_____________.
A. water and alkene C. ethanol and acid
B. ketone and aldehyde D. glycerol and soap
10. This type of isomerism occurs when different compounds are formed due to the
different arrangements of their atoms leading to different functional groups.
A. chain C. positional
B. functional D. geometric

230. 1
What I Need to Know
Organic compounds are carbon-containing compounds that do not only pertain
to hydrocarbons but also compounds with any number of other elements, including
hydrogen (most compounds contain at least one carbon–hydrogen bond), oxygen,
nitrogen, phosphorus, halogens, silicon, and sulfur. In this lesson, you should be able
to familiarize and describe the structures and properties of hydrocarbons and organic
functional groups.
What’s New
Activity 1: Matching Type
Directions: Match column A with the corresponding item in column B. Write the letter
of your answer for each number.
1
Hydrocarbons and Functional
Groups
1. ketone a. carbonyl group
2. carboxylic acid b. hydroxyl group
3. alkane c. general formula R3N
4. alkyne d. triple bond
5. ester e. smell of fruits
6. alcohol f. fuels
7. amine g. vinegar
8. hydrocarbon h. carbon and hydrogen

231. 2
What Is It
From the discussion on the bonding models in the previous module, it can be
seen that carbon has a unique nature. Carbon completes its octet by sharing electrons
with other carbon atoms forming single, double, and triple bonds. It also readily forms
bonds with atoms of other elements like H, O, N, and the halogens. Carbon can form
millions of different compounds and can form more compounds than any other element
in the periodic table.
Organic Compounds: Hydrocarbons
A hydrocarbon is a compound composed of only carbon and hydrogen atoms
and is considered as one of the major groups of organic compounds. Based on
structure, hydrocarbons are divided into two main classes—aliphatic (those that do not
contain a benzene ring) and aromatic (those that contain a benzene ring).
Figure 1. Classification of Hydrocarbons
(https://www.tes.com/lessons/hUjQYBl3Z4qV_w/hydrocarbons)
a. Alkanes- are aliphatic, saturated hydrocarbons and has only single carbon-
carbon bonds in the molecule. They are referred to as saturated
hydrocarbons because they hold the maximum number of hydrogen atoms
that can bond to the carbon atoms present. Some alkanes are gases and
are used directly as fuels. Saturated hydrocarbons that forms a single ring
are called cycloalkanes. Alkanes are nonpolar, not very reactive, insoluble
in water and have little biological activity. They are all colorless and
odorless.
Figure 2. Some common alkanes and cycloalkanes
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/12-2-structures-and-names-of-alkanes/)
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/12-9-cycloalkanes/)

232. 3
a. Alkenes- also called as olefins. These are unsaturated hydrocarbons that
contain at least one carbon-carbon double bond. They are all colorless and
odorless in nature except ethene. They are insoluble in water due to their
nonpolar feature but are more reactive than alkanes.
Figure 3. Short chain alkenes
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/alkenes-structures-and-names/)
b. Alkynes- contain at least one carbon-carbon triple bond, hence
unsaturated. They have physical properties similar to alkanes and alkenes.
They dissolve in organic solvents, slightly soluble in polar solvents, and are
insoluble in water. Alkynes are more reactive and have slightly higher boiling
points compared to alkanes and alkenes.
Figure 4. Short chain alkynes
(https://www.toppr.com/content/story/amp/physical-properties-of-alkynes-37358/)
Organic Compounds: Functional Groups
A group of atoms that is mainly responsible for the chemical behavior of the
parent molecule is called functional group. Organic compounds may also be classified
according to the functional groups they contain. Compounds with the same functional
groups undergo like reactions.
a. Alcohols- (R-OH; R=hydrocarbon)
All alcohols contain the hydroxyl functional group, -OH. Ethyl alcohol, or
ethanol, which is produced biologically by the fermentation of sugar or starch,
is by far the best known. The alcohols are very weakly acidic and are soluble in
water because of their polar nature. Most alcohols are highly flammable.
Methanol is highly toxic.
Ethyne Propyne Butyne

