Chapter 6B
Chemical
Quantities
1
CHAPTER OUTLINE
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The Mole Concept
Molar Mass
Calculations Using the Mole
Stoichiometry & Molar Ratios
Mole-Mole & Mass-Mole Calculations
Mass-Mass Calculations
Limiting Reactant
Percent Yield
Energy Changes in Chemical Reactions
2
THE MOLE
CONCEPT
 Chemists find it more convenient to use mass
relationships in the laboratory, while chemical
reactions depend on the number of atoms
present.
 In order to relate the mass and
number of atoms, chemists use
the SI unit mole (abbreviated
mol).
3
THE MOLE
CONCEPT
 The number of particles in a mole is called
Avogadro’s number and is 6.02x1023.
1 mole
equals to
6.02 x 1023
Avogadro’s
number (NA)
4
THE MOLE
CONCEPT
A mole is a very large quantity
6.02x1023
If 10,000 people started to count
Avogadro’s number and counted at the
rate of 100 numbers per minute each
minute of the day, it would take over 1
trillion years to count the total number.
5
THE MOLE
CONCEPT
1 mole H atoms = 6.02x1023 H atoms
1 mole H2 molecules = 6.02x1023 H2 molecules
= 2 x (6.02x1023) H atoms
1 mole H2O molecules = 6.02x1023 H2O molecules
= 2 x (6.02x1023) H atoms
= 6.02x1023 O atoms
1 mole Na+ ions = 6.02x1023 Na+ ions
6
THE MOLE
CONCEPT
 The atomic mass of one atom expressed in amu
is numerically the same as the mass of 1 mole of
atoms of the element expressed in grams.
Mass
Mass
Mass
ofof
of
1 1H
1Mg
Cl
atom
atom
atom
= 1.008
== 35.45
24.31
amu
amu
amu
Mass
Mass
Mass
ofof
of
1 mol
11mol
mol
H Mg
atoms
Cl atoms
atoms
= 1.008
== 35.45
24.31
grams
gg
7
MOLAR MASS
 The mass of one mole of a substance is called
molar mass, and is measured in grams.
Mass of one mole of H2O
2 mol H atoms = 2 (1.01 g) = 2.02 g
1 mol O atom = 1 (16.00 g) = 16.00 g
Molar mass
18.02 g
8
MOLAR MASS
Mass of one mole of Ca(OH)2
1 mol Ca atom = 1 (40.08 g) = 40.08 g
2 mol O atoms = 2 (16.00 g) = 32.00 g
2 mol H atoms = 2 (1.01 g) = 2.02 g
Molar mass
74.10 g
9
Example 1:
Calculate the molar mass of each compound shown
below:
Lithium carbonate Li2CO3
Li
2 x 6.94 = 13.88 g
C
1 x 12.01 = 12.01 g
O
3 x 16.00 = 48.00 g
Molar mass
= 73.89 g/mol
10
Example 2:
Calculate the molar mass of each compound shown
below:
Salicylic acid
C7 H6O3
C
7 x 12.01 = 84.07 g
H
6 x 1.01
O
3 x 16.00 = 48.00 g
= 6.06 g
Molar mass = 138.13 g/mol
11
CALCULATIONS
USING THE MOLE
 Conversions between mass, mole and particles
can be done using molar mass and Avogadro’s
number.
Avogadro’s
Molar
number
mass
Mass of a
substance
MM
Moles of a
substance
NA
Particles of
a substance
12
Example 1:
How many moles of iron are present in 25.0 g of iron?
25.0 g Fe x
1 mole
55.85 g
= 0.448 mol Fe
Molar mass
3 significant figures
13
Example 2:
What is the mass of 5.00 mol of water?
