CHAPTER THREE
STOICHIOMETRY
For Review
1.
Counting by weighing utilizes the average mass of a particular unit of substance. For marbles,
a large sample size will contain many different individual masses for the various marbles.
However, the large sample size will have an average mass so that the marbles behave as if
each individual marble has that average mass. This assumption is valid as long as the sample
size is large. When a large sample of marbles is weighed, one divides the total mass of
marbles by the average mass of a marble and this will equal the number of marbles present.
For atoms, because we can’t count individual atoms, we “count” the atoms by weighing them
and convert the mass in grams to the quantity of atoms in the sample. The mole scale of
atoms is a huge number (6.022 × 1023 atoms = 1 mol), so the assumption that a weighable
sample size behaves as a bunch of atoms each with an average mass is valid and very useful.
2.
The masses of all the isotopes are relative to a specific standard. The standard is one atom of
the carbon-12 isotope weighing exactly 12 .000 0 amu. One can determine from experiment
how much heavier or lighter any specific isotope is than 12C. From this information, we
assign an atomic mass value to that isotope. For example, experiment tells one that 16O is
about 4/3 heavier than 12C, so a mass of 4/3(12.00) = 16.00 amu is assigned to 16O.
3.
The two major isotopes of boron are 10B and 11B. The listed mass of 10.81 is the average
mass of a very large number of boron atoms.
4.
There are several ways to do this. The three conversion factors to use are Avogadro’s
number, the molar mass, and the chemical formula. Two ways to use these conversions to
convert grams of aspirin to number of H atoms are given below:
molar mass aspirin = 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol
1.00 g C9H8O4 ×
1 mol C9 H 8O 4
8 mol H
6.022  10 23 atoms H


180 .15 g C9 H 8O 4
mol C 9 H 8 O 4
mol H
= 2.67 × 1022 H atoms
or
1.00 g C9H8O4 ×
1 mol C9 H 8O 4
6.022  10 23 molecules C9 H 8 O 4

×
180 .15 g C 9 H 8O 4
mol C 9 H 8O 4
8 atoms H
= 2.67 × 1022 H atoms
molecule C9 H8O 4
Of course, the answer is the same no matter the order of the conversion factors.
40
CHAPTER 3
5.
STOICHIOMETRY
41
CxHyOz + oxygen → x CO2 + y/2 H2O
From the equation above, the only reactant that contains carbon is the unknown compound
and the only product that contains carbon is CO2. From the mass of CO2 produced, one can
calculate the mass of C present which is also the mass of C in CxHyOz. Similarly, all the
hydrogen in the unknown compound ends up as hydrogen in water. From the mass of H2O
produced, one can calculate the mass of H in CxHyOz. Once the mass of C and H are known,
the remainder of the compound is oxygen. From the mass of C, H, and O in the compound,
one can then go on to determine the empirical formula.
6.
The molecular formula tells us the actual number of atoms of each element in a molecule (or
formula unit) of a compound. The empirical formula tells only the simplest whole number
ratio of atoms of each element in a molecule. The molecular formula is a whole number
multiple of the empirical formula. If that multiplier is one, the molecular and empirical
formulas are the same. For example, both the molecular and empirical formulas of water are
H2O. For hydrogen peroxide, the empirical formula is OH; the molecular formula is H2O2.
7.
The product of the reaction has two A atoms bonded to a B atom for a formula of A2B. The
initial reaction mixture contains 4 A2 and 8 AB molecules and the final reaction mixture
contains 8 A2B molecules. The reaction is:
8 AB(g) + 4 A2(g) → 8 A2B(g)
Using the smallest whole numbers, the balanced reaction is:
2 AB(g) + A2(g) → 2 A2B(g)
2.50 mol A2 ×
2 mol A 2 B
= 5.00 mol A2B
mol A 2
The atomic mass of each A atom is 40.0/2 = 20.0 amu and the atomic mass of each B atom is
30.0  20.0 = 10.0 amu. The mass of A2B = 2(20.0) + 10.0 = 50.0 amu.
15.0 g AB ×
1 mol A 2
40.0 g A 2
1 mol AB
= 10.0 g A2


30.0 g AB 2 mol AB
mol A 2
From the law of conservation of mass, the mass of product is:
10.0 g A2 + 15.0 g AB = 25.0 g A2B
or by stoichiometric calculation.
15.0 g AB ×
1 mol AB 1 mol A 2 B 50.0 g A 2 B
= 25.0 g A2B


30.0 g AB
mol AB
mol A 2 B
or
10.0 g A2 ×
1 mol A 2
2 mol A 2 B 50.0 g A 2 B
= 25.0 g A2B


