Me cchapter 4

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Chapter 4
Calculations and the Chemical
Equation

Denniston
Topping
Caret
7th Edition

4.1 The Mole Concept and Atoms
• Atoms are exceedingly small
– Unit of measurement for mass of an atom is
atomic mass unit (amu) – unit of measure for
the mass of atoms
• carbon-12 assigned the mass of exactly 12 amu
• 1 amu = 1.66 x 10-24 g
• Periodic table gives atomic weights in amu

4.1 The Mole Concept and
Mass of Atoms
• What is the atomic
weight of one atom
of fluorine?
Answer: 19.00 amu
Atoms

• What would be −the
19.00 amu F 1.661× 10 -24 g 3.156 ×10 23 g F
× mass= this one
of
F atom 1 amuatom in grams?
F F atom

• Chemists usually
work with much

4.1 The Mole Concept and The Mole and Avogadro’s
Number
• A practical unit for defining a collection
of atoms is the mole
Atoms

1 mole of atoms = 6.022 x 1023 atoms

• This is called Avogadro’s number
– This has provided the basis for the concept
of the mole

The Mole

• To make this connection we must define
the mole as a counting unit
– The mole is abbreviated mol
Atoms

• A mole is simply a unit that defines an
amount of something
– Dozen defines 12
– Gross defines 144

4.1 The Mole Concept and Atomic Mass
• The atomic mass of one atom of an element
corresponds to:
– The average mass of a single atom in amu
Atoms

– The mass of a mole of atoms in grams
– 1 atom of F is 19.00 amu 19.00 amu/atom F
– 1 mole of F is 19.00 g 19.00 g/mole F

19.00 amu F 1.66 × 10 −24 g F 6.022 × 10 23 atom F
× ×
1 atom F 1 amu F 1 mol F

=19.00 g F/mol F or 19.00 g/mol F

Molar Mass
• Molar mass - The mass in grams of 1 mole of
atoms
• What is the molar mass of carbon?
Atoms

12.01 g/mol C

• This means counting out a mole of Carbon
atoms (i.e., 6.022 x 1023) they would have a mass
of 12.01 g
• One mole of any element contains the same
number of atoms, 6.022 x 1023, Avogadro’s number

Calculating Atoms, Moles, and Mass

• We use the following conversion factors:
• Density converts grams – milliliters
Atoms

• Atomic mass unit converts amu –
grams
• Avogadro’s number converts moles –
number of atoms
• Molar mass converts grams – moles

4.1 The Mole Concept and Strategy for Calculations
• Map out a pattern for the required
conversion
• Given a number of grams and asked for
Atoms

number of atoms

• Two conversions are required
• Convert grams to moles
1 mol S/32.06 g S OR 32.06 g S/1 mol S
• Convert moles to atoms
mol S x (6.022 x 1023 atoms S) / 1 mol S

4.1 The Mole Concept and Practice Calculations
1. Calculate the number of atoms in 1.7
moles of boron.
2. Find the mass in grams of 2.5 mol Na
Atoms

(sodium).
3. Calculate the number of atoms in 5.0 g
aluminum.
4. Calculate the mass of 5,000,000 atoms
of Au (gold)

Interconversion Between Moles,
Particles, and Grams
Atoms

4.2 The Chemical Formula,
Formula Weight, and Molar
Mass
• Chemical formula - a combination of
symbols of the various elements that make up
the compound
• Formula unit - the smallest collection of
atoms that provide two important pieces of
information
– The identity of the atoms
– The relative number of each type of atom

Formula Weight and Molar Mass Chemical Formula
Consider the following formulas:
• H2 – 2 atoms of hydrogen are chemically bonded
forming diatomic hydrogen, subscript 2
• H2O – 2 atoms of hydrogen and 1 atom of
oxygen, lack of subscript means one atom
• NaCl – 1 atom each of sodium and chlorine
• Ca(OH)2 – 1 atom of calcium and 2 atoms each
of oxygen and hydrogen, subscript outside
parentheses applies to all atoms inside

Formula Weight and Molar Mass Chemical Formula
Consider the following formulas:
• (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion
– Ammonium ion contains 1 nitrogen and 4 hydrogen
– Sulfate ion contains 1 sulfur and 4 oxygen
– Compound contains 2 N, 8 H, 1 S, and 4 O

• CuSO4.5H2O
– This is an example of a hydrate - compounds containing
one or more water molecules as an integral part of their
structure
– 5 units of water with 1 CuSO4

