The document discusses quantitative chemistry, focusing on percent composition, empirical and molecular formulas. It outlines key concepts such as calculating percentage composition from chemical formulas, determining empirical formulas from mass data, and the mole concept. Additionally, it includes lesson objectives and various worked examples and activities for practical application.

1. QUANTITATIVE
CHEMISTRY
Percent composition and chemical formulae

2. “
”
1. What is a chemical
formula?
2. Write the types of
chemical formula do you
know and give their names.
3. How do you determine a
molecular formula?
2

3. By the end of the lesson learners should be able to
1. Calculate percentage composition of a substance
2. Define empirical formula and molecular formula.
3. Differentiate an empirical formula from a molecular formula.
4. Determine an empirical formula from given percentages or mass.
5. Determine a molecular formula
3
Lesson objectives

4. Percent Composition of Compounds
 The percent composition of a compound is the percent by mass of each element in the compound.
 The composition of compounds is given by their formulas.
 For example, the formula of carbon dioxide (𝐶𝑂2), tells us it is composed of carbon and oxygen.
 Every molecule of 𝐶𝑂2 is composed of one atom of carbon and two atoms of oxygen.
Does knowing the number of each atom in the chemical formula give their mass proportion in the formula?
 For example, every sample of sulfur trioxide (𝑆𝑂3), is approximately 60% sulfur and 40% oxygen, by mass.
• This means that a 100 gram sample of sulfur trioxide contains 60 g of sulfur and 40 g of oxygen.
 Sodium chloride (NaCl) is 39.3% Na and 60.7% Cl, by mass.
• One hundred grams of NaCl contain 39.3 g of sodium and 60.7 g of chlorine.
In terms of real-life situations, why is knowing the percent composition of substances important?

5. Calculating the Percent Composition of Compounds
 The percent composition of a compound can be determined from its formula and the molar masses of the atoms that make it up.
 There are a few steps to calculating the percent composition of a compound (Let’s practice using H2O):
𝑯𝟐𝑶 𝑯 𝑶
𝒏 (𝒎𝒐𝒍)
𝑴 (𝒈. 𝒎𝒐𝒍−𝟏
)
𝒎 (𝒈)
1 mol of H2O contains 2 mol of hydrogen and 1 mol of oxygen.
𝒎 = 𝒏 × 𝑴
3. Determine the percent composition of water by comparing
the masses of hydrogen and oxygen atoms in water to the
mass of water.
% 𝐻 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻
𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻2𝑂
× 100
=
2
18
× 100
= 11.11%
% 𝑂 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑂
𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻2𝑂
× 100
=
16
18
× 100
= 88.89%
2. Calculate the mass of the compounds and the constituent atoms.
1 2 1
18 1 16
18 2 16
1. Assume you have 1 mole of the compound.
Therefore, every water molecule contains about 11% of hydrogen
and 89% of Oxygen.

6. Worked example
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Determine the percent composition of
TNT, 𝐶7𝐻5(𝑁𝑂2)3.
STEP 1: Assume you have 1 mol of TNT
Calculate the mass of TNT and the atoms in TNT
STEP 2:
O C H N TNT
𝒏 (𝒎𝒐𝒍) 1
𝑴 (𝒈. 𝒎𝒐𝒍−𝟏
)
𝒎 (𝒈)
6 7 5 3
16 12 1 14 227
96 84 5 42 227

7. O C H N TNT
𝒏 (𝒎𝒐𝒍) 1
𝑴 (𝒈. 𝒎𝒐𝒍−𝟏
)
𝒎 (𝒈)
6 7 5 3
16 12 1 14 227
96 84 5 42 227
STEP 3: Calculate the percentage of each atom in TNT
% 𝑂 =
96
227
× 100
= 42.29%
% 𝐶 =
84
227
× 100
= 37.00%
% 𝐻 =
5
227
× 100
= 2.20%
% 𝑁 =
42
227
× 100
= 18.50%

8. Activity
1. Calculate the percent composition of aluminum oxide, 𝐴𝑙2𝑂3.
2. What is the percent composition of diphosphorus pentoxide, 𝑃2𝑂5?
3. The molecular formula of the insecticide DDT is 𝐶14𝐻9𝐶𝑙5. Calculate the molar mass of the
compound and the percent by mass of each element.
4. Determine the percentage by mass of silver in silver sulphide (𝐴𝑔2𝑆).
5. How many grams of iron are in 5 kg of 𝐹𝑒2𝑂3?
6. How many grams of oxygen are in 200 g of 𝐴𝑙2 𝑆𝑂4 3?
7. Chalcopyrite (𝐶𝑢𝐹𝑒𝑆2) is a principal mineral of copper. Calculate the number of kilograms of Cu in
3.71 tons of chalcopyrite.

9. Empirical Formulas
List the main types of chemical formulae:
Structural formula
Molecular formula Empirical formula?
The empirical formula of a compound shows the simplest ratio of ions or atoms in the compound.
𝐇𝐎
Empirical formulae use the smallest possible set of subscript numbers.
Molecular formula Empirical formula
𝐶6𝐻6 𝐶𝐻
𝐶8𝐻18 𝐶4𝐻9
𝐶6𝐻12𝑂6 𝐶𝐻2𝑂
Sometimes the molecular and the empirical formulae of a molecular compound are the same, as they are for 𝐻2𝑂, 𝐶𝑂2, 𝐵𝐶𝑙3, 𝑒𝑡𝑐.
Therefore, the molecular formula of a compound can either be the same as or a whole-number multiple of its empirical formula.