40.0 g A 2
mol A 2
mol A 2 B
42
CHAPTER 3
STOICHIOMETRY
Generally, there are several ways to correctly do a stoichiometry problem. You should choose
the method you like best.
8.
A limiting reactant problem gives you initial masses of at least two of the reactants and then
asks for the amount of product that can form. Because one doesn’t know which reactant runs
out first and hence determines the mass of product formed, this is a more difficult problem.
The first step in solving the problem is to figure which reactant runs out first (is limiting).
The strategy outlined in the text is to calculate the mole ratio of reactants actually present and
compare this mole ratio to that required from the balanced reaction. Whichever ratio is larger
allows one to deduce the identity of the limiting reactant and can, in turn, be used to calculate
the amount of product formed. Another strategy is to pick one of the reactants and then
calculate the mass of the other reactant required to react with it. By comparing the calculated
mass to the actual mass present in the problem, one can deduce the identity of the limiting
reactant and go on to solve the problem. A third common strategy is to assume each reactant
is limiting and then calculate for each reactant the amount of product that could form. This
gives two or more possible answers. The correct answer is the mass of product that is
smallest. Even though there is enough of the other reactant to form more product, once the
smaller amount of product is formed, the limiting reactant has run out.
9.
Balanced reaction: 2 SO2(g) + O2(g) → 2 SO3(g)
We have 6 SO2 and 6 O2 molecules present. If all six of the SO2 molecules react, then 3
molecules of O2 will react producing 6 molecules of SO3. These numbers were determined
using the balanced reaction. Since 6 molecules of O2 are present, and only 3 react when the
SO2 reacts completely, SO2 is limiting. The product mixture will contain 6 – 6 = 0 SO2
molecules, 6 – 3 = 3 O2 molecules in excess, and 6 molecules of SO3 formed.
96.0 g SO2 ×
1 mol SO 2
1 mol O 2
32.00 g O 2
= 24.0 g O2


64.07 g SO 2
2 mol SO 2
mol O 2
Because 32.0 g O2 are actually present, O2 is in excess and SO2 is the limiting reactant. Note
that if the calculated amount of O2 was greater than 32.0 g, then we would have deduced that
O2 is limiting. Solving the rest of the problem:
CHAPTER 3
STOICHIOMETRY
96.0 g SO2 ×
10.
43
2 mol SO3 80.07 g SO3
1 mol SO 2
= 120. g SO3


64.07 g SO 2
2 mol SO 2
mol SO3
Side reactions may occur. For example, in the combustion of CH4 (methane) to CO2 and
H2O, some CO may also form. Also, reactions only go part way to completion, instead
reaching a state of equilibrium where both reactants and products are present (see Ch. 13).
Questions
19.
isotope
12
C
13
C
mass
12 .000 0 amu
13.034 amu
abundance
98.89%
1.11%
average mass = 0.9889 (12.0000) + 0.0111(13.034) = 12.01 amu
From the relative abundances, there would be 9889 atoms of 12C and 111 atoms of 13C in the
10,000 atom sample. The average mass of carbon is independent of the sample size; it will
always be 12.01 amu.
total mass = 10,000 atoms ×
12.01 amu
= 1.201 × 105 amu
atom
For one mol of carbon (6.0221 × 1023 atoms C), the average mass would still be 12.01 amu.
The number of 12C atoms would be 0.9889 (6.0221 × 1023) = 5.955 × 1023 atoms 12C and the
number of 13C atoms would be 0.0111 (6.0221 × 1023) = 6.68 × 1021 atoms 13C.
total mass = 6.0221 × 1023 atoms ×
12.01 amu
= 7.233 × 1024 amu
atom
total mass in g = 6.0221 × 1023 atoms ×
12.01 amu
1g
×
= 12.01 g/mol
atom
6.0221  10 23 amu
By using the carbon-12 standard to define the relative masses of all of the isotopes as well as
to define the number of things in a mole, then each element’s average atomic mass in units of
grams is the mass of a mole of that element as it is found in nature.
20.
Consider a sample of glucose, C6H12O6. The molar mass of glucose is 180.16 g/mol. The
chemical formula allows one to convert from molecules of glucose to atoms of carbon,
hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole
relationship in the formula. One mol of glucose contains 6 mol C, 12 mol H, and 6 mol O.
Thus, mole conversions between molecules and atoms are possible using the chemical formula. The molar mass allows one to convert between mass and moles of compound and
Avogadro’s number (6.022 × 1023) allows one to convert between moles of compound and
number of molecules.
44
21.
CHAPTER 3
STOICHIOMETRY
6.022  10 23 dollars
mol dollars
23
6.022  10 dollars
1 mol dollars 
mol dollars
= 1 × 1014 dollars/person
9
6  10 people
Avogadro’s number of dollars =
1 trillion = 1,000,000,000,000 = 1 × 1012; Each person would have 100 trillion dollars.
22.
The molar mass is the mass of 1 mol of the compound. The empirical mass is the mass of 1
mol of the empirical formula. The molar mass is a whole number multiple of the empirical
mass. The masses are the same when the molecular formula = empirical formula, and the
masses are different when the two formulas are different. When different, the empirical mass
must be multiplied by the same whole number used to convert the empirical formula to the
molecular formula. For example, C6H12O6 is the molecular formula for glucose and CH2O is
the empirical formula. The whole number multiplier is 6. This same factor of 6 is the
multiplier used to equate the empirical mass (30 g/mol) of glucose to the molar mass (180
g/mol).
23.
The mass percent of a compound is a constant no matter what amount of substance is present.
Compounds always have constant composition.
24.
A balanced reaction starts with the correct formulas of the reactants and products. The coefficients necessary to balance the reaction give molecule relationships as well as mole
relationships between reactants and products. The state (phase) of the reactants and products
is also given. Finally, special reaction conditions are sometimes listed above or below the
arrow. These can include special catalysts used and/or special temperatures required for a
reaction to occur.
25.
The specific information needed is mostly the coefficients in the balanced equation and the
molar masses of the reactants and products. For percent yield, we would need the actual yield
of the reaction and the amounts of reactants used.
a. mass of CB produced = 1.00 × 104 molecules A2B2 ×
1 mol A 2 B2
2 mol CB
molar mass of CB