4.2 The Chemical Formula, Comparison of Hydrated and
Formula Weight and Molar Mass
Anhydrous Copper Sulfate

Hydrated copper sulfate Anhydrous copper sulfate

Marked color difference illustrates the fact
that these are different compounds

Formula Weight and Molar Mass Formula Weight and Molar Mass

• Formula weight - the sum of the atomic weights
of all atoms in the compound as represented by its
correct formula
– expressed in amu
• What is the formula weight of H2O?
– 16.00 amu + 2(1.008 amu) = 18.02 amu
• Molar mass – mass of a mole of compound in
grams / mole
– Numerically equal to the formula weight in amu
• What is the molar mass of H2O?
– 18.02 g/mol H2O

4.2 The Chemical Formula, Formula
Formula Unit
Weight and Molar Mass
• Formula unit – smallest
collection of atoms from which
the formula of a compound can
be established
• When calculating the formula
weight (or molar mass) of an
ionic compound, the smallest
unit of the crystal is used
What is the molar mass of (NH4)3PO4?
3(N amu) + 12(H amu) + P amu + 4(O amu)=
3(14.01) + 12(1.008) + 30.97 + 4(16.00)=
149.10 g/mol (NH4)3PO4

4.3 The Chemical Equation and the
Information It Conveys
A Recipe For Chemical Change
• Chemical equation - shorthand notation of a
chemical reaction
– Describes all of the substances that react and all
the products that form, physical states, and
experimental conditions
– Reactants – (starting materials) – the substances
that undergo change in the reaction
– Products – substances produced by the reaction

and the Information It Conveys
4.3 The Chemical Equation Features of a Chemical Equation
1. Identity of products and reactants must
be specified using chemical symbols
2. Reactants are written to the left of the
reaction arrow and products are written
to the right
3. Physical states of reactants and products
may be shown in parentheses
4. Symbol ∆ over the reaction arrow
means that energy is necessary for the
reaction to occur
5. Equation must be balanced

4.3 The Chemical Equation Features of a Chemical Equation
∆
2HgO( s ) → 2Hg( l ) + O 2 (g )

Products and reactants must be
specified using chemical symbols
Reactants – written on the left of arrow
Products – written on the right
∆ – energy is needed
Physical states are shown in parentheses

4.3 The Chemical Equation The Experimental Basis of a
Chemical Equation

We know that a chemical equation
represents a chemical change
• One or more substances changed into
new substances
• Different chemical and physical
properties

and the Information It Conveys Evidence of a Reaction Occurring
4.3 The Chemical Equation
The following can be visual evidence of a reaction:
•Release of a gas
– CO2 is released when acid is placed in a solution
containing CO32- ions
•Formation of a solid (precipitate)
– A solution containing Ag+ ions mixed with a solution
containing Cl- ions
•Heat is produced or absorbed
– Acid and base are mixed together
•Color changes

Subtle Indications of a Reaction
• Heat or light is absorbed or emitted
• Changes in the way the substances
behave in an electrical or magnetic
field before and after a reaction
• Changes in electrical properties

Writing Chemical Reactions

• We will learn to identify the following
patterns of chemical reactions:
– combination
– decomposition
– single-replacement
– double-replacement
• Recognizing the pattern will help you
write and understand reactions

Combination Reactions
• The joining of two or more elements or
compounds, producing a product of
different composition
A + B → AB
• Examples:
2Na(s) + Cl2(g) → 2NaCl(s)
MgO(s) + CO2(g) → MgCO3(s)

4.3 The Chemical Equation Types of Combination
Reactions
1. Combination of a metal and a nonmetal
to form a salt
2. Combination of hydrogen and chlorine
molecules to produce hydrogen chloride
3. Formation of water from hydrogen and
oxygen molecules
4. Reaction of magnesium oxide and
carbon dioxide to produce magnesium
carbonate

4.3 The Chemical Equation Decomposition Reactions
• Produce two or more products from a
single reactant
• Reverse of a combination reaction
AB → A + B

• Examples:
2HgO(s) → 2Hg(l) + O2(g)
CaCO3(s) → CaO(s) + CO2(g)

Types of Decomposition
Reactions
1. Heating calcium carbonate to produce
calcium oxide and carbon dioxide
2. Removal of water from a hydrated
material

Replacement Reactions
1. Single-replacement
• One atom replaces another in the
compound producing a new compound

A + BC → B + AC

• Examples:
Cu(s)+2AgNO3(aq) → 2Ag(s)+Cu(NO3)2(aq)
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

4.3 The Chemical Equation Types of Replacement
Reactions

1. Replacement of copper by
zinc in copper sulfate
2. Replacement of aluminum
by sodium in aluminum
nitrate

Replacement Reactions
2. Double-replacement
• Two compounds undergo a “change
of partners”
• Two compounds react by
exchanging atoms to produce two
new compounds