233. 4
Figure 5. Some common alcohols
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/alcohols-nomenclature-and-classification/)
b. Ethers- (R-O-R’; R, R’=hydrocarbon)
Ethers contain the R-O-R’ linkage, where R and R’ are hydrocarbon.
They are slightly polar and slightly soluble in water, but they are extremely
flammable. They tend to slowly form explosive peroxides when left standing in
air. Diethyl ether, one of the common ethers, was used as an anesthetic for
many years but also known for its irritating effects on the respiratory system
and the incidence of postanesthetic vomiting and nausea.
Figure 6. Some common ethers
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/ethers/)
c. Aldehydes and Ketones – ( , )
The carbonyl group ( ) is the functional group of aldehydes and
ketones. At least one hydrogen atom is bonded to the carbon in the carbonyl
group in an aldehyde. The carbon atom in the carbonyl group is bonded to two
hydrocarbon groups in a ketone. Aldehydes and ketones are highly polar
molecules. The small aldehydes and ketones are soluble in water, but solubility
falls with chain length. They also have slightly higher boiling and melting points
than alkanes. Formaldehyde, the simplest aldehyde, is used in the laboratory
to preserve animal specimens. Acetone, the simplest ketone, is primarily used
as solvent for organic compounds and as nail polish remover.
Figure 7. Some common aldehydes and ketones
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/aldehydes-and-ketones-structure-and-names/)
Methanol Ethanol Propanol Phenol

234. 5
d. Carboxylic acids- ( )
The functional group in carboxylic acids is the carboxyl group,
These acids are weak in nature and are widely found in both plant and animal
kingdoms. Carboxylic acids are polar and are soluble in water. Acetic acid, one
of the common carboxylic acids, is also known as vinegar.
Figure 8. Some common carboxylic acids
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/carboxylic-acids-structures-and-names/)
e. Esters- ( )
Esters have the general formula R’-COOR, where R’ can be a
hydrocarbon group or Hydrogen and R is a hydrocarbon group. They are used
in the production of perfumes and as flavouring agents. The smell and flavour
of many fruits come from the presence of small quantities of esters. Oranges
contain octyl acetate (CH3COOCHCH3C6H13), and apples contain methyl
butyrate (CH3CH2CH2COOCH3). Esters are polar and water soluble.
Figure 9. Esters in some common fruits
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/esters-structures-and-names/)
f. Amines – ( )
Amines are organic bases having the general formula R3N, where R may
be Hydrogen or a hydrocarbon group. When amines are allowed to react with
acids, they form colourless and odourless salts. When all Rs are hydrogens,
the resulting compound is ammonia, NH3.
Figure 10. Some common amines
(https://courses.lumenlearning.com/suny-orgbiochemistry/chapter/amines-structures-and-names/)
Methylamine Diethylamine Purine Pyrimidine

235. 6
What’s More
Activity 2: Identification
Directions: Identify what functional group the structural compound represents. If the
compound represents two or more functional groups, list all groups being
represented. If the compound is a hydrocarbon, write the specific type of
hydrocarbon class where it belongs. Write your answer on your gen chem
journal.
1. ____________ 2. ____________ 3. _____________
4. _____________ 5. ____________ 6. ______________
7.______________ 8. ____________ 9. ______________
What I Have Learned
Activity 3: Take your pick!
Directions: Choose one functional group which you like most and fill out the table
below with the needed details. Write your answer on your Gen Chem
Journal.
The most
interesting
functional group.
Why you find it
more interesting
than other
Functional
group?
Give a sample
structure of
your chosen
functional
compound
Give its Special
Properties and
Common Uses
Name Products
or things in which
this functional group
is present.

236. 7
What’s In
In lesson 1, you have learned that organic compounds are carbon-containing
compounds which do not only pertain to hydrocarbons but also compounds with any
number of other elements. Moreover, you also examined the structures and properties
of hydrocarbons and some functional groups. In this lesson, you will learn the
structural isomerism, simple reactions and preparation of selected organic
compounds.
What’s New
Activity 1: Clones or isomers
Directions: Identify if the two structures are the same compounds or isomers. Write
“isomers” if the compounds have the same chemical formula but exactly
different structure or compound and write “clones” if the compounds are
exactly the same.
1.
2. .
3.
4.
Structural Isomerism and
Organic Reactions
2

237. 8
STRUCTURAL ISOMERISM
CHAIN POSITIONAL FUNCTIONAL METAMERISM TAUTOMERISM RING-CHAIN
What Is It
I. STRUCTURAL ISOMERISM
When two or more organic compounds have the same molecular formula but
different properties due to their difference in arrangement of atoms along the carbon,
they are called structural isomers. There are six forms of structural isomerism and
the most common are chain, positional and functional isomerism.
Figure 1. Forms of Structural Isomerism
(http://padakshep.org/otp/subjects/chemistry/organic-chemistry/structural-isomerism/)
a. Chain Isomerism
Chain isomers are organic compounds with the same formula, but the
arrangements of their carbon skeleton are different.
Figure 2. Example of chain isomerism
(http://padakshep.org/otp/subjects/chemistry/organic-
chemistry/structural-isomerism/)
b. Positional Isomerism
Positional isomerism occurs when the same functional groups are attached to
different positions on a carbon chain.
Figure 3. Examples of positional isomerism
(http://padakshep.org/otp/subjects/chemistry/organic-chemistry/structural-isomerism/)
FORMULA: C4H10
FORMULA: C4H8 FORMULA: C3H7OH