23
mol CB
6.022  10 molecules A 2 B2 1 mol A 2 B2
b. atoms of A produced = 1.00 × 104 molecules A2B2 ×
c. mol of C reacted = 1.00 × 104 molecules A2B2 ×
2 atoms A
1 molecule A 2 B2
1 mol A 2 B2
×
6.022  10 23 molecules A 2 B2
2 mol C
1 mol A 2 B2
actual mass
× 100; The theoretical mass of CB produced was calculated
theoretical mass
in part a. If the actual mass of CB produced is given, then the percent yield can be
determined for the reaction using the % yield equation.
d. % yield =
CHAPTER 3
26.
STOICHIOMETRY
45
One method is to determine the actual mole ratio of XY to Y2 present and compare this ratio
to the required 2:1 mole ratio from the balanced equation. Which ratio is larger will allow one
to deduce the limiting reactant. Once the identity of the limiting reactant is known, then one
can calculate the amount of product formed. A second method would be to pick one of the
reactants and then calculate how much of the other reactant would be required to react with it
all. How the answer compares to the actual amount of that reactant present allows one to
deduce the identity of the limiting reactant. Once the identity is known, one would take the
limiting reactant and convert it to mass of product formed.
When each reactant is assumed limiting and the amount of product is calculated, there are
two possible answers (assuming two reactants). The correct answer (the amount of product
that could be produced) is always the smaller number. Even though there is enough of the
other reactant to form more product, once the small quantity is reached, the limiting reactant
runs out and the reaction cannot continue.
Exercises
Atomic Masses and the Mass Spectrometer
27.
A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)
A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; From the periodic table, the element is Pb.
28.
A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) +
0.0540(49.944792) = 47.88 amu
This is element Ti (titanium).
29.
Let A = mass of 185Re:
186.207 = 0.6260(186.956) + 0.3740(A), 186.207  117.0 = 0.3740(A)
A=
30.
69.2
= 185 amu (A = 184.95 amu without rounding to proper significant figures.)
0.3740
abundance 28Si = 100.00 (4.70 + 3.09) = 92.21%; From the periodic table, the average
atomic mass of Si is 28.09 amu.
28.09 = 0.9221(27.98) + 0.0470 (atomic mass 29Si) + 0.0309(29.97)
atomic mass 29Si = 29.01
The mass of 29Si is actually a little less than 29 amu. There are other isotopes of silicon that
are considered when determining the 28.09 amu average atomic mass of Si listed in the
atomic table.
31.
There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with
two isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br 2
molecule composed of two atoms of the lighter isotope. This isotope has mass equal to
46
CHAPTER 3
STOICHIOMETRY
157.84/2 or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to
161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in
order of increasing mass. The intensities of the highest and lowest mass tell us the two
isotopes are present in about equal abundance. The actual abundance is 50.69% 79Br and
49.31% 81Br. The calculation of the abundance from the mass spectrum is beyond the scope
of this text.
32.
GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have 2 peaks at 144
(= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60:40 or 3:2.
144
146
Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have 3 peaks at
288, 290, and 292 with intensities in the ratio of 36:48:16 or 9:12:4. We get this ratio from
the following probability table:
69
Ga (0.60)
71
Ga (0.40)
69
0.36
0.24
71
0.24
0.16
Ga (0.60)
Ga (0.40)
288
290
292
Moles and Molar Masses
33.
When more than one conversion factor is necessary to determine the answer, we will usually
put all the conversion factors into one calculation instead of determining intermediate
answers. This method reduces round-off error and is a time saver.
500. atoms Fe 
34.
500.0 g Fe ×
1 mol Fe
55.85 g Fe
= 4.64 × 10 20 g Fe

23
mol Fe
6.022  10 atoms Fe
1 mol Fe
= 8.953 mol Fe
55.85 g Fe
8.953 mol Fe ×
6.022  10 23 atoms Fe
= 5.391 × 1024 atoms Fe
mol Fe
CHAPTER 3
STOICHIOMETRY
0.200 g C 1 mol C
6.022  10 23 atoms C