10. Determination of Empirical Formulae from Composition Data
We can determine the empirical formula of a compound from its composition data.
We can determine the mole ratio of each element from the mass to determine the formula of the compound.
Consider a 2.476 g sample of an oxide of copper which is found to contain 2.199 g of copper.
Solution:
Cu O
𝑚 2.199 0.277
𝑀
𝑛 0.0346 0.0173
3. Determining the empirical formula:
Since we conventionally use whole number subscripts, we
must simplify the ratio.
• Divide each number by the smallest number in ratio.
𝐶𝑢2𝑂
1. Determine the mass of oxygen in the sample.
2. Determine the moles of Cu and O in the sample.
3. Determine the empirical formula.
1. 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑂 = 2.476 − 2.199 = 0.277 𝑔
2.
63.5 16
𝐶𝑢: 𝑂 = 0.0346 ∶ 0.0173
𝐶𝑢: 𝑂 ≈ 𝟐: 𝟏

11. Worked example
Determine the empirical formula of a compound of phosphorus and oxygen that contains 43.64%
phosphorus by mass.
1. Determine the percentage of oxygen in the compound.
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓𝑂 = 56.36%
2. Assumed you have 100 g of the compound.
In 100 g of the compound, there is 43.64 g of phosphorus and 56.36 g of oxygen.
3. Determine the moles of the atoms in the compound.
P O
𝑚 43.64 56.36
𝑀
𝑛 1.4077 3.5225
31 16
4. Determine the empirical formula.
𝑷𝟐𝑶𝟓
𝑃 ∶ 𝑂 = 1.4077 ∶ 3.5225
= 1 ∶ 2.5 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 1.4077
= 2 ∶ 5 (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑦 2)

12. Activity
1. A sample of a compound containing boron (B) and hydrogen (H) contains 6.444 g of B and 1.803 g of H.
Determine its empirical formula.
2. A 1.500 g sample of a compound contains 0.467 g sulfur and 1.033 g chlorine. What is the empirical
formula of the compound?
3. Determine the empirical formula of a compound that is 63.6% nitrogen and 36.4% oxygen.
4. The percentage of carbon, hydrogen and nitrogen in an unknown compound is found to be 23.30%,
4.85% and 40.78% respectively. Determine the empirical formula of the compound.
5. 2.40 g of element Z combines exactly with 1.60 g of oxygen to form a compound with the formula 𝑍𝑂2.
Determine the relative atomic mass of Z.
6. A sample of a brown-black compound is composed of 2.477 g manganese and 1.323 g oxygen. What is the
empirical formula of the compound?

13. Molecular formula and Empirical formula
For each compound:
• Calculate the molar mass of both the molecular formula and empirical formula,
• Then divide the molar mass of the molecular formula by the molar mass of the empirical formula.
As a general statement:
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝒏 × 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
 n represents any natural number.
 n is a factor by which the subscripts of the atoms in the empirical formula are multiplied.
What do you notice?

14. Exit ticket

15. QUANTITATIVE
CHEMISTRY
The mole concept

16. “
”
PDN
1. What do you think is a
molecular formula?
2.Where have you come
across the term mole?
3. What was the meaning of
this term?
16

17. By the end of the lesson learners should be able to
-Define the terms molecular formula and mole.
-Determine a molecular formula from percent composition.
-State the relationship between the number of elementary particles and the mole.
-Calculate the number of particles from the number of moles and the number of moles from the
number of particles.
17
Lesson objectives

18. How is the value of n determined?
𝑛 =
𝑀(𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎)
𝑀 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
Worked example
The empirical formula of a molecular compound of boron is 𝐵𝐻3.
The molar mass of a molecule of the boron compound is found to
be 28 𝑔. 𝑚𝑜𝑙−1
. Determine the molecular formula of the
compound.
Solution:
𝑀 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 28 𝑔. 𝑚𝑜𝑙−1
𝑀 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝑀 𝐵𝐻3
= 1 11 + 3 1 = 14 𝑔. 𝑚𝑜𝑙−1
𝑛 =
𝑀(𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎)
𝑀 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
=
28
14
= 2
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 2 × (𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎)
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 2(𝐵𝐻3) = 𝐵2𝐻6
Activity
1. A compound has been isolated and analyzed. The empirical
formula of the compound is 𝐶𝐻2. Its molecular mass is
42 𝑔. 𝑚𝑜𝑙−1
. What is the molecular formula of the
compound?
2. The empirical formula of colorless gas is 𝐶𝑂𝐶𝑙2. Its
molecular mass is 99 𝑔. 𝑚𝑜𝑙−1
. What is the molecular
formula of the compound?
3. compound used to whiten teeth has the empirical formula
of HO. The molecular mass of this compound is
34 𝑔. 𝑚𝑜𝑙−1
. What is its molecular formula?
4. Xylene, a solvent used in industry, has the empirical
formula of 𝐶4𝐻5. The molecular mass of xylene is
106 𝑔. 𝑚𝑜𝑙−1
. What is the molecular formula of xylene?