18.02 g
5.00 mol H 2O x
= 90.1 g H 2O
1 mol
3 significant figuresMolar mass
14
Example 3:
How many Mg atoms are present in 5.00 g of Mg?
mass  mol  atoms
5.00 g Mg x
Molar mass
1 mol
24.30 g
x
atoms
=
mol
6.02 x 1023
1
1.24x1023 atoms Mg Avogadro’s
3 significant figures
number
15
Example 4:
How many molecules of HCl are present in 25.0 g
of HCl?
mass  mol  molecules
25.0 g HCl x
1 mol
36.46 g
x
6.02 x 1023 molecules
1
mol
=
4.13 x 1023 molecules HCl
3 significant figures
16
MOLES OF ELEMENTS
IN A FORMULA
 The subscripts in a chemical formula of a
compound indicate the number of atoms of
each type of element.
 For example, 1 molecule of aspirin contains:
C9H8O4
9 C atoms
8H
atoms
4 O atoms
17
MOLES OF ELEMENTS
IN A FORMULA
 The subscripts also indicate the number of
moles of each element in one mole of the
compound.
 For example, 1 mole of aspirin contains:
C9H8O4
9 mole C
atoms
8 mole
H atoms
4 mole O
atoms
18
MOLES OF ELEMENTS
IN A FORMULA
 Using the subscripts from the aspirin formula,
one can write the following conversion factors
for each of the elements in 1 mole of aspirin.
C9H8O4
9 moles C
8 moles H
4 moles O
1 mole C9H8O4
1 mole C9H8O4
1 mole C9H8O4
19
Example 1:
Determine the number of moles of C atoms in 1
mole of each of the following substances:
a) Acetoaminophen used in Tylenol C8 H9NO2
8 moles C
1 mole C8H9 NO2
b) Zinc dietary supplement
Zn(C2 H3O2)2
4 moles C
1 mole Zn(C2 H3O2)2
20
Example 2:
How many carbon atoms are present in 1.50 moles
of aspirin, C9H8O4?
1.50 mol C9H8O4 x
9 moles C
1 mole C9H8 O4
x
6.02x1023 atoms
1 mole C
= 8.13x1024 atoms
21
SUMMARY OF
MASS-MOLE CALCULATIONS
22
STOICHIOMETRY
 Stoichiometry is the quantitative relationship
between the reactants and products in a
balanced chemical equation.
 A balanced chemical equation provides several
important information about the reactants and
products in a chemical reaction.
23
MOLAR
RATIOS
For example:
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
This
is the molar 2 molecules
3 molecules
ratios between
100 molecules
300 molecules
the
reactants and200 molecules
products
6 molecules
106 molecules
3x10
2x106 molecules
1 molecule
1 mole
3 moles
2 moles
24
Example 1:
Determine each mole ratio below based on the
reaction shown:
2 C4H10 + 13 O2  8 CO2 + 10 H2O
mol O 2
13
=
mol CO 2
8
mol C4 H10
2
=
mol H 2O
10
25
STOICHIOMETRIC
CALCULATIONS
 Stoichiometric
calculations can be classified as
Mass-mass
one of the following:
calculations
MASS of
compound A
Mass-mole
MASS of
Mole-mole
calculations
compound B
calculations
MM
MM
MOLES of
compound A
molar ratio
MOLES of
compound B
26
MOLE-MOLE
CALCULATIONS
 Relates moles of reactants and products in a
balanced chemical equation
MOLES of
compound A
molar ratio
MOLES of
compound B
27
Example 1:
How many moles of nitrogen will react with 2.4
moles of hydrogen to produce ammonia as shown in
the reaction below?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
2.4 mol H2 x
1 mol N 2
= 0.80 mol N2
3 mol H 2
Mole ratio
28
Example 2:
How many moles of ammonia can be produced from
32 moles of hydrogen? (Assume excess N2 present)
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
32 mol H2 x
2 mol NH 3
3 mol H 2 = 21 mol NH3
Mole ratio
29
Example 3:
In one experiment, 6.80 mol of ammonia are
prepared. How many moles of hydrogen were used
up in this experiment?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
