Chapter 3
Stoichiometric
Atomic Masses, Mole concept, and Molar
Mass (Average atomic mass).
Number of atoms per amount of element.
Percent composition and Empirical formula
of molecules.
Chemical equations, Balancing equations,
and Stoichiometric calculations including
limiting reagents.
1
Chemical Stoichiometry
Stoichiometry - The study of quantities of
materials consumed and produced in
chemical reactions.
2
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
Copyright©2000 by Houghton Mifflin Company.
All rights reserved.
= 16.00 amu
3
Atomic Masses
Elements occur in nature as mixtures of
isotopes
Carbon =
98.89% 12C
1.11% 13C
<0.01% 14C
Carbon atomic mass = 12.01 amu
4
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
5
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022  1023 units of
that thing
6
Avogadro’s number
equals
23
6.022  10 units
7
Molar Mass
A substance’s molar mass (molecular weight) is
the mass in grams of one mole of the compound.
C=12
O=16
CO2 = 44.01 grams per mole
8
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
9
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
10
Measuring Atomic Mass
Figure 3.1: (left) A scientist injecting a sample into a
mass spectrometer. (right) Schematic diagram of a mass
spectrometer.
11
Spectrum
Most Abundant Isotope
12
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
13
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
14
Determining Elemental Composition
(Formula)
Figure 3.5:
A schematic diagram of the
combustion device used to analyze substances for
carbon and hydrogen.
15
The masses obtained (mostly CO2 and
H2O and sometimes N2)) will be used to
determine:
1. % composition in compound
2. Empirical formula
3. Chemical or molecular formula if the
Molar mass of the compound is known or
given.
16
Example of Combustion
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
17
Collect 22.0 g CO2 and 13.5 g H2O
16 x 2
12
1mol CO2
44gCO2
1mol C
22gCO2
mc
22gCO2 x 12gC
mc =
12g C
% C in CO2
collected
= 6g C
44gCO2
18
Convert g to mole:
1mol C
nmol C
12g C
6g
nc =
6g x 1molC
12 gC
= 0.5 mol
Repeat the same for H from H2O
1mol H2O
18g H2O
2 mol H
13.5g H2O
nmol H
nmol H =
2x13.5
2g H
mH
= 1.5 mol H
18 mol H
Faster H but
still need O 19
mH =
2x13.5
18
mO = 11.5g – mC – mH
= 11.5 – 6 – 1.5 = 4g
= 1.5 g H
nO =
m
MM
=
4
16
= 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
20
OR
for C
mC
1. Fraction of C in CO2 =
12.01
=
MMCO2
=
44.01
2. Mass of C in compound = mass of CO2 x Fraction of C
mC
3. % C in compound =
x 100
msample
4. If N then % N = 100 - % C - % H
21
Then
Empirical Formula
Using the previously calculated % in compound:
% in gram
a. Number of mole of C =
Atomic mass of C
% in gram
b. Number of mole of H =
Atomic mass of H
a
Then divide by the smallest number:
smallest
:
b
smallest
:
c
smallest
22
Note
If results are :
0.99 : 2.01 : 1.00
Then you have to convert to whole numbers:
1 :2
:1
CH2N
If results are :
1.49 : 3.01 : 0.99
Then you have to multiply by 2:
3
:6
:2
C3H6N2
Hence, empirical formula is the simplest formula of a compound
23
Formulas
molecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Then
Molecular Mass
=
n
Empirical Mass
24
Figure 3.6:
Examples of substances whose empirical and molecular formulas differ. Notice
that molecular formula = (empirical formula)n, where n is a integer.
25
26
27
28
Chemical Equations
Chemical change involves a
reorganization of the atoms in one or
more substances.
29
Chemical Equation
A representation of a chemical reaction:
C2H5OH + O2  CO2 + H2O
reactants
products
Unbalanced !
30
Chemical Equation
C2H5OH + 3O2  2CO2 + 3H2O
The equation is balanced.
1 mole of ethanol reacts with 3 moles of
oxygen
to produce
2 moles of carbon dioxide and 3 moles of
water
31
32
Calculating Masses of Reactants
and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to grams, if necessary.
33
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
34
OR
2CH3OH + 3O2
2 mol
2CO2 + 4H2O
4 mol
2x(12+4+16) g
209 g
m=
4x(2+16) g
m
209 x 4(2+16)
2(12+4+16)
35
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
36
Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
37
Limiting Reagents
6 red
green
leftused
overup
38
Limiting Reactant Calculations
What weight of molten iron is produced by 1 kg each of
the reactants?