19. Introduction to quantitative chemistry
 What is matter?
 Quantity is defined as an amount, measure or number.
 What do you understand by the word “Quantity”?
Matter is anything that has mass and takes up space.
 At a minimum, matter requires at least one subatomic particle, although most matter consist of atoms.
 Matter can include molecules, compounds, and elements.
 Physical quantities used to measure matter:
Mass:
Volume:
Mole:
is a measure of the amount of matter in a substance or an object.
is a measure of the amount of space that a substance or an object takes up.
one mole of a substance contains 6.02 × 1023
of its elementary particles.
 The mole allows scientists to calculate the number of elementary entities (usually atoms or molecules ) in a
certain mass of a given substance.
Type of substance Elementary particles
Molecular substance molecules
Ionic substance formula units
Element atoms

20. The Mole Concept
 The mole (abbreviated mol) is the measure of quantity of the number of elementary particles
in a substance.
 1 mol of a substance contains 6.02 × 1023 of its elementary particles.
 How many elementary particles are in 0.5 mol of a substance?
 0.5 mol of a substance contains 3.01 × 1023 of its elementary particles.
 How many moles of a substance do 24.08 × 1023 elementary particles make up?
 24.08 × 1023 elementary particles make up 4 𝑚𝑜𝑙 of a substance.
𝒏 =
𝑵
𝑵𝑨
 The number of moles a particular number of elementary particles make up is calculated a follows:
𝒏 represents the number of moles of the substance.
𝑵 represents the number of elementary particles in the substance.
𝑵𝑨 is called the Avogadro’s number
 Avogadro’s number (NA) is the numerical
value of particles in a mole.
𝑁𝐴 = 6.02 × 1023
𝑚𝑜𝑙−1

21. Worked Example 1
How many molecules are present in 2.76 𝑚𝑜𝑙 of
𝑤𝑎𝑡𝑒𝑟? How many atoms are in this?
𝒏 =
𝑵
𝑵𝑨
𝑵 = 𝒏 × 𝑵𝑨
𝑵 = 𝟐. 𝟕𝟔 𝒎𝒐𝒍 × 6.02 × 1023
𝑚𝑜𝑙−1
𝑁 = 1,66 × 1024
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
To determine the total number of atoms, we have to
note that there are 3 atoms in a single molecule of
water.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 = 3 × 1.66 × 1024
= 4,98 × 1024
EXIT TICKET
1. How many molecules are present in 4.61 × 10−2 mol of 𝑂2?
2. How many moles of a substance do 3.35 × 1022 molecules
make?

22. QUANTITATIVE
CHEMISTRY
Relationship between the mole concept

23. “
”
PDN
1. Is there a relationship
between the mass of a
substance and number of
moles?
2. If yes, what is the
relationship and if no, give a
reason.
23

24. By the end of the lesson learners should be able to
-Define the term molar mass.
-Calculate the molar mass of any given substance.
- State temperature and pressure at STP.
-Calculate number of moles from given mass or volume at STP and vice versa.
24
Lesson objectives

25.  Molar mass of a specific substance is the mass in grams of one mole of that substance.
 Molar mass is represented by the symbol 𝑴.
 The unit for molar mass is 𝒈. 𝒎𝒐𝒍−𝟏
.
 The molar mass of water is 𝟏𝟖 𝒈. 𝒎𝒐𝒍−𝟏
.
Worked example
Determine the molar mass of 𝑆𝑂3.
Solution:
In 𝑆𝑂3, there is 1 Sulphur atom and 3 oxygen atoms.
To calculate the molar mass, we will add the molar masses of the
atoms in the formula.
𝑀 𝑆𝑂3 = 1 32 + 3 16
= 80 𝑔. 𝑚𝑜𝑙−1
 Note: the value of the molar mass of a substance is equal to the value of the formula mass of its elementary particle.
Activity
Which of the following compounds have the greatest
molar mass?
A. The herbicide paraquat (𝐶12𝐻14𝑁2𝐶𝑙2)
B. Urea (𝐶𝑂 𝑁𝐻2 2)
C. The anaesthetic haloethane (𝐶2𝐻𝐵𝑟𝐶𝑙𝐹3)
D. Caffeine (𝐶8𝐻10𝑁4𝑂2)
E. A typical soap, 𝐶17𝐻35𝐶𝑂2𝑁𝑎

26. The relationship between the number of moles in a particular mass of a substance is as follows:
𝒏 =
𝒎
𝑴
𝒏 represents number of moles of the substance (in mol)
𝒎 represents mass of the substance (in grams)
𝑴 is the molar mass of the substance (in 𝑔. 𝑚𝑜𝑙−1
)
Worked example
1. Calculate the number of moles in 50 g of oxygen gas (𝑂2)
2. Calculate the mass of 3.5 moles of 𝐶𝑂2.
Solution:
1. 𝑀 𝑂2 = 2 16
= 32 𝑔. 𝑚𝑜𝑙−1
𝑚 = 50 𝑔
𝑛 =
𝑚
𝑀
=
50
32
= 1.56 𝑚𝑜𝑙
Activity
1. Is 6.07 g of 𝐶𝐻4 equal, greater or less than 1 mole of 𝐶𝐻4?
2. Calculate the number of moles of chloroform (𝐶𝐻𝐶𝑙3 ) in
198 g of chloroform.
3. Calculate the molar mass of a compound if 0.372 mole of it
has a mass of 152 g.
4. How many molecules of ethane (𝐶2𝐻6) are present in 0.334
g of 𝐶2𝐻6?
5. Calculate the number of C, H, and O atoms in 1.50 g of
glucose (𝐶6𝐻12𝑂6), a sugar.
2. 𝑀 𝐶𝑂2 = 1 12 + 2 16
= 44 𝑔. 𝑚𝑜𝑙−1
𝑛 = 3.5 𝑚𝑜𝑙
𝑚 = 𝑛 × 𝑀
= 3.5 × 44
= 154 𝑔