= 1.00 × 1022 atoms C
carat
12.01 g C
mol C
35.
1.00 carat ×
36.
5.0 × 1021 atoms C ×
8.3 × 10 3 mol C ×
37.
47
1 mol C
6.022  10 23 atoms C
= 8.3 × 10 3 mol C
12.01 g C
= 0.10 g C
mol C
Al2O3: 2(26.98) + 3(16.00) = 101.96 g/mol
Na3AlF6: 3(22.99) + 1(26.98) + 6(19.00) = 209.95 g/mol
38.
HFC  134a, CH2FCF3: 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/mol
HCFC 124, CHClFCF3: 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol
39.
a. The formula is NH3. 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol
b. The formula is N2H4. 2(14.01) + 4(1.008) = 32.05 g/mol
c. (NH4)2Cr2O7: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol
40.
a. The formula is P4O6. 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol
b. Ca3(PO4)2: 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol
c. Na2HPO4: 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol
41.
a. 1.00 g NH3 ×
b. 1.00 g N2H4 ×
1 mol NH3
= 0.0587 mol NH3
17.03 g NH3
1 mol N 2 H 4
= 0.0312 mol N2H4
32.05 g N 2 H 4
c. 1.00 g (NH4)2Cr2O7 ×
42.
1 mol ( NH4 ) 2 Cr2O7
= 3.97 × 10 3 mol (NH4)2Cr2O7
252.08 g ( NH4 ) 2 Cr2O7
1 mol P 4 O 6
= 4.55 × 10 3 mol P4O6
219 .88 g
1 mol Ca 3 (PO4 ) 2
b. 1.00 g Ca3(PO4)2 ×
= 3.22 × 10 3 mol Ca3(PO4)2
310 .18 g
a. 1.00 g P4O6 ×
c. 1.00 g Na2HPO4 ×
1 mol Na 2 HPO4
= 7.04 × 10 3 mol Na2HPO4
141 .96 g
48
43.
CHAPTER 3
a. 5.00 mol NH3 ×
b. 5.00 mol N2H4 ×
17.03 g NH 4
= 85.2 g NH3
mol NH 3
32.05 g N 2 H 4
= 160. g N2H4
mol N 2 H 4
c. 5.00 mol (NH4)2Cr2O7 ×
44.
a. 5.00 mol P4O6 ×
STOICHIOMETRY
252.08 g ( NH 4 ) 2 Cr2 O 7
= 1260 g (NH4)2Cr2O7
1 mol ( NH 4 ) 2 Cr2 O 7
219 .88 g
= 1.10 × 103 g P4O6
1 mol P 4 O 6
310 .18 g
= 1.55 × 103 g Ca3(PO4)2
mol Ca 3 (PO4 ) 2
141 .96 g
c. 5.00 mol Na2HPO4 ×
= 7.10 × 102 g Na2HPO4
mol Na 2 HPO4
b. 5.00 mol Ca3(PO4)2 ×
45.
Chemical formulas give atom ratios as well as mol ratios.
a. 5.00 mol NH3 ×
1 mol N 14.01 g N
= 70.1 g N

mol NH3
mol N
2 mol N 14.01 g N
= 140. g N

mol N 2 H 4
mol N
2 mol N
14.01 g N
c. 5.00 mol (NH4)2Cr2O7 ×
= 140. g N

mol ( NH4 ) 2 Cr2O7
mol N
b. 5.00 mol N2H4 ×
46.
47.
a. 5.00 mol P4O6 ×
4 mol P 30.97 g P
= 619 g P

mol P4 O 6
mol P
b. 5.00 mol Ca3(PO4)2 ×
2 mol P
30.97 g P
= 310. g P

mol Ca 3 (PO4 ) 2
mol P
c. 5.00 mol Na2HPO4 ×
1 mol P
30.97 g P
= 155 g P

mol Na 2 HPO4
mol P
a. 1.00 g NH3 ×
b. 1.00 g N2H4 ×
1 mol NH3
6.022  10 23 molecules NH3

= 3.54 × 1022 molecules NH3
17.03 g NH3
mol NH3
1 mol N 2 H 4
6.022  10 23 molecules N 2 H 4

32.05 g N 2 H 4
mol N 2 H 4
= 1.88 × 1022 molecules N2H4
CHAPTER 3
STOICHIOMETRY
c. 1.00 g (NH4)2Cr2O7 ×

48.
49
1 mol ( NH4 ) 2 Cr2O7
252.08 g ( NH4 ) 2 Cr2O7
6.022  10 23 formula units ( NH 4 ) 2 Cr2 O 7
= 2.39 × 1021 formula units (NH4)2Cr2O7
mol ( NH 4 ) 2 Cr2 O 7
a. 1.00 g P4O6 ×
1 mol P4 O 6 6.022  10 23 molecules

= 2.74 × 1021 molecules P4O6
219 .88 g
mol P4 O 6
b. 1.00 g Ca3(PO4)2 ×
1 mol Ca 3 (PO 4 ) 2 6.022  10 23 formula units

310 .18 g
mol Ca 3 (PO 4 ) 2
= 1.94 × 1021 formula units Ca3(PO4)2
c. 1.00 g Na2HPO4 ×
1 mol Na 2 HPO4 6.022  10 23 formula units