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
1 mol
2 mol
6.26mol
18.52
Ratio:
0.160
Limiting
>
0.108
Excess
The 6.26 mol Fe2O3 will
Disappear first
39
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted. Its amount is
Calculated using the balanced equation.
Actual Yield is the amount of product actually obtained
from a reaction. It is usually given.
40
Percent Yield
Actual yield = quantity of product actually
obtained
Theoretical yield = quantity of product
predicted by stoichiometry
41
Percent Yield Example
14.4 g
excess
Actual yield = 6.26 g
42
43
Sample Exercise
Titanium tetrachloride, TiCl4, can be made by
combining titanium-containing ore (which is often
impure TiO2) with carbon and chlorine TiO2(s) + 2 Cl2(g) + C(s)
TiCl4(l) + CO2(g)
If one begins with 125 g each of Cl2 and C, but
plenty of titanium-containing ore, which is the
limiting reagent in the reaction? What quantity of
TiCl4 can be produced?
44
Virtual Laboratory Project
45
Practice Example 1
A compound contains C, H, N. Combustion of
35.0mg of the compound produces 33.5mg CO2
and 41.1mg H2O. What is the empirical formula of
the compound?
Solution:
1. Determine C and H, the rest from 33.5mg is N.
2. Determine moles from masses.
3. Divide by smallest number of moles.
46
Practice Example 2
Caffeine contains 49.48% C, 5.15% H, 28.87%
N and 16.49% O by mass and has a molar mass
of 194.2 g/mol. Determine the molecular
formula.
Solution:
1. Convert mass to moles.
2. Determine empirical formula.
3. Determine actual formula.
C8H10N4O2
47
Practice Example 3
Nitrogen gas can be prepared by passing
gaseous ammonia over solid copper(II)
oxide at high temperatures. The other
products of the reaction are solid copper and
water vapor. If a sample containing 18.1g of
NH3 is reacted with 90.4g of CuO, which is
the limiting reactant? How many grams of
N2 will be formed.
48
Practice Example 4
Methanol can be manufactured by
combination of gaseous carbon monoxide
and hydrogen. Suppose 68.5Kg CO(g) is
reacted with 8.60Kg H 2(g). Calculate the
theoretical yield of methanol. If 3.57x104g
CH3OH is actually produced, what is the
percent yield of methanol?
49
Practice Example 5
SnO2(s) + 2 H2(g)  Sn(s) + 2 H2O(l)
a) the mass of tin produced from 0.211 moles of hydrogen
gas.
b) the number of moles of H2O produced from 339 grams of
SnO2.
c) the mass of SnO2 required to produce 39.4 grams of tin.
d) the number of atoms of tin produced in the reaction of 3.00
grams of H2.
e) the mass of SnO2 required to produce 1.20 x 1021
molecules of water.
50
QUESTION
Bromine exists naturally as a mixture of
bromine-79 and bromine-81 isotopes. An atom
of bromine-79 contains:
1) 35 protons, 44 neutrons, 35 electrons.
2) 34 protons and 35 electrons, only.
3) 44 protons, 44 electrons, and 35 neutrons.
4) 35 protons, 79 neutrons, and 35 electrons.
5) 79 protons, 79 electrons, and 35 neutrons.
51
ANSWER
1)
35 protons, 44 neutrons, 35 electrons.
Section 2.5 The Modern View of Atomic
Structure: An Introduction (p. 54)
The 79 in bromine-79 is the mass number (the
number of protons plus neutrons). The atomic
number for bromine is 35. The number of
neutrons is 79 – 35 = 44. The number of
electrons equals the number of protons.
52
QUESTION
The atomic mass of rhenium is 186.2. Given that
37.1% of natural rhenium is rhenium-185, what
is the other stable isotope?
1)
2)
3)
183
75
187
Re
4)
Re
5)
75
189
Re
75
181
75
190
75
Re
Re
53
ANSWER
2)
187
Re
75
Section 3.1 Atomic Masses (p. 82)
(0.371  185) + (0.629  M) = 186.2. Solve for M
(the isotopic mass). Since neutrons are slightly
heavier than protons, this calculation is only an
approximation, but is accurate enough to
determine the correct answer.
54
QUESTION
54
Naturally occurring iron contains 5.82% Fe,
26
56
57
58
91.66% Fe, 2.19% Fe, and 0.33% Fe.