27. Moles, mass and number of particles
 The molar mass of a substance is used to convert between grams of a substance and moles of a substance.
 How do we convert between moles and mass of a substance?
 How do we convert between number of particles in a substance and mass of a substance?
 If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass.
𝒏 =
𝒎
𝑴
𝒏 =
𝑵
𝑵𝑨
then 𝒏 =
𝒎
𝑴
OR
𝑵
𝑵𝑨
=
𝒎
𝑴
leading to 𝒎 =
𝑵
𝑵𝑨
× 𝑴 OR 𝑵 =
𝒎
𝑴
× 𝑵𝑨

28.  At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 𝑙 (or 22.4 𝑑𝑚3
).
 The volume occupied by 1 mole of gas (22.4 L) is called the molar volume.
Molar Volume
 Standard temperature and pressure are respectively 0C (273.15 K) and 1 atm (101.3 kPa).
 The symbol for molar volume is 𝑉
𝑚.
 Molar volume has units of 𝑑𝑚3
. 𝑚𝑜𝑙−1
.
 The moles in a particular volume of gas at STP can be calculated as follows:
𝒏 =
𝑽
𝑽𝒎
𝒏 represents the moles of the gas at STP (in moles).
𝑽 represents the volume of the gas at STP (𝑑𝑚3
).
𝑽𝒎 represents the molar Volume at STP (𝑑𝑚3
. 𝑚𝑜𝑙−1
).

29. Worked Example
A sample of methane, CH4, occupies 4.50 𝑙 at
STP. How many moles of methane are present?
Solution
𝑉 = 4.5 𝑑𝑚3
𝑉
𝑚 = 22.4 𝑑𝑚3
. 𝑚𝑜𝑙−1
𝑛 =
𝑉
𝑉
𝑚
=
4.5
22.4
= 0.20 𝑚𝑜𝑙
EXIT TICKET
1. How many moles of gas occupy 50 𝑑𝑚3
at STP?
2. Determine the volume occupied by 0.85 mol of gas at STP.
3. 1.96 g of an unknown gas occupies 1000 𝑐𝑚3
at STP. What
is the molar mass of the gas?
4. What is the mass of 3.36 𝑑𝑚3
of ozone gas (𝑂3) at STP?
5. How many molecules of propane gas (𝐶3𝐻8) occupy 500 𝑚𝑙
at 0 ℃ and 101.3 𝑘𝑃𝑎?

30. Calculating moles of atoms in a compound:
1. Calculate the number of nitrogen and oxygen atoms in 1 mole of 𝑁2𝑂5.
2. Convert the number of nitrogen and oxygen atoms to moles of the atoms.
3. How many moles of N and O are in 1 mole of 𝑁2𝑂5
Oxygen atoms:
𝑁 = 3.01 × 1024
𝑛 =
𝑁
𝑁𝐴
=
3.01×1024
6.02×1023
= 5 𝑚𝑜𝑙
Nitrogen atoms:
𝑁 = 12.04 × 1023
𝑛 =
𝑁
𝑁𝐴
=
12.04×1023
6.02×1023
= 2 𝑚𝑜𝑙
There are 2 moles of N and 5 moles of O in 1 mole of 𝑵𝟐𝑶𝟓.
Activity
Determine the number of moles of constituent atoms in
1 mole of the following compounds.
1. 𝐻2𝑂
2. 𝐻𝑁𝑂3
3. 𝐻2𝐶2𝑂4
4. 𝐴𝑙 𝑀𝑛𝑂4 3
5. 𝑁𝐻4 2𝐻2𝑃𝑂4
 There are 6.02 × 1023
molecules of 𝑁2𝑂5 in 1 mol of 𝑁2𝑂5.
 In each molecule, there are 2 nitrogen atoms and 5 oxygen atoms.
 Therefore, in 1 mole of 𝑁2𝑂5, there are 1.204 × 1024
nitrogen atoms, and 3.01 × 1024
oxygen atoms.
Note: the subscript numbers in the formula are equal to the
number of moles of each atom in one mole of the compound.

31. QUANTITATIVE
CHEMISTRY
The mole concept and solutions

32. “
”
PDN
1. What is a solution?
2. What is the difference
between a concentrated
solution and a diluted
solution?
32

33. By the end of the lesson learners should be able to
-Define the term concentration.
-Calculate concentration of a solution given mass of solute and volume of solvent .
- Calculate number of moles of the solute in a given solution
-
33
Lesson objectives

34. Solution:
Solutions
a homogenous mixture of solute and solvent
Solute: the substance that is dissolved in the solution
Solvent: the substance in which another substance is dissolved, forming a solution.
The solvent is the major component of the solution and the solute is the minor component.
Forms of liquid solutions:
 Aqueous solution:
 Non-aqueous solution:
the solution in which water is the solvent.
 E.g. salt in water, sugar in water and copper sulfate in water.
 E.g., iodine in carbon tetrachloride, sulphur in carbon disulfide, phosphorus in ethyl alcohol.
the solution in which a solute is dissolved in a solvent other than water.
• When a gaseous or solid material dissolves in a liquid, the gas or solid material is called the solute.
• When two liquids dissolve in each other, the major component is called the solvent and the minor component is called
the solute.
Types of Solutions
1. gaseous solutions (air, natural gas, LPG, etc)
2. liquid solutions (salt water, tea, etc)
3. solid solutions (brass, bronze, steel, etc)
Why is a solution is considered a mixture?
the ratio of solute to solvent can vary.
Why is a solution is considered homogeneous?
it has the same composition throughout.