6.80 mol NH3 x
3 mol H 2
2 mol NH 3 = 10.2 mol H2
Mole ratio
30
MASS-MOLE
CALCULATIONS
 Relates moles and mass of reactants or products
in a balanced chemical equation
MASS of
compound A
MM
MOLES of
compound A
molar ratio
MOLES of
compound B
31
Example 1:
How many grams of ammonia can be produced
from the reaction of 1.8 moles of nitrogen with
excess hydrogen as shown below?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
1.8 mol N2 x
Mole ratio
2 mol NH 3 x 17.04 g NH 3
= 61 g NH3
1 mol N 2
1 mol NH 3
Molar mass
32
Example 2:
How many moles of hydrogen gas are required to
produce 75.0 g of ammonia?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
1 mol NH 3
3 mol H 2
75.0 g NH3 x
x
= 6.60 mol H2
g
NH
17.04
2 mol NH 3
3
Molar mass
Mole ratio
33
Example 3:
How many moles of ammonia can be produced from
the reaction of 125 g of nitrogen?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
1 mol N 2
125 g N2 x
28.02 g N 2
Molar mass
2 mol NH 3
x
1 mol N 2 = 8.92 mol NH3
Mole ratio
34
MASS -MASS
CALCULATIONS
 Relates mass of
reactants and products
in a balanced chemical
equation
35
Example 1:
What mass of oxygen will be required to react
completely with 96.1 g of propane, C3H8, according
to the equation below?
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Mass of
propane
Moles of
propane
Moles of
oxygen
Mass of
oxygen
36
Example 1:
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
1 mol C3 H 8
5 mol O 2
x
96.1 g C3H8 x
44.11 g C3 H 8
1 mol C3 H 8
32.00 g O 2
x
= 349 g O2
1 mol O 2
Molar mass
Molar mass
Mole ratio
37
Example 2:
What mass of carbon dioxide will be produced from
the reaction of 175 g of propane, as shown?
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Mass of
propane
Moles of
propane
Moles of
carbon
dioxide
Mass of
carbon dioxide
38
Example 2:
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
175 g C3H8
1 mol C3 H 8
3 mol CO 2
x
x
44.11 g C3 H 8
1 mol C3 H 8
g CO 2
x 44.01
= 524 g CO2
mol
CO
1
2
Molar mass
Molar mass
Mole ratio
39
LIMITING
REACTANT
 When 2 or more reactants are combined in nonstoichiometric ratios, the amount of product
produced is limited by the reactant that is not in
excess.
 This reactant is referred to as limiting reactant.
 When doing stoichiometric problems of this
type, the limiting reactant must be determined
first before proceeding with the calculations.
40
LIMITING REACTANT
ANALOGY
Consider the following recipe for a sundae:
41
LIMITING REACTANT
ANALOGY
The number
How
many sundaes
of sundaes
can possible
be prepared
is limited
from the
by the
followingofingredients:
amount
syrup, the limiting reactant.
Limiting
reactant
Excess
reactants
42
GUIDE TO
LIMITING REACTANT CALC.
Assume reactant 1
is limiting
Assume reactant 2
is limiting
Compare
43
Example 1:
How many moles of H2O can be produced by
reacting 4.0 mol of hydrogen and 3.0 mol of oxygen
gases as shown below:
2 H2 (g) + 1 O2 (g)  2 H2O (g)
Mole-mole
calculations
Limiting
reactant
44
Example 1:
Correct
2 H2 (g) + 1 O2 (g)  2 H2O (g)
answer
Assume H2 is LR
2 mol H 2O
4.0 mol H2 x
= 4.0 mol H2O
2 mol H 2
Correct
Assume O2 is LR
assumption
2 mol H 2O
3.0 mol O2 x
= 6.0 mol H2O
1 mol O 2
45
Example 2:
A fuel mixture used in the early days of rocketry
was a mixture of N2H4 and N2O4, as shown below.
How many grams of N2 gas is produced when 100. g
of N2H4 and 200. g of N2O4 are mixed?