26
26
26
The respective atomic masses are 53.940 amu,
55.935 amu, 56.935 amu, and 57.933 amu.
Calculate the average atomic mass of iron.
55
ANSWER
55.85 amu
Section 3.1 Atomic Masses (p. 83)
(.0582  53.940 amu) + (.9166  55.935 amu) +
(.0219  56.935 amu) + (.0033  57.933 amu) =
55.85 amu
56
QUESTION
What is the mass of one atom of copper in
grams?
1) 63.5 g
2) 52.0 g
3) 58.9 g
4) 65. 4 g
–22
5) 1.06  10 g
57
ANSWER
5)
–22
1.06  10
g
Section 3.2 The Mole (p. 87)
A mole of copper atoms has a mass of 63.55 g.
The mass of 1 copper atoms is
23
–22
63.55 g/(6.022  10 ) = 1.06  10 g.
58
ANSWER
3)
46.07
Section 3.3 Molar Mass (p. 90)
The molar mass is the sum of masses of all the
atoms in the molecule.
2  12.01 + 6  1.008 + 1  16.00 = 46.07
59
QUESTION
What is the molar mass of ethanol (C2H5OH)?
1) 45.07
2) 38.90
3) 46.07
4) 34.17
5) 62.07
60
QUESTION
For which compound does 0.256 mole weigh
12.8 g?
1) C2H4O
2) CO2
3) CH3Cl
4) C2H6
5) None of these
61
ANSWER
3)
CH3Cl
Section 3.3 Molar Mass (p. 90)
The molar mass has units of g/mol.
(12.8 g/0.256 mol) = 50.0 g/mol. The molecule
with the closest molar mass is CH3Cl.
62
QUESTION
Phosphorus has the molecular formula P4 and
sulfur has the molecular formula S8. How many
grams of phosphorus contain the same number
of molecules as 6.41 g of sulfur?
1) 3.10 g
2) 3.21 g
3) 6.19 g
4) 6.41 g
5) None of these
63
ANSWER
1)
3.10 g
Section 3.3 Molar Mass (p. 90)
6.41 g of S8 converts to 0.0250 mol. Therefore
0.0250 mol of phosphorus is needed to have the
same number of atoms as sulfur. 0.0250 mol P4 
(123.88 g/mol P4) = 3.10 g.
64
QUESTION
How many grams are in a 6.94-mol sample of
sodium hydroxide?
1) 40.0 g
2) 278 g
3) 169 g
4) 131 g
5) 34.2 g
65
ANSWER
2)
278 g
Section 3.3 Molar Mass (p. 91)
The molar mass of sodium hydroxide, NaOH, is
22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00
g/mol. Convert to grams: 6.94 mol  (40.00 g/mol)
= 278 g.
66
QUESTION
How many grams of potassium are in 12.5 g of
K2CrO7?
1) 2.02 g
2) 8.80 g
3) 4.04 g
4) 78.2 g
5) 25.0 g
67
ANSWER
3)
4.04 g
Section 3.4 Percent Composition of Compounds
(p. 94)
The molar mass of K2CrO7 is 2  39.10 + 52.00 +
7  16.00 = 242.2. The mass fraction of potassium
is (2  39.10)/242.2 = 0.3229.
0.3229  12.5 g = 4.04 g.
68
QUESTION
In balancing an equation, we change the
__________ to make the number of atoms on
each side of the equation balance.
1) formulas of compounds in the reactants
2) coefficients of compounds
3) formulas of compounds in the products
4) subscripts of compounds
5) none of these
69
ANSWER
2)
coefficients of compounds
Section 3.7 Balancing Chemical Equations
(p. 104)
Changing the subscripts of a compound or the
addition of a compound to a chemical equation
changes the reaction. Only the coefficients can
be adjusted without altering the reaction.
70
QUESTION
Suppose the reaction
Ca3(PO4)2 + 3H2SO4  3CaSO4 + 2H3PO4
is carried out starting with 103 g of Ca3(PO4)2
and 75.0 g of H2SO4. How much phosphoric acid
will be produced?
1) 74.9 g
2) 50.0 g
3) 112 g
4) 32.5 g
71
5) 97.6 g
ANSWER
2)
50.0 g
Section 3.9 Calculations Involving a Limiting
Reactant (p. 113)
This is a limiting reactant problem. The first step
is determining which reactant will run out first.