35. Concentration of Solutions
The concentration of a solution expresses the amount of solute present in a solution.
Concentration: the amount of solute per unit volume of solution.
𝑐 =
𝑛
𝑉
𝑛 represents the moles of the solute (in mol)
𝑉represents the volume of the solution (in 𝑑𝑚3
)
𝐶 represents the concentration of solution (in 𝑚𝑜𝑙. 𝑑𝑚−3
)
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Dilute solution:
Concentrated solution: contains a relatively large amount of solute dissolved in a given amount of solution.
contains a relatively small amount of solute dissolved in a given amount of solution.
NOTE: Acids and bases are purchased from suppliers as concentrated solutions.
 They are then diluted with water to form dilute solutions.

36. Worked example
Calculate the concentration of the solution formed when
4 moles of glucose are dissolved in 5 𝑑𝑚3
of water.
Solution:
𝑛 = 4 𝑚𝑜𝑙
𝑉 = 5 𝑑𝑚3
𝐶 =?
𝐶 =
𝑛
𝑉
=
4 𝑚𝑜𝑙
5 𝑑𝑚3
= 0.8 𝑚𝑜𝑙. 𝑑𝑚−3
Activity
1. Determine the concentration of 3.0 moles of nitric acid in 4.0 𝑑𝑚3
of solution.
2. Calculate the number of moles of hydrochloric acid present in 0.80 𝑑𝑚3
of a solution
with a concentration of 0.40 𝑚𝑜𝑙. 𝑑𝑚−3
.
3. Calculate the concentration of a solution that contains 50.0 g of NaOH in 850 𝑐𝑚3
of
solution.
4. What volume of a 1.50 𝑚𝑜𝑙. 𝑑𝑚−3
copper (II) sulphate solution contains 35.0 g of
𝐶𝑢𝑆𝑂4?
5. How many grams of NaCl are in 2.53 𝑑𝑚3
of a 0.750 𝑚𝑜𝑙. 𝑑𝑚−3
solution of NaCl?
Combining the equations 𝑪 =
𝒏
𝑽
and 𝒏 =
𝒎
𝑴
𝑔𝑖𝑣𝑒𝑠
𝑪 =
𝒎
𝑴𝑽

37. Dilution of Solutions
 Many chemical reagents are supplied in solutions of much higher concentration than their use requires.
 The concentrated solutions need to be diluted with solvent to prepare solutions of lower concentration.
Concentrated solution:
𝑛1 = 𝐶1 × 𝑉1
 How can a solution be diluted? Adding MORE SOLVENT dilutes a solution
 State how adding more solvent affects the volume of solution, moles of solute and concentration of solution.
• the volume of the solution increases without changing the number of moles of solute.
• the concentration of the solution decreases.
Diluted solution:
𝑛2 = 𝐶2 × 𝑉2
𝑉1 and 𝐶1 are the volume and concentration of the concentrated solution
𝑉2 and 𝐶2 are the volume and concentration of the diluted solution
Since the number of moles of solute remains constant with dilution:
𝑛1 = 𝑛2
This yields the dilution equation:
𝐶1 × 𝑉1 = 𝐶2 × 𝑉2
If any three of the four terms are known, the fourth can be calculated.
Note that C and V can be in any units provided that the same units are used
on both sides of the equation.

38. Worked example
Hydrochloric acid is obtained commercially at a concentration of
12.5 𝑚𝑜𝑙. 𝑑𝑚−3
. How many milliliters of the 12.1 𝑚𝑜𝑙. 𝑑𝑚−3
solution of HCl must be used to prepare 2000 𝑚𝑙 of
0.500 𝑚𝑜𝑙. 𝑑𝑚−3
solution of HCl?
Solution:
𝑉2 = 2000 𝑚𝑙
𝐶2 = 0.5 𝑚𝑜𝑙. 𝑑𝑚−3
𝐶1 = 12.5 𝑚𝑜𝑙. 𝑑𝑚−3
𝑉1 =?
𝐶1𝑉1 = 𝐶2𝑉2
(12.5)𝑉1 = (0.5)(2000)
12.5𝑉1 = 1000
𝑉1 = 80 𝑚𝑙
Activity
1. Calculate the volume to which 20.0 𝑐𝑚3
of 7.63 𝑚𝑜𝑙. 𝑑𝑚−3
hydrochloric acid must be
diluted to produce a solution with a concentration of exactly 5.00 𝑚𝑜𝑙. 𝑑𝑚−3
.
2. How would you prepare 1.2 𝑑𝑚3
of a 0.40 𝑚𝑜𝑙. 𝑑𝑚–3
solution of hydrochloric acid
starting from a 2.0 𝑚𝑜𝑙. 𝑑𝑚−3
solution?
3. 9.50 𝑐𝑚3
of 0.500 𝑚𝑜𝑙. 𝑑𝑚–3
NaCl is added to 500 𝑐𝑚3
of a 1.00 𝑚𝑜𝑙. 𝑑𝑚–3
𝑁𝑎2𝐶𝑂3
solution. Calculate the concentration of 𝑁𝑎+
ions in the final solution.

39. Chemical Equations
Chemical reaction: a process in which one or more substances are changed into one or more new substances.
Chemical equation: a standard way to represent and communicate chemical reactions.
Types of chemical species in chemical reactions:
Reactants: the starting materials in a chemical reaction.
Products: the substances formed as a result of a chemical reaction.
A chemical equation:
• shows the formulae (molecular formulae or formula units) of all the reactants and all the products.
• also gives the number of each species that are required/produced from a complete reaction.