238. 9
c. Functional Isomerism
Functional isomerism occurs if different compounds are formed due to the
different arrangements of their atoms leading to different functional groups. As
functional groups are usually the reactive centre of a molecule that leads entirely
different properties.
Figure 4. Examples of positional isomerism
(http://padakshep.org/otp/subjects/chemistry/organic-chemistry/structural-isomerism/)
II. Simple Reactions of Organic Compounds
a. Combustion Reactions
When a substance reacts with oxygen gas, releasing energy in the form of light
and heat, a combustion reaction occurs. This type of reaction must involve O2 or
oxygen gas as one reactant. One of the most notable combustion reactions is the
combustion of organic fuels. Fuels rapidly react with oxygen to produce energy.
These fuels include: coal, high molecular weight hydrocarbons, methane, propane,
and butane.
Reaction 1. Complete combustion of butane
(https://www.clutchprep.com/chemistry/practice-problems/84902/calculate-the-enthalpy-of-combustion-of-butane-c4h10-g-for-
the-formation-of-h2o-)
b. Addition Reactions
Multiple bonds, such as a double or a triple bond, can be converted into other
functional groups using addition reactions. Other elements such as hydrogen,
halogens, compounds like water and functional groups such as the hydroxyl group
can be attached or added to one or both of the carbons involved in the multiple
bond.
(https://socratic.org/questions/how-can-alkenes-be-used-to-
make-ethanol)
FORMULA: C2H6O
FORMULA: C3H6O
FORMULA: C2H4O2

239. 10
(http://lucychemistry.blogspot.com/2013/05/38-
describe-addition-reaction-of.htm)l
(https://www.masterorganicchemistry.com/2013/05/24/alkyne-
reaction-patterns-the-carbocation-pathway/)
Reaction2. Addition reactions of alkenes and alkynes
c. Condensation Reactions
A condensation reaction takes place when two or more molecules combine to
form a larger and new molecule, with the simultaneous loss of a small molecule
such as water and a formation of a new bond.
(https://www.saddleback.edu/faculty/jzoval/chem108_lab/lab_6_ https://www.chegg.com/homework-help/questions-
carboxylic%20acids/lab_6_Pre_lab_carboxylic_acids_current.pdf) and-answers/reaction-based-nmr-provided-identify-
aldehyde-ketone-possible-list-aldehyde-ketone-list-1—
q27958045
Reactions 3. Examples of condensation reactions
d. Saponification Reactions
Saponification reaction takes place when a fat, oil or lipid is cleaved and
converted into soap and alcohol by the action of heat and with the presence of
water and a base. Fats are typically in the form of esters. When esters undergo
saponification, carboxylate (soap) and an alcohol (glycerol) functional groups are
produced.
(https://sites.google.com/site/chemistry_olp/formation-of-esters)
Loss of water
New oxygen-carbon bond is formed Water is lost as a product
ketone aldehyde

240. 11
Reaction 4. Saponification reaction of a fat
(https://nsb.wikidot.com/c-9-5-5-1)
III. Preparation of Selected Organic Compounds
Organic compounds are usually synthesized from other groups of organic
compounds. Two or more organic compounds react to form a new organic compound
with characteristic properties. Sometimes, a single type of organic compound
undergoes a certain reaction, producing a new compound as a product. Heat,
temperature and pressure, acids, bases and water are some of the essentials needed
to make a chemical reaction feasible. There are several ways to prepare a certain
organic compound; the examples given below are the most common and just few of
the many preparation routes.
1. Preparation of Alkanes
1a. Catalytic Cracking- large hydrocarbon molecules are broken or
fragmented into smaller and more useful bits of hydrocarbons using high
pressures and temperatures.
Reaction 5. Catalytic cracking of a large alkane
(https://getrevising.co.uk/revision-notes/uses-and-cracking-of-crude-oil)
1b. Alkanes from Unsaturated Hydrocarbon.
Alkane can be prepared from alkene and alkyne through addition of H2
gas or the process called hydrogenation. In this process, dihydrogen gas is
added to alkynes and alkenes in the presence of a catalyst - substance that
makes the reaction proceed faster.
Reaction 6. Hydrogenation of an alkyne to produce an alkane
https://study.com/academy/lesson/catalytic-hydrogenation-of-alkynes-mechanism-explanation.html
Heat, pressure