141 .96 g
mol Na 2 HPO4
= 4.24 × 1021 formula units Na2HPO4
49.
Using answers from Exercise 47:
a. 3.54 × 1022 molecules NH3 ×
b. 1.88 × 1022 molecules N2H4 ×
1 atom N
= 3.54 × 1022 atoms N
molecule NH3
2 atoms N
= 3.76 × 1022 atoms N
molecule N 2 H 4
c. 2.39 × 1021 formula units (NH4)2Cr2O7 ×
2 atoms N
formula unit ( NH 4 ) 2 Cr2 O 7
= 4.78 × 1021 atoms N
50.
Using answers from Exercise 48:
a. 2.74 × 1021 molecules P4O6 ×
51.
4 atoms P
= 1.10 × 1022 atoms P
molecule P4 O 6
b. 1.94 × 1021 formula units Ca3(PO4)2 ×
2 atoms P
= 3.88 × 1021 atoms P
formula unit Ca 3 (PO4 ) 2
c. 4.24 × 1021 formula units Na2HPO4 ×
1 atom P
= 4.24 × 1021 atoms P
formula unit Na 2 HPO4
Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol
500.0 mg ×
1g
1 mol
= 2.839 × 10 3 mol

1000 mg 176 .12 g
50
CHAPTER 3
2.839 × 10-3 mol ×
52.
6.022  10 23 molecules
= 1.710 × 1021 molecules
mol
a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol
b. 500. mg ×
1g
1 mol
= 2.78 × 10 3 mol

1000 mg 180 .15 g
2.78 × 10 3 mol ×
53.
a. 150.0 g Fe2O3 ×
b. 10.0 mg NO2 ×
6.022  10 23 molecules
= 1.67 × 1021 molecules
mol
1 mol
= 0.9393 mol Fe2O3
159.70 g
1g
1 mol
= 2.17 × 10 4 mol NO2

1000 mg 46.01 g
c. 1.5 × 1016 molecules BF3 ×
54.
STOICHIOMETRY
1 mol
= 2.5 × 10 8 mol BF3
23
6.02  10 molecules
1g
1 mol
= 1.03 × 10 4 mol C8H10N4O2

1000 mg 194.20 g
1 mol
b. 2.72 × 1021 molecules C2H5OH ×
6.022  10 23 molecules
a. 20.0 mg C8H10N4O2 ×
= 4.52 × 10 3 mol C2H5OH
c. 1.50 g CO2 ×
55.
1 mol
= 3.41 × 10 2 mol CO2
44.01 g
a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to
show how these conversion factors can be used.
Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol
5.00 g C2H5O2N ×
1 mol C 2 H 5 O 2 N
6.022  10 23 molecules C 2 H 5 O 2 N

×
75.07 g C 2 H 5 O 2 N
mol C 2 H 5 O 2 N
1 atom N
= 4.01 × 1022 atoms N
molecule C 2 H 5 O 2 N
b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol
5.00 g Mg3N2 ×
1 mol Mg 3 N 2
6.022  10 23 formula units Mg 3 N 2
2 atoms N


100 .95 g Mg 3 N 2
mol Mg 3 N 2
mol Mg 3 N 2
= 5.97 × 1022 atoms N
CHAPTER 3
STOICHIOMETRY
51
c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol
5.00 g Ca(NO3)2 ×
2 mols N
1 mol Ca ( NO 3 ) 2
6.022  10 23 atoms N


164 .10 g Ca ( NO 3 ) 2
mol Ca ( NO 3 ) 2
mol N
= 3.67 × 1022 atoms N
d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol
5.00 g N2O4 ×
56.
4.24 g C6H6 ×
2 mol N
1 mol N 2 O 4
6.022  10 23 atoms N


92.02 g N 2 O 4 mol N 2 O 4
mol N
= 6.54 × 1022 atoms N
1 mol
= 5.43 × 10 2 mol C6H6
78.11 g
6.022  10 23 molecules
= 3.27 × 1022 molecules C6H6
mol
5.43 × 10 2 mol C6H6 ×
Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 total atoms.
3.27 × 1022 molecules C6H6 ×
12 atoms total
= 3.92 × 1023 atoms total
molecule
0.224 mol H2O ×
18.02 g
= 4.04 g H2O
mol
0.224 mol H2O ×
6.022  10 23 molecules
= 1.35 × 1023 molecules H2O
mol
1.35 × 1023 molecules H2O ×
3 atoms total
= 4.05 × 1023 atoms total
molecule
2.71 × 1022 molecules CO2 ×
1 mol
= 4.50 × 10 2 mol CO2
23
6.022  10 molecules
4.50 × 10 2 mol CO2 ×
44.01 g
= 1.98 g CO2
mol
2.71 × 1022 molecules CO2 ×
3.35 × 1022 atoms total ×
3 atoms total
= 8.13 × 1022 atoms total
molecule CO2
1 molecule
= 5.58 × 1021 molecules CH3OH
6 atoms total
5.58 × 1021 molecules CH3OH ×
1 mol
= 9.27 × 10 3 mol CH3OH
6.022  10 23 molecules
52
CHAPTER 3
9.27 × 10 3 mol CH3OH ×
57.
STOICHIOMETRY
32.04 g
= 0.297 g CH3OH
mol
 12.01 g 
 + 18 mol H
a. 14 mol C 
 mol C 
 1.008 g 