40. EXIT TICKET
Activity
1. Write the formulae of the following
common compounds:
a) Carbon monoxide
b) Sulphur dioxide
c) Hydrochloric acid
d) Copper (II) sulphate
e) Sulphuric acid
f) Aluminium oxide
g) Magnesium nitrate
h) Iron (III) sulphide
i) Calcium phosphate
j) Ammonium carbonate
k) Nitrous acid
l) Potassium acetate
m) Calcium chlorate
n) Chromium (VI) oxalate
o) Manganese (IV) nitride
2. Write balanced equations for the following reactions.
a) Iron and bromine reacting to form iron (III) bromide.
b) Calcium carbonate + Hydrochloric acid → Calcium chloride + Carbon dioxide + Water
c) Sodium oxide reacting with sulphuric acid to form sodium sulphate and water.
d) Iron (II) chloride reacting with chlorine to form iron (III) chloride.
e) Sulfur dioxide reacting with hydrogen sulfide to form sulfur and water.
f) Silver reacting with nitric acid to form silver nitrate, nitrogen dioxide and water.
g) Manganese (IV) oxide reacting with hydrochloric acid to form manganese (II)
chloride, chlorine and water.
h) Ammonia reacting with oxygen to form nitrogen monoxide and water.
i) Lead(IV) oxide and carbon monoxide forming lead metal and carbon dioxide.
j) Ethanol burning in air to form carbon dioxide and water.

41. QUANTITATIVE
CHEMISTRY
Interpretation of chemical equations

42. “
”
1. What is a chemical
equation?
2. Are chemical equation
important? If yes, given a
reason for your answer and
if no also given a reason for
your answer
42

43. By the end of the lesson learners should be able to
- State the law of conservation of mass.
- Interpret a chemical equation in terms of moles, mass and volume.
- Use chemical equations to calculate unknown quantities.
-
43
Lesson objectives

44. Stoichiometry
 Stoichiometry is the study of quantitative aspects of chemical equations.
 The amounts of substances in a balanced equation are known as the stoichiometry of the reaction.
• For example, consider the industrial synthesis of ammonia from hydrogen and nitrogen:
 The coefficients in balanced equations indicate the number of each species involved in a single chemical reaction.
• How many of each species are required/produced if 6.02 × 1023
reactions are to take place?
• Determine the number of moles of each species that are required/produced in these reactions:
• Therefore, the stoichiometric coefficients in a chemical
equation can be interpreted as the ratio of the number
of moles of substances in a reaction.
• In the ammonia synthesis reaction above, one molecule of 𝑁2 reacts with three molecules of 𝐻2 to form two molecules of 𝑁𝐻3.

45. Quantitative Proof of the Law of Conservation of Mass
𝑪𝟑𝑯𝟖(𝒈) + 𝟓 𝑶𝟐(𝒈) → 𝟑 𝑪𝑶𝟐(𝒈) + 𝟒 𝑯𝟐𝑶(𝒈)
Consider the combustion of propane:
1 mol 5 mol 3 mol 4 mol
Calculate the mass of each species in the reaction:
𝑪𝟑𝑯𝟖 𝑶𝟐 𝑪𝑶𝟐 𝑯𝟐𝑶
𝒏 (𝒎𝒐𝒍)
𝑴(𝒈. 𝒎𝒐𝒍−𝟏
)
𝒎 (𝒈)
3 4
44 32 44 18
44 160 132 72
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 = 𝑚𝐶3𝐻8
+ 𝑚𝑂2
= 44 + 160
= 204 𝑔
𝑚𝑎𝑠𝑠 𝑜𝑓𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 = 𝑚𝐶𝑂2
+ 𝑚𝐻2𝑂
= 132 + 72
= 204 𝑔
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
Therefore, mass is conserved
State the law of conservation of mass in words:
1 5
Matter cannot be created or destroyed during a chemical or physical change.

46. Stoichiometry and the Mole Concept
1. the amount of one reactant needed to completely react with a known amount of another reactant.
2. the amount of product that will be formed from specific amounts of reactants.
3. the amount of reactants that must be used to obtain a specific amount of product.
The knowledge of stoichiometry is essential in interpreting a reaction quantitatively.
The coefficients in a balanced equation can be interpreted as mole ratios of the respective substances.
Consider the reaction between methane gas and chlorine gas:
𝐶𝐻4(𝑔) + 𝟒 𝐶𝑙2(𝑔) → 𝐶𝐶𝑙4(𝑙) + 𝟒 𝐻𝐶𝑙(𝑔)
In terms of moles of each substance:
𝟏 𝒎𝒐𝒍𝒆 𝐶𝐻4(𝑔) + 𝟒 𝒎𝒐𝒍𝒆𝒔 𝐶𝑙2(𝑔) → 𝟏 𝒎𝒐𝒍𝒆 𝐶𝐶𝑙4(𝑙) + 𝟒 𝒎𝒐𝒍𝒆𝒔 𝐻𝐶𝑙(𝑔)
Interpreting a balanced equation in terms of moles allows the possibility of determining