241. 12
2. Preparation of Alkenes
2a. Alkenes from the Dehydration of Alcohols
Alkenes are generally prepared through dehydration of an alcohol or
removal of water from an alcohol compound. Water is removed from the alcohol
compound and is freed as a product.
Reaction 7. Removal of water from an alcohol to produce an alkene
(http://www.mendelset.com/articles/687/dehydration_alcohols
3. Preparation of Alcohols
3a. Alcohols from the Hydration of Alkenes
Alcohols are usually obtained by the net addition of water across the
double bond of an alkene. This reaction uses an acid as a catalyst.
Reaction 8. Acid-catalysed hydration of an alkene to produce an alcohol
(https://socratic.org/questions/how-can-alkenes-be-used-to-make-ethanol)
4. Preparation of Esters
4a. Esters from the Reaction of Alcohol and Carboxylic acid
Esters are produced when carboxylic acids are heated with alcohols
in the presence of a catalyst (usually an acid) and this process is called
esterification.
Reaction 9. Esterification reaction of ethanoic acid and ethanol to produce an
ester named ethyl ethanoate
(http://www.passmyexams.co.uk/GCSE/chemistry/carboxylic-acids-reaction-alcohol.html)
1. Preparation of Carboxylic Acids
5a. Carboxylic acids from the Oxidation of Primary Alcohols
The oxidation of primary alcohols is a common method for the synthesis
of carboxylic acids which requires a strong oxidizing agent. Primary alcohols
are those alcohols where the carbon atom of the hydroxyl group (OH) is
attached to only one single hydrocarbon or R group like ethanol, propanol, etc.
acid
acid
Heat,acid

242. 13
Reaction 10. Oxidation of ethanol to ethanoic acid (carboxylic acid)
(https://www.chemistryscl.com/organic/oxidation-of-alcohols/index.php)
What’s More (A)
Activity 2.1: Structural Isomerism Identification
Directions: Identify what type of structural isomerism is being exhibited by each pair
of compounds. Choose from the three most common types of isomerism
(e.g. chain, position and functional isomerism).
1. 3.
2. 4.
What’s More (B)
Activity 2.2: Organic Reaction Identification
Directions: Identify what type of organic reaction is being represented by each item.
Choose from the basic types of organic reactions (e.g. combustion,
addition, condensation, and saponification reaction).
_____________1. Hydrocarbons reacts with oxygen gas producing water and carbon
dioxide as products.
_____________2. Reaction of water to an alkene leading to the removal of the double
bond and production of an alcohol.
Strong
oxidizing
agent
Heat
Ethanol, primary alcohol

243. 14
_____________3. A carboxylic acid and an alcohol react forming a new bond with the
simultaneous loss of water molecule, hence the formation of a new
compound.
_____________4. The production of soap from fats.
_____________5. The production of esters from alcohol and carboxylic acid.
What I Have Learned
Activity 3: Think Like a Chemist!
Directions: Level up your way of thinking and put yourself in a chemist’s perspective.
Answer and explain each item briefly and concisely.
1. Structural isomers have the same molecular formula but have different
properties. How is this possible?
2. How is positional and functional isomerism differ from each other?
3. Ethanol and dimethyl ether have the same molecular formula C2H6O.
Ethanol is liquid at room temperature while dimethyl ether is gas. How is this
possible?
4. Why short-chain alkanes like methane and butane serve as good fuels?
5. Does the compound below undergo saponification reaction? Why?
6. Describe how you will prepare an alcohol from an alkene.
7. Describe how you will prepare an alkene from an alcohol.
8. Describe how you will prepare an ester.
What I Can Do (A)
Activity 4.1: Meet My Isomers
Directions: Do a research about the structural isomers of the compound C4H10O.
Identify at least three structural isomers, provide their structures and their
individual properties, and uses. Write your answers in each cell of the
table below.