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Limiting
reactant
Mass-mass
calculations
46
Example 2:
Assumes
N2H4 is LR
Assumes
N2O4 is LR
47
Example 2:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2H4 is LR
1 mol N 2 H 4 x 3 mol N 2 =
100. g N2H4 x
2 mol N 2 H 4
32.06 g N 2 H 4
4.68 mol N2
48
Example 2:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2O4 is LR
1 mol N 2O 4 x 3 mol N 2 =
200. g N2O4 x
1 mol N 2O 4
92.02 g N 2O4
6.52 mol N2
49
Example 2:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2H4 is LR
4.68 mol N2
Assume N2O4 is LR
6.52 mol N2
N2H4 is
LR
Correct
amount
50
Example 2:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Calculate mass of N2
4.68 mol N2
28.02 g N 2
x
1 mol N 2 = 131 g N2
51
Example 3:
How many moles of Fe3O4 can be produced by
reacting 16.8 g of Fe with 10.0 g of H2O as shown
below:
3 Fe (s) + 4 H2O (l)  Fe3O4 (s) + 4 H2 (g)
Limiting Mass-mol
reactantcalculations
52
Example 3:
3 Fe (s) + 4 H2O (l)  Fe3O4 (s) + 4 H2 (g)
Assume Fe is LR
mol Fe
mol Fe 3O 4
1
1
16.8 g Fe x
x
=
55.85 g Fe
3 mol Fe
0.100 mol Fe3O4
53
Example 3:
3 Fe (s) + 4 H2O (l)  Fe3O4 (s) + 4 H2 (g)
Assume H2O is LR
10.0 g H2O
x
1 mol H 2O x 1 mol Fe 3O 4 =
18.02 g H 2O
4 mol H 2O
0.139 mol Fe3O4
54
Example 3:
3 Fe (s) + 4 H2O (l)  Fe3O4 (s) + 4 H2 (g)
Assume Fe is LR
0.100 mol Fe3O4
Assume H2O is LR
0.139 mol Fe3O4
Fe is
LR
Correct
amount
55
Example 4:
How many grams of AgBr can be produced when
50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as
shown below:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Limiting
Reactant
56
Example 4:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume MgBr2 is LR
1 mol MgBr2 x 2 mol AgBr
50.0 g MgBr2 x
184.10 g MgBr2 1 mol MgBr2
g AgBr
187.77
x
= 102 g AgBr
1 mol AgBr
57
Example 4:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume AgNO3 is LR
mol AgBr
mol AgNO 3
1
2
x
100.0 g AgNO3 x
169.88 g AgNO 3 2 mol AgNO 3
g AgBr
187.77
x
= 111 g AgBr
1 mol AgBr
58
Example 4:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume MgBr2 is LR
102 g AgBr
Assume AgNO3 is LR
111 g AgBr
MgBr2
is LR
Correct
amount
59
PERCENT YIELD
 The amount of product calculated through
stoichiometric ratios are the maximum amount
product that can be produced during the
reaction, and is thus called theoretical yield.
 The actual yield of a product in a chemical
reaction is the actual amount obtained from the
reaction.
60
PERCENT YIELD
 The percent yield of a reaction is obtained as
follows:
Actual yield
x100 = Percent yield
Theoretical yield
61
Example 1:
In an experiment forming ethanol, the theoretical
yield is 50.5 g and the actual yield is 46.8 g. What is
the percent yield for this reaction?
46.8 g
Actual yield
x100 = 92.7 %
% yield =
x100 =
50.5 g
Theoretical yield
62
Example 2:
Silicon carbide can be formed from the reaction of
sand (SiO2) with carbon as shown below:
1 SiO2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g)
When 125 g of sand are processed, 68.4 of SiC is
produced. What is the percent yield of SiC in this
reaction?