47. Application of Stoichiometry
 ONLY MOLES are used in STOICHIOMETRY.
Worked example
1. When butane is burnt in excess air, the following reaction takes
place:
𝟐 𝐶4𝐻10 𝑔 + 𝟏𝟑 𝑂2 𝑔 → 8 𝐶𝑂2 𝑔 + 10 𝐻2𝑂 𝑙
Calculate the number moles of oxygen that are required to completely
react 5 moles of butane.
Solution:
The ratio of 𝐶4𝐻10 to 𝑂2 is
𝟐 ∶ 𝟏𝟑
For 5 mol:
Determine the value of x 2𝑥 = 65
𝑥 = 32.5 𝑚𝑜𝑙
32.5 moles of oxygen are required to completely react 5 moles of butane.
𝟓 ∶ 𝑥
2. When magnesium is added to aqueous silver nitrate, the
following reaction takes place
𝑀𝑔(𝑠) + 2𝐴𝑔𝑁𝑂3(𝑎𝑞) → 2𝐴𝑔(𝑠) + 𝑀𝑔 𝑁𝑂3 2(𝑎𝑞)
What mass of silver is formed when 3.6 g of magnesium is added to
an excess of aqueous silver nitrate?
Solution:
We must first calculate the moles of magnesium.
𝑛𝑀𝑔 =
𝑛
𝑀
=
3.6
24
= 0.15 𝑚𝑜𝑙
𝑛𝑀𝑔 ∶ 𝑛𝐴𝑔 = 1 ∶ 2
0.15 ∶ 𝑥
𝑥 = 0.3 𝑚𝑜𝑙
0.3 mol of silver is formed form 0.15 mol of magnesium
Determining the mass of silver:
𝑀 = 108 𝑔. 𝑚𝑜𝑙−1
𝑛 = 0.3 𝑚𝑜𝑙
Therefore, 32.4 g of silver is formed from 3.6 g of magnesium
𝑚 = 𝑛 × 𝑀
= 0.3 × 108
= 32.4 𝑔

48. EXIT TICKET
1. When heated potassium chlorate decomposes to form potassium chloride and oxygen. The unbalanced
equation is:
𝐾𝐶𝑙𝑂3 𝑠 → 𝐾𝐶𝑙 𝑠 + 𝑂2(𝑔)
a) Balance the equation.
b) Calculate how many moles of 𝐾𝐶𝑙𝑂3 are needed to produce 0.60 moles of oxygen.
c) What mass of 𝐾𝐶𝑙𝑂3 is needed to produce 0.02 moles of 𝐾𝐶𝑙?
2. When heated sodium hydrogen carbonate decomposes to form sodium carbonate, carbon dioxide and water.
a) Write a balanced chemical equation for this decomposition reaction.
b) What mass of sodium hydrogen carbonate must be heated to give 8.80 g of carbon dioxide?
3. Calculate the mass of 𝑂2 required for the complete combustion of 0.25 mol of propane gas, 𝐶3𝐻8 (g).
4. How many moles of 𝐶𝑂2 are produced in the complete combustion of 0.75 mole of hexane, 𝐶6𝐻14?
5. At high temperatures, aluminum reacts with oxygen to form aluminum oxide. The balanced equation is:
4𝐴𝑙 𝑠 + 3𝑂2 𝑔 → 2𝐴𝑙2𝑂3 𝑠
a) How many moles of 𝑂2 are consumed for each 0.45 mole of Al?
b) How many moles of 𝐴𝑙2𝑂3 will be obtained when 0.45 mole Al is consumed?

49. QUANTITATIVE
CHEMISTRY
The limiting reagent

50. “
”
1. What stops you from
buying a car?
2. Write down what you
think is meant by the term
limiting reagent
50

51. By the end of the lesson learners should be able to
- Define the term limiting reagent.
-Determine a limiting reagent and an excess reagent in a chemical reaction.
-Use the limiting reagent to calculate the amount of excess reagent used or unreacted and
quantities of products formed.
-Use the limiting reagent to determine the percentage yield
-
51
Lesson objectives

52. Limiting Reagents
 When a chemist carries out a reaction, the reactants are usually not present in exact stoichiometric amounts.
 Stoichiometric amounts mean that the reagents are in the proportions indicated by the balanced equation.
The goal of a reaction is to produce the maximum quantity of a useful substance from the starting materials.
Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted to
the desired product.
Consequently, some reactant will be left over at the end of the reaction.
The reactant used up first in a reaction is called the limiting reagent (or reactant)
When the limiting reactant is used up, no more product can be formed.
Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent.
LIMITING REAGENTS determine the AMOUNT OF PRODUCT formed.
PHeT Simulation: Limiting reagents

53. In the reaction of sodium hydroxide, NaOH, with hydrochloric acid, HCl(aq), the balanced equation shows that
each mole of NaOH reacts with the exact same number of moles of HCl(aq):
𝑁𝑎𝑂𝐻(𝑎𝑞) + 𝐻𝐶𝑙 𝑎𝑞 → 𝑁𝑎𝐶𝑙 𝑎𝑞 + 𝐻2𝑂(𝑙)
 NaOH and HCl react in a 1 to 1 mole ratio.
If NaOH and HCl(aq) are combined in any ratio other than a 1 to 1 mole ratio, one reactant will be in excess.
Whenever starting amounts of the reactants are given, first determine which reactant is the limiting reactant before
calculating amounts of product.

54. Worked example
Consider the reaction:
𝐻2𝑂2(𝑎𝑞) + 2𝐾𝐼(𝑎𝑞) + 𝐻2𝑆𝑂4(𝑎𝑞) → 𝐼2(𝑠) + 𝐾2𝑆𝑂4(𝑎𝑞) + 2𝐻2𝑂(𝑙)
What mass of iodine is produced when 100.00 g of KI is added to a solution containing 12.00 g of 𝐻2𝑂2 and 50.00 g 𝐻2𝑆𝑂4?
Solution:
The mole ratio from the equation is
𝐻2𝑂2 ∶ 𝐾𝐼 ∶ 𝐻2𝑆𝑂4
1 ∶ 2 ∶ 1
The actual mole ratio of reagents present is
12.00
34
:
100.00
166
∶
50.00
98
= 0.3529 ∶ 0.6024 ∶ 0.5102
Simplify the ratio: = 1 ∶ 1.707 ∶ 1.446
Ratio divided by coefficient: = 1 ∶ 0.834 ∶ 1.446
Potassium iodide (KI) is the limiting reagent
Amount of iodine produced:
𝐾𝐼 ∶ 𝐼2 = 2 ∶ 1
0.6024 ∶ 𝑥
2𝑥 = 0.6024
𝑥 = 0.3012 𝑚𝑜𝑙
Determining mass of iodine:
𝑚𝐼2
= 𝑛 × 𝑀 = 0.30 × 254
= 76.51 𝑔
76.51 g of iodine is produced from
100 g of KI