244. 15
STRUCTURAL ISOMERS OF C4H10O
Name of the isomer Structure of the isomer Properties and uses
1.
2.
3.
What I Can Do (B)
Activity 4.2: The Grandma Ester Corporation
Directions: Using the link: https://jameskennedymonash.wordpress.com/2013/
12/13/infographic-table-of-esters-and-their-smells/ to look for the
infographic of the ester compounds which is responsible for the smell of
different fruits, perfumes and flowers. The goal is to convince an investor
to invest in your company so that you can build a factory of an ester
compounds. For example, you could make hexyl pentanoate, and sell it
as perfume ingredient. Give at least five ester compounds and filled out
the table below as your guide.
Ester
Compound
Name
Structure
How to
prepare the
compound?
Properties Uses

245. 16
SUMMARY
• Alcohols contain the hydroxyl functional group, -OH.
• Alkanes are aliphatic, saturated hydrocarbons and has only single carbon-
carbon bonds in the molecule.
• Alkenes are also called olefins. These are unsaturated hydrocarbons that
contain at least one carbon-carbon double bond.
• Alkynes contain at least one carbon-carbon triple bond, hence unsaturated.
• Amines are organic bases having the general formula R3N, where R may be
Hydrogen or a hydrocarbon group.
• Carbon completes its octet by sharing electrons with other carbon atoms
forming single, double, and triple bonds. It readily forms bonds with atoms of
other elements like H, O, N, and the halogens.
• Carbonyl group is the functional group of aldehydes and ketones. At least one
hydrogen atom is bonded to the carbon in the carbonyl group in an aldehyde.
The carbon atom in the carbonyl group is bonded to two hydrocarbon groups in
a ketone.
• Carboxyl group is the functional group of carboxylic acids.
• Combustion reaction occurs when a substance reacts with oxygen gas,
releasing energy in the form of light and heat
• Condensation reaction takes place when two or more molecules combine to
form a larger and new molecule, with the simultaneous loss of a small molecule
such as water and a formation of a new bond.
• Esters have the general formula R’-COOR, where R’ can be a hydrocarbon
group or Hydrogen and R is a hydrocarbon group.
• Ethers contain the R-O-R’ linkage, where R and R’ are hydrocarbon.
• Functional group is a group of atoms that is mainly responsible for the
chemical behavior of the parent molecule.
• Hydrocarbon is a compound composed of only carbon and hydrogen atoms
and is considered as one of the major groups of organic compounds.
• Multiple bonds, such as a double or a triple bond, can be converted into other
functional groups using addition reactions.
• Organic compounds are usually synthesized from other groups of organic
compounds. Two or more organic compounds react to form a new organic
compound with characteristic properties. Sometimes, a single type of organic
compound undergoes a certain reaction, producing a new compound as a
product. Heat, temperature and pressure, acids, bases and water are some of
the essentials needed to make a chemical reaction feasible.
• Saponification reaction takes place when a fat, oil or lipid is cleaved and
converted into soap and alcohol by the action of heat and with the presence of
water and a base.
• Structural isomers when two or more organic compounds have the same
molecular formula but different properties due to their difference in arrangement
of atoms along the carbon.

246. 17
Assessment: (Post-Test)
Multiple Choice: Answer the questions that follow. Choose the best letter among
the given choices for each item.
1. Which of the following compounds is a functional group isomer of C2H5OH, ethanol
(ethyl alcohol)?
A. ethanal, CH3CHO C. acetic acid, CH3COOH
B. dimethyl ether, (CH3)2O D. diethyl ether, (C2H5)2O
2. Two isomeric forms of a saturated hydrocarbon _____________
A. have the same structure.
B. have different compositions of elements.
C. have the same molecular formula.
D. react vigorously with one another.
3. The reaction of water with alkene to produce an alcohol is a/an ______ reaction.
A. Saponification C. Combustion
B. Condensation D. Addition
4. The organic starting materials for the preparation of an ester could be_________
A. an acid and alcohol C. a ketone and alcohol
B. water and oxygen D. alkane and aldehyde
5. What is the product of the following reaction?
A. ester C. alcohol
B. aldehyde D. alkyne
6. The isomerism shown is _____________.
A. chain C. positional
B. functional D. geometric
7. The isomerism shown is _____________.
A. chain C. positional
B. functional D. geometric
8. Short chain alkanes are prepared through __________
A. Saponification C. Catalytic cracking
B. Hydration D. Oxidation
9. Saponification of fats needs a strong __________ to proceed to completion.
A. halogen C. acid
B. catalyst D. base
10. Fruits have pleasant odour because of the presence of _______________.
A. ethers C. ethyl
B. esters D. aldehyde

247. 18
Key to Answers

248. 19
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