 + 2 mol N
 mol H 
 14.01 g 

 + 5 mol O
 mol N 
 16.00 g 


 mol O 
= 294.30 g/mol
b. 10.0 g aspartame ×
c. 1.56 mol ×
d. 5.0 mg ×
1 mol
= 3.40 × 10 2 mol
294 .30 g
294 .30 g
= 459 g
mol
1g
1 mol
6.02  10 23 molecules


= 1.0 × 1019 molecules
1000 mg 294 .30 g
mol
e. The chemical formula tells us that 1 molecule of aspartame contains two atoms of N.
The chemical formula also says that 1 mol of aspartame contains two mol of N.
1 mol aspartame
2 mol N
6.02  10 23 atoms N


294 .30 g aspartame mol aspartame
mol N
1.2 g aspartame ×
= 4.9 × 1021 atoms of nitrogen
f.
1.0 × 109 molecules ×
g. 1 molecule aspartame ×
58.
1 mol
294.30 g
= 4.9 × 10 13 g or 490 fg

23
mol
6.02  10 molecules
1 mol
6.022  10
23
molecules

294.30 g
= 4.887 × 10 22 g
mol
a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b. 500.0 g ×
1 mol
= 3.023 mol
165.39 g
d. 5.0 g C2H3Cl3O2 ×
c.
2.0 × 10 2 mol ×
165.39 g
= 3.3 g
mol
1 mol
6.02  10 23 molecules 3 atoms Cl


165 .39 g
mol
molecule
= 5.5 × 1022 atoms of chlorine
e. 1.0 g Cl ×
f.
1 mol Cl 1 mol C2 H3Cl3O2 165.39 g C2 H3Cl3O2
= 1.6 g chloral hydrate


35.45 g
3 mol Cl
mol C2 H3Cl3O 2
500 molecules ×
1 mol
165.39 g
= 1.373 × 10 19 g

23
mol
6.022  10 molecules
CHAPTER 3
STOICHIOMETRY
53
Percent Composition
59.
a. C3H4O2: Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06
g/mol
36.03 g C
4.032 g H
× 100 = 50.00% C; %H =
× 100
72.06 g compound
72.06 g compound
= 5.595% H
32.00 g
%O = 100.00 - (50.00 + 5.595) = 44.41% O or %O =
× 100 = 44.41% O
72.06 g
%C =
b. C4H6O2: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00
= 86.09 g/mol
%C =
48.04 g
6.048 g
× 100 = 55.80% C; %H =
× 100 = 7.025% H
86.09 g
86.09 g
%O = 100.00 - (55.80 + 7.025) = 37.18% O
c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01
= 53.06 g/mol
60.
%C =
36.03 g
× 100 = 67.90% C;
53.06 g
%N =
14.01 g
× 100 = 26.40% N or %N = 100.00  (67.90 + 5.699) = 26.40% N
53.06 g
%H =
3.024 g
× 100 = 5.699% H
53.06 g
molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol
%C =
20(12.01) g C
× 100 = 71.40% C
336 .43 g compound
%H =
29(1.008) g H
× 100 = 8.689% H
336 .43 g compound
%F =
19.00 g F
× 100 = 5.648% F
336 .43 g compound
%O = 100.00  (71.40 + 8.689 + 5.648) = 14.26% O or:
%O =
61.
3(16.00) g O
× 100 = 14.27% O
336 .43 g compound
a. NO: %N =
14.01 g N
× 100 = 46.68% N
30.01 g NO
54
CHAPTER 3
b. NO2: %N =
14.01 g N
× 100 = 30.45% N
46.01 g NO2
c. N2O4: %N =
28.02 g N
× 100 = 30.45% N
92.02 g N 2 O 4
d. N2O: %N =
28.02 g N
× 100 = 63.65% N
44.02 g N 2 O
STOICHIOMETRY
The order from lowest to highest mass percentage of nitrogen is: NO2 = N2O4 < NO < N2O.
62.
C8H10N4O2: molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol
%C =
8(12.01) g C
96.08
× 100 =
194.20 g C 8 H10 N 4 O 2
194 .20
× 100 = 49.47% C
C12 H22O11: molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol
%C =
12(12.01) g C
× 100 = 42.10% C
342 .30 g C12 H 22 O11
C2H5OH: molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol
%C =
2(12.01) g C
× 100 = 52.14% C
46.07 g C 2 H 5 OH
The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11)
< caffeine (C8H10N4O2) < ethanol (C2H5OH)
63.
There are many valid methods to solve this problem. We will assume 100.00 g of compound,
then determine from the information in the problem how many mol of compound equals
100.00 g of compound. From this information, we can determine the mass of one mol of
compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:
mol cyanocobalamin = 4.34 g Co ×
1 mol Co
1 mol cyanocobalamin

58.93 g Co
mol Co
= 7.36 × 10 2 mol cyanocobalamin
x g cyanocobalamin
100 .00 g
=
, x = molar mass = 1360 g/mol
1 mol cyanocobalamin
7.36  10 2 mol
64.
There are 0.390 g Cu for every 100.00 g of fungal laccase. Assuming 100.00 g fungal
laccase:
CHAPTER 3
STOICHIOMETRY
mol fungal laccase = 0.390 g Cu ×
55
1 mol Cu
1 mol fungal laccase
= 1.53 × 10 3 mol