55. Activity
1. Consider the reaction:
2 𝐹𝑒2𝑂3 𝑠 + 3 𝐶 𝑠 → 4 𝐹𝑒 𝑠 + 3 𝐶𝑂2 𝑔
Three starting mixtures of 𝐹𝑒2𝑂3 and C are listed below. Determine the limiting reagent in each mixture.
a) 2.0 moles 𝐹𝑒2𝑂3 + 4.0 moles C
b) 2.0 moles 𝐹𝑒2𝑂3 + 3.0 moles C
c) 2.0 moles 𝐹𝑒2𝑂3 + 2.0 moles C
2. 6.00 moles of Al are combined with 6.00 moles of S and heated to form 𝐴𝑙2𝑆3.
2𝐴𝑙 𝑠 + 3𝑆(𝑠) → 𝐴𝑙2𝑆3(𝑠)
How many moles of 𝐴𝑙2𝑆3 will be produced?
3. 4.00 moles of 𝐶2𝐻6 and 12.0 moles of 𝑂2 are reacted according to the reaction
2 𝐶2𝐻6(𝑔) + 7𝑂2(𝑔) → 4𝐶𝑂2(𝑔) + 6𝐻2𝑂(𝑔)
How many moles of 𝐶𝑂2 will form?

56. Reaction Yield
Chemical reactions in the real world don't always go exactly as planned on paper.
Factors that could contribute to less product formation:
 spills and other experimental errors
 undesirable side reactions
 reversibility of reactions
Chemists need a measurement that indicates how successful a reaction has been.
Types of reaction yields
• theoretical yield:
• actual yield:
the amount of product that would result if all the limiting reagent reacted.
the amount of product actually obtained from a reaction.
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 =
𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
× 100%
• The actual yield and theoretical yield should be in the same units (e.g. grams,
moles, etc).
• Low percent yields can cause a large waste of reactants and unnecessary expenses.
• Thus, chemists strive to maximize the percent yield in a reaction.
• This measurement is called the percent yield:
• Predicted amounts of product are sometimes not attained.
Yield is a measure of the extent of a reaction, generally measured by comparing
the amount of product against the amount of product that is possible.

57. Suppose in a synthesis reaction it was calculated that 50.0 grams of product could be obtained. But when
the reaction is performed, only 45.5 grams was obtained. The percent yield would be:
% 𝑦𝑖𝑒𝑙𝑑 =
45.5 𝑔
50.0 𝑔
× 100%
= 91%
Worked example
0.800 mole of phosphorus was reacted with an excess of
sulphur to form diphosphorus pentasulphide, 𝑃2𝑆5.
2 𝑃(𝑠) + 5𝑆(𝑠) → 𝑃2𝑆5(𝑠)
At the end of the reaction, 82.7 g of 𝑃2𝑆5 was produced.
What is the percent yield of this reaction?
Solution:
First, calculate the moles of 𝑃2𝑆5 that could be obtained from
0.8 mol of phosphorus.
𝑃2𝑆5 ∶ 𝑃 = 1 ∶ 2
0.4 mol of 𝑃2𝑆5 could be obtained from 0.8 mol of 𝑃.
Determining mass of 𝑷𝟐𝑺𝟓:
𝑛 = 0.4 𝑚𝑜𝑙
𝑀 = 222 𝑔. 𝑚𝑜𝑙−1
𝑚 = 𝑛 × 𝑀
= 0.4 × 222
= 88.8 𝑔
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = 88.8 𝑔
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 = 82.7 𝑔
% 𝑦𝑖𝑒𝑙𝑑 =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
× 100%
=
82.7 𝑔
88.8 𝑔
× 100% = 93.13%

58. Activity
1. Potassium chlorate decomposes according to the following reaction:
2𝐾𝐶𝑙𝑂3(𝑠) → 2𝐾𝐶𝑙(𝑠) + 3𝑂2(𝑔)
In a certain experiment, 40 g 𝐾𝐶𝑙𝑂3 is heated until it completely decomposes.
a) What is the theoretical yield of oxygen gas?
b) The experiment is performed and the oxygen gas is collected and its mass is found to be 14.9g . What is the
percent yield for this reaction?
2. Mercury(II) oxide decomposes with heating to form elemental mercury, Hg, and oxygen gas, 𝑂2.
2 𝐻𝑔𝑂(𝑠) → 2 𝐻𝑔(𝑙) + 𝑂2(𝑔)
The decomposition of 0.058 mole of HgO produced 0.724 g of 𝑂2 . What is the percent yield of this reaction?
3. When solutions of calcium chloride and silver nitrate are mixed, a precipitate of silver chloride forms.
𝐶𝑎𝐶𝑙2(𝑎𝑞) + 2 𝐴𝑔𝑁𝑂3(𝑎𝑞) → 2 𝐴𝑔𝐶𝑙(𝑠) + 𝐶𝑎 𝑁𝑂3 2(𝑎𝑞)
A solution containing 0.32 mole of 𝐴𝑔𝑁𝑂3 was treated with an excess amount of 𝐶𝑎𝐶𝑙2. After removing and drying
the precipitate, 44.1 g of AgCl was obtained. What is the percent yield of this reaction?