AB + CD → AD + CB

4.3 The Chemical Equation Types of Double-Replacement
• Reaction of an acid with a base to
produce water and salt
HCl(aq)+NaOH(aq) →NaCl(aq)+H2O(l)
• Formation of solid lead chloride from
lead nitrate and sodium chloride
Pb(NO3)2(aq) + 2NaCl(aq) →
PbCl2(s) + 2NaNO3(aq)
AB + CD → AD + CB

Types of Chemical Reactions
Precipitation Reactions
• Chemical change in a solution that
results in one or more insoluble products
• To predict if a precipitation reaction can
occur it is helpful to know the
solubilities of ionic compounds

4.3 The Chemical Equation Predicting Whether Precipitation
Will Occur
• Recombine the ionic compounds to
have them exchange partners
• Examine the new compounds formed
and determine if any are insoluble
according to the rules in Table 4.1
• Any insoluble salt will be the precipitate
Pb(NO3)2(aq) + NaCl(aq) →
PbCl2 (s) + NaNO3 ( ?)
(?) (aq)

4.3 The Chemical Equation Predict Whether These Reactions
Form Precipitates
• Potassium chloride and silver nitrate

• Potassium acetate and silver nitrate

4.3 The Chemical Equation Reactions with Oxygen
• Reactions with oxygen generally release
energy
• Combustion of natural gas
– Organic compounds CO2 and H2O are
usually the products

CH4+2O2→CO2+2H2O
• Rusting or corrosion of iron
4Fe + 3O2 → 2Fe2O3

4.3 The Chemical Equation Acid-Base Reactions
• These reactions involve the transfer of a
hydrogen ion (H+) from one reactant (acid)
to another (base)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

The H+ on HCl was
transferred to the oxygen
in OH-, giving H2O

4.3 The Chemical Equation Oxidation-Reduction Reactions
• Reaction involves the transfer of one or
more electrons from one reactant to
another
Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+(aq)

Two electrons are
transferred from Zn to
Cu2+

4.3 The Chemical Equation Writing Chemical Reactions
Consider the following reaction:
hydrogen reacts with oxygen to produce water
• Write the above reaction as a chemical
equation
H2 + O2 → H2O
• Don’t forget the diatomic elements

Law of Conservation of Mass

• Law of conservation of mass - matter
cannot be either gained or lost in the
process of a chemical reaction
– The total mass of the products must equal
the total mass of the reactants

A Visual Example of the Law of
4.4 Balancing Chemical
Equations Conservation of Mass

4.4 Balancing Chemical Equations
• A chemical equation shows the molar
quantity of reactants needed to produce a
particular molar quantity of products
• The relative number of moles of each
product and reactant is indicated by
placing a whole-number coefficient
before the formula of each substance in
the chemical equation

4.4 Balancing Chemical Balancing
Coefficient - how many of that substance
are in the reaction
Equations

∆
2HgO( s ) 
→ 2Hg(l ) + O 2 ( g )
• The equation must be balanced
– All the atoms of every reactant must also
appear in the products
• Number of Hg on left? 2
– on right 2
• Number of O on left? 2
– on right 2

Examine the Equation

H 2 + O 2 → H 2O
• Is the law of conservation of mass obeyed
Equations

as written? NO
• Balancing chemical equations uses coefficients
to ensure that the law of conservation of mass is
obeyed
• You may never change subscripts!
• WRONG: H2 + O2 → H2O2

4.4 Balancing Chemical Steps in Equation Balancing
H2 + O2 → H2O
The steps to balancing:
Equations

Step 1. Count the number of moles of
atoms of each element on both
product and reactant sides

Reactants Products
2 mol H 2 mol H
2 mol O 1 mol O

H2 + O2 → H2O
Step 2. Determine which elements are not
balanced – do not have same number on
Equations

both sides of the equation
– Oxygen is not balanced
Step 3. Balance one element at a time by
changing the coefficients
H2 + O2 → 2H2O

This balances oxygen, but is hydrogen
still balanced?

H2 + O2 → 2H2O
How will we balance hydrogen?
Equations

2H2 + O2 → 2H2O
Step 4. Check! Make sure the law of
conservation of mass is obeyed

Reactants Products
4 mol H 4 mol H
2 mol O 2 mol O

Equations
Balancing an Equation

Practice Equation Balancing

Balance the following equations:
Equations

1. C2H2 + O2 → CO2 + H2O

2. AgNO3 + FeCl3 → Fe(NO3)3 + AgCl

3. C2H6 + O2 → CO2 + H2O

4. N2 + H2 → NH3

4.5 Calculations Using the
Chemical Equation
• Calculation quantities of reactants and
products in a chemical reaction has many
applications
• Need a balanced chemical equation for the
reaction of interest
• The coefficients represent the number of
moles of each substance in the equation

4.5 Calculations Using the General Principles
1. Chemical formulas of all reactants and
Chemical Equation
products must be known
2. Equation must be balanced to obey the
law of conservation of mass
• Calculations of an unbalanced equation
are meaningless
1. Calculations are performed in terms of
moles
• Coefficients in the balanced equation
represent the relative number of moles of
products and reactants

4.5 Calculations Using the Using the Chemical Equation
Chemical Equation
• Examine the reaction:
2H2 + O2 → 2H2O
• Coefficients tell us?
– 2 mol H2 reacts with 1 mol O2 to produce 2
mol H2O

• What if 4 moles of H2 reacts with 2 moles of
O2?