63.55 g Cu
4 mol Cu
x g fungal laccase
100 .00 g
=
, x = molar mass = 6.54 × 104 g/mol
1 mol fungal laccase 1.53  10 3 mol
Empirical and Molecular Formulas
65.
 12.01 g C 
 + 2 mol H
a. Molar mass of CH2O = 1 mol C 
 mol C 
 1.008 g H 


 mol H 
 16.00 g O 
 = 30.03 g/mol
+ 1 mol O 
 mol O 
%C =
12.01 g C
2.016 g H
× 100 = 39.99% C; %H =
× 100 = 6.713% H
30.03 g CH 2 O
30.03 g CH 2 O
%O =
16.00 g O
× 100 = 53.28% O or %O = 100.00 - (39.99 + 6.713) = 53.30%
30.03 g CH 2 O
b. Molar Mass of C6H12O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol
%C =
76.06 g C
12.(1.008) g
× 100 = 40.00%; %H =
× 100 = 6.714%
180.16 g C 6 H12 O 6
180 .16 g
%O = 100.00 - (40.00 + 6.714) = 53.29%
c. Molar mass of HC2H3O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol
%C =
24.02 g
4.032 g
× 100 = 40.00%; %H =
× 100 = 6.714%
60.05 g
60.05 g
%O = 100.00 - (40.00 + 6.714) = 53.29%
66.
All three compounds have the same empirical formula, CH2O, and different molecular
formulas. The composition of all three in mass percent is also the same (within rounding
differences). Therefore, elemental analysis will give us only the empirical formula.
67.
a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the
empirical formula) is NO2.
b. Molecular formula: C3H6; empirical formula = CH2
c. Molecular formula: P4O10; empirical formula = P2O5
d. Molecular formula: C6H12O6; empirical formula = CH2O
56
68.
CHAPTER 3
STOICHIOMETRY
a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g
188.35 g
= 4.000; So the molecular formula is (SNH)4 or S4N4H4.
47.09 g
b. NPCl2: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol
347 .64 g
= 3.0000; Molecular formula is (NPCl2)3 or N3P3Cl6.
115 .88 g
c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol
341 .94 g
= 2.0000; Molecular formula: Co2C8O8
170 .97 g
184.32 g
d. SN: 32.07 + 14.01 = 46.08 g/mol;
= 4.000; Molecular formula: S4N4
46.08 g
69.
Out of 100.00 g of adrenaline, there are:
56.79 g C ×
1 mol C
1 mol H
= 4.729 mol C; 6.56 g H ×
= 6.51 mol H
12.01 g C
1.008 g H
28.37 g O ×
1 mol O
1 mol N
= 1.773 mol O; 8.28 g N ×
= 0.591 mol N
16.00 g O
14.01 g N
Dividing each mol value by the smallest number:
4.729
6.51
1.773
0.591
= 8.00;
= 11.0;
= 3.00;
= 1.00
0.591
0.591
0.591
0.591
This gives adrenaline an empirical formula of C8H11O3N.
70.
Assuming 100.00 g of nylon-6:
63.68 g C ×
9.80 g H ×
1 mol C
1 mol N
= 5.302 mol C; 12.38 g N ×
= 0.8837 mol N
12.01 g C
14.01 g N
1 mol H
1 mol O
= 9.72 mol H; 14.14 g O ×
= 0.8838 mol O
1.008 g H
16.00 g O
Dividing each mol value by the smallest number:
5.302
9.72
= 6.000;
= 11.0
0.8837
0.8837
The empirical formula for nylon-6 is C6H11NO
CHAPTER 3
71.
STOICHIOMETRY
57
Compound I: mass O = 0.6498 g HgxOy  0.6018 g Hg = 0.0480 g O
0.6018 g Hg ×
0.0480 g O ×
1 mol Hg
= 3.000 × 10 3 mol Hg
200.6 g Hg
1 mol O
= 3.00 × 10 3 mol O
16.00 g O
The mol ratio between Hg and O is 1:1, so the empirical formula of compound I is HgO.
Compound II: mass Hg = 0.4172 g HgxOy  0.016 g O = 0.401 g Hg
1 mol Hg
1 mol O
= 2.00 × 10 3 mol Hg; 0.016 g O ×
200.6 g Hg
16.00 g O
0.401 g Hg ×
= 1.0 × 10 3 mol O
The mol ratio between Hg and O is 2:1, so the empirical formula is Hg2O.
72.
1.121 g N ×
0.480 g C ×
1 mol N
1 mol H
= 8.001 × 10 2 mol N; 0.161 g H ×
= 1.60 × 10 1 mol H
14.01 g N
1.008 g H
1 mol C
1 mol O
= 4.00 × 10 2 mol C; 0.640 g O ×
= 4.00 × 10 2 mol O
12.01 g C
16.00 g O
Dividing all mol values by the smallest number:
8.001  10 2
= 2.00;
4.00  10  2
1.60  10 1
4.00  10 2
=
4.00;
= 1.00
4.00  10  2
4.00  10  2
Empirical formula = N2H4CO
73.
Out of 100.0 g, there are:
69.6 g S ×
1 mol S
1 mol N
= 2.17 mol S; 30.4 g N ×
= 2.17 mol N
32.07 g S
14.01 g N
Empirical formula is SN since mol values are in a 1:1 mol ratio.
The empirical formula mass of SN is ~ 46 g. Because 184/46 = 4.0, the molecular formula is
S4N4.
74.
Assuming 100.0 g of compound:
26.7 g P ×
1 mol P
30.97 g P
= 0.862 mol P; 12.1 g N ×
1 mol N
14.01 g N
= 0.864 mol N
58
CHAPTER 3
61.2 g Cl ×
1 mol Cl
35.45 g Cl
STOICHIOMETRY
= 1.73 mol Cl
1.73
= 2.01; Empirical formula = PNCl2
0.862
The empirical formula mass is ~31.0 + 14.0 + 2(35.5) = 116
molar mass
580
= 5; molecular formula = (PNCl2)5 =P5N5Cl10