72
ANSWER (continued)
Use the coefficients of the balanced reaction to
determine the amount of phosphoric acid that
will be produced by the reaction of 103 g of
Ca3(PO4)2 and 75.0 g H2SO4. Which ever
produces the lesser amount is the limiting
reactant.
73
QUESTION
Which of the following compounds has the same
percent composition by mass as styrene, C8H8?
1) Acetylene, C2H2
2) Benzene, C6H6
3) Cyclobutadiene, C4H4
4) -ethyl naphthalene, C12H12
5) All of these
74
ANSWER
5)
All of these
Section 3.4 Percent Composition of Compounds
(p. 93)
The ratio of C to H in C8H8 is 1:1. This is the
same ratio found for each of the compounds, so
all have the same percent composition by mass.
75
QUESTION
The empirical formula of styrene is CH; its molar
mass is 104.1. What is the molecular formula of
styrene?
1) C2H4
2) C8H8
3) C10H12
4) C6H6
5) None of these
76
ANSWER
2)
C8H8
Section 3.5 Determining the Formula of a
Compound (p. 96)
The mass of CH is 13.018. 13.018 divides into
104.1 about 8 times. Therefore there are 8 CH
groups in this compound. C8H8 is the molecular
formula.
77
QUESTION
Balanced chemical equations imply which of the
following?
1) Numbers of molecules are conserved in
chemical change.
2) Numbers of atoms are conserved in
chemical change.
3) Volume is conserved in chemical change.
4) 1 and 2
5) 2 and 3
78
ANSWER
2)
Numbers of atoms are conserved in
chemical change.
Section 3.7 Balancing Chemical Equations
(p. 104)
This is another way of stating the Law of
Conservation of Mass.
79
QUESTION
How many of the following statements are true
concerning chemical equations?
a. Coefficients can be fractions.
b. Subscripts can be fractions.
c. Coefficients represent the relative
masses of the reactants and/or products.
d. Changing the subscripts to balance an
equation can only be done once.
e. Atoms are conserved when balancing
chemical equations.
80
QUESTION
(continued)
1)
2)
3)
4)
5)
a
b
c
d
e
81
ANSWER
2)
b
Section 3.7 Balancing Chemical Equations
(p. 104)
Coefficients can be fractions, but usually
fractions are multiplied by the appropriate
number to make them whole numbers. Atoms
are conserved according to the Law of
Conservation of Mass.
82
QUESTION
Determine the coefficient for O2 when the
following equation is balanced in standard form
(smallest whole number integers)
C4H10(g) + O2(g)  CO2(g) + H2O(g)
1) 4
2) 8
3) 10
4) 13
5) 20
83
ANSWER
4)
13
Section 3.7 Balancing Chemical Equations
(p. 104)
O2 should be balanced last since it contains only
one type of element and balancing it will not
cause an imbalance in another element.
84
QUESTION
What is the subscript of aluminum in the formula
of aluminum phosphate?
1) 1
2) 2
3) 3
4) 4
5) 0
85
ANSWER
1)
1
Section 3.7 Balancing Chemical Equations
(p. 104)
The phosphate ion has a charge of –3, while the
aluminum ion takes a charge of +3. Therefore
the formula of aluminum phosphate is AlPO4.
86
QUESTION
How many grams of Ca(NO3)2 can be produced
by reacting excess HNO3 with 7.40 g of
Ca(OH)2?
1) 10.2 g
2) 16.4 g
3) 32.8 g
4) 8.22 g
5) 7.40 g
87
ANSWER
2)
16.4 g
Section 3.8 Stoichiometric Calculations:
Amounts of Reactants and Products (p. 108)
The balanced equation is Ca(OH)2 + 2HNO3 
Ca(NO3)2 + 2H2O. Water is the other product for
an acid/base reaction.
88
QUESTION
The limiting reactant in a reaction:
1) is the reactant for which there is the least
amount in grams.
2) is the reactant which has the lowest
coefficient in a balanced equation.
3) is the reactant for which there is the most
amount in grams.
4) is the reactant for which there is the fewest
number of moles.
5) none of these
89
ANSWER
5)
none of these
Section 3.9 Calculations Involving a Limiting
Reactant (p. 113)
The limiting reactant in a reaction is the reactant
that produces the least number of grams of any
product.
90
QUESTION
The molecular formula always represents the
total number of atoms of each element present
in a compound.
1) True
2) False
91
ANSWER
1) True
Section 3.5 Determining the Formula of a
Compound (p. 98)
The molecular formula contains all the
information about the compound.
92