– It yields 4 moles of H2O

Using the Chemical Equation
2H2 + O2 → 2H2O
Chemical Equation
• The coefficients of the balanced equation
are used to convert between moles of
substances
• How many moles of O2 are needed to
react with 4.26 moles of H2?
• Use the factor-label method to perform
this calculation

Use of Conversion Factors
Chemical Equation
2H2 + O2 → 2H2O
__mol O 2
1
4.26 mol H 2 × = 2.13 mol O2
__ mol H 2
2
• Digits in the conversion factor come
from the balanced equation

4.5 Calculations Using the Conversion Between Moles
Chemical Equation and Grams
• Requires only the formula weight
• Convert 1.00 mol O2 to grams
moles of grams of
– Plan the path Oxygen Oxygen
– Find the molar mass of oxygen
• 32.0 g O2 = 1 mol O2
– Set up the equation
– Cancel units 1.00 mol O2 x 32.0 g O2
1 mol O2
– Solve equation 1.00 x 32.0 g O2 = 32.0 g O2

Conversion of Mole Reactants to
Chemical Equation Mole Products
• Use a balanced equation
• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
• 1 mol C3H8 results in:
– 5 mol O2 consumed 1 mol C3H8 /5 mol O2
– 3 mol CO2 formed 1 mol C3H8 /3 mol CO2
– 4 mol H2O formed 1 mol C3H8 /4 mol H2O
• This can be rewritten as conversion
factors

Chemical Equation Calculating Reacting Quantities
• Calculate grams O2 reacting with 1.00 mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles O2
– Moles of O2 to grams O2
moles moles grams
C3 H8 Oxygen Oxygen

– Set up the equation and cancel units
– 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 =
1 mol C3H8 1 mol O2
– 1.00 x 5 x 32.0 g O = 1.60 x 102 g O

4.5 Calculations Using the Calculating Grams of Product
Chemical Equation from Moles of Reactant
• Calculate grams CO2 from combustion of 1.00
mol C3H8
– Moles C3H8 to moles CO2
– Moles of CO2 to grams CO2
moles moles grams
C3 H8 CO2 CO2

– 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 =
1 mol C3H8 1 mol CO2
– 1.00 x 3 x 44.0 g CO = 1.32 x 102 g CO

4.5 Calculations Using the Relating Masses of Reactants
Chemical Equation and Products
• Calculate grams C3H8 required to produce
36.0 grams of H2O
– Grams H2O to moles H2O
– Moles H2O to moles C3H8
– Moles of C3H8 to grams C3H8
grams moles moles grams
H2 O H2 O C3 H8 C3 H8

36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8
18.0 g H2O 4 mol H2O 1 mol C3H8
– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8

Calculating a Quantity of Reactant
• Ca(OH)2 neutralizes HCl
Chemical Equation
• Calculate grams HCl neutralized by 0.500 mol
Ca(OH)2
– Write chemical equation and balance
• Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)
– Plan the path
moles moles grams
Ca(OH)2 HCl HCl

0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl

Chemical Equation General Problem-solving Strategy

Sample Calculation
Na + Cl2 → NaCl
Chemical Equation
1. Balance the equation 2Na + Cl2 → 2NaCl
2. Calculate the moles Cl2 reacting with 5.00
mol Na
3. Calculate the grams NaCl produced when
5.00 mol Na reacts with an excess of Cl2
4. Calculate the grams Na reacting with
5.00 g Cl2

Theoretical and Percent Yield
Chemical Equation
• Theoretical yield - the maximum amount of
product that can be produced
– Pencil and paper yield

• Actual yield - the amount produced when
the reaction is performed
– Laboratory yield
actual yield
• Percent yield: % yield = × 100%
theoretical yield

= 125 g CO2 actual x 100% = 97.4%
132 g CO2 theoretical

Sample Calculation

If the theoretical yield of iron was 30.0 g
Chemical Equation
and actual yield was 25.0 g, calculate the
percent yield:
2 Al(s) + Fe2O3(s) → Al2O3(aq) + 2Fe(aq)
• [25.0 g / 30.0 g] x 100% = 83.3%
• Calculate the % yield if 26.8 grams iron
was collected in the same reaction

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