Actual
yield
63
Example 2:
1 SiO2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g)
Calculate theoretical yield
125 g SiO2
1 mol SiO 2 x 1 mol SiC 40.10 g SiC
x
x
=
60.09 g SiO 2 1 mol SiO 2
1 mol SiC
83.4 g SiC
64
Example 2:
Calculate percent yield
Actual yield
68.4 g
% yield =
x100 =
x100 = 82.0 %
Theoretical yield
83.4 g
65
Example 3:
In an experiment to prepare aspirin, the theoretical
yield is 153.7 g and the actual yield is 124.0 g. What
is the percent yield of this reaction?
124.0 g
Actual yield
x100 = 80.68 %
% yield =
x100 =
153.7 g
Theoretical yield
66
Example 4:
Carbon tetrachloride (CCl4) was prepared by
reacting 100.0 g of Cl2 with excess carbon disulfide
(CS2), as shown below. If 65.0 g was prepared in
this reaction, calculate the percent yield.
CS2 + 3 Cl2  CCl4 + S2Cl2
67
Example 4:
CS2 + 3 Cl2  CCl4 + S2Cl2
Calculate theoretical yield
100.0 g Cl2
1 mol Cl 2 x 1 mol CCl 4
x
3 mol Cl 2
70.90 g Cl 2
g CCl 4
153.81
x
= 72.31 g
1 mol CCl 4
68
Example 4:
Calculate percent yield
Actual yield
65.0 g
% yield =
x100 =
x100 = 89.9 %
Theoretical yield
72.31 g
69
ENERGY CHANGES
IN CHEMICAL REACTIONS
 Heat is energy change that is lost or gained when a
chemical reaction takes place.
 The system is the reactants and products that we
are observing. The surroundings are all the things
that contain and interact with the system, such as
the reaction flask, the laboratory room and the air
in the room.
 The direction of heat flow depends whether the
products in a reaction have more or less energy
than the reactants.
70
ENERGY CHANGES
IN CHEMICAL REACTIONS
 For a chemical reaction to occur, the molecules of the
reactants must collide with each other with the proper
energy and orientation.
 The minimum amount of energy
required for a chemical reaction
to occur is called the activation
energy.
 The heat of reaction is the
amount of heat absorbed or
released during a reaction and is
designated by the symbol H.
71
ENERGY CHANGES
IN CHEMICAL REACTIONS
 When heat is released
during a chemical reaction,
it is said to be exothermic.
 For exothermic reactions,
H is negative and is
included on the right side of
the equation.
3 H2 (g) + N2 (g)
2 NH3 (g) + 22.0 kcal
H= 22.0 kcal
72
ENERGY CHANGES
IN CHEMICAL REACTIONS
 When heat is gained during
a chemical reaction, it is
said to be endothermic.
 For endothermic reactions,
H is positive and is
included on the left side of
the equation.
2 H2O (l) + 137 kcal
2 H2 (g) + O2 (g)
H= +137 kcal
73
CALCULATING HEAT
IN A REACTION
 The value of H in a chemical reaction refers to the
heat lost or gained for the number of moles of
reactants and products in a balanced chemical
equation.
 For example, based on the chemical equation shown
below:
2 H2O (l)
137 kcal
2 mol H2O
2 H2 (g) + O2 (g)
or
137 kcal
2 mol H2
H= +137 kcal
or
137 kcal
1 mol O2
74
Example 1:
Given the reaction shown below, how much heat, in kJ,
is released when 50.0 g of NH3 form?
3 H2 (g) + N2 (g)
2 NH3 (g)
H= 92.2 kJ
1 mol NH3 x 92.2 kJ
50.0 g NH3 x
= 135 kJ
17.04 g NH3 2 mol NH3
75
Example 2:
How many kJ of heat are needed to react 25.0 g of HgO
according to the reaction shown below:
2 HgO (s)
25.0 g HgO x
2 Hg (l) + O2 (g)
H= 182 kJ
1 mol HgO x 182 kJ
= 10.5 kJ
216.59 g HgO 2 mol HgO
76
THE END
77