empirical formula mass
116
75.
Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 49.31 g C - 43.79 g O
= 6.90 g H):
49.31 g C ×
1 mol C
12.01 g C
= 4.106 mol C; 6.90 g H × ×
43.79 g O ×
1 mol O
16.00 g O
= 2.737 mol O
1 mol H
= 6.85 mol H
1.008 g H
Dividing all mole values by 2.737 gives:
4.106
6.85
2.737
= 1.500;
= 2.50;
= 1.000
2.737
2.737
2.737
Since a whole number ratio is required, the empirical formula is C3H5O2.
The empirical formula mass is: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol
molar mass
146.1
=
= 1.999; molecular formula = (C3H5O2)2 = C6H10O4
empirical formula mass
73.07
76.
Assuming 100.00 g of compound (mass oxygen = 100.00 g  41.39 g C  3.47 g H
= 55.14 g O):
41.39 g C ×
55.14 g O ×
1 mol C
12.01 g C
1 mol O
16.00 g O
= 3.446 mol C; 3.47 g H ×
1 mol H
1.008 g H
= 3.44 mol H
= 3.446 mol O
All are the same mol values so the empirical formula is CHO. The empirical formula mass is
12.01 + 1.008 + 16.00 = 29.02 g/mol.
molar mass =
15.0 g
0.129 mol
= 116 g/mol
CHAPTER 3
77.
STOICHIOMETRY
59
molar mass
116
= 4.00; molecular formula = (CHO)4 = C4H4O4

empirical mass 29.02
When combustion data are given, it is assumed that all the carbon in the compound ends up
as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the
sample of propane combusted, the moles of C and H are:
mol C = 2.641 g CO2 ×
1 mol CO 2
1 mol C
= 0.06001 mol C

44.01 g CO 2
mol CO 2
mol H = 1.442 g H2O ×
1 mol H 2 O
2 mol H
= 0.1600 mol H

18.02 g H 2 O
mol H 2 O
mol H 0.1600
= 2.666

mol C 0.06001
Multiplying this ratio by three gives the empirical formula of C3H8.
78.
This compound contains nitrogen, and one way to determine the amount of nitrogen in the
compound is to calculate composition by mass percent. We assume that all of the carbon in
33.5 mg CO2 came from the 35.0 mg of compound and all of the hydrogen in 41.1 mg H2O
came from the 35.0 mg of compound.
3.35 × 10 2 g CO2 ×
%C =
9.14  10 3 g C
3.50  10  2 g compound
4.11 × 10 2 g H2O ×
%H =
1 mol CO 2
1 mol C 12.01 g C
= 9.14 × 10 3 g C


44.01 g CO 2
mol CO 2
mol C
× 100 = 26.1% C
1 mol H 2 O
2 mol H
1.008 g H
= 4.60 × 10 3 g H


18.02 g H 2 O
mol H 2 O
mol H
4.60  10 3 g H
3.50  10  2 g compound
× 100 = 13.1% H
The mass percent of nitrogen is obtained by difference:
%N = 100.0 - (26.1 + 13.1) = 60.8% N
Now perform the empirical formula determination by first assuming 100.0 g of compound.
Out of 100.0 g of compound, there are:
26.1 g C ×
1 mol C
= 2.17 mol C; 13.1 g H ×
12.01 g C
60.8 g N ×
1 mol N
= 4.34 mol N
14.01 g N
1 mol H
= 13.0 mol H
1.008 g H
60
CHAPTER 3
Dividing all mol values by 2.17 gives:
79.
STOICHIOMETRY
2.17
13.0
4.34
= 1.00;
= 5.99;
= 2.00
2.17
2.17
2.17
The empirical formula is CH6N2.
The combustion data allow determination of the amount of hydrogen in cumene. One way to
determine the amount of carbon in cumene is to determine the mass percent of hydrogen in
the compound from the data in the problem; then determine the mass percent of carbon by
difference (100.0 - mass %H = mass %C).
42.8 mg H2O ×
%H =
1g
1000 mg

2.016 g H
1000 mg
= 4.79 mg H

18.02 g H 2 O
g
4.79 mg H
× 100 = 10.1% H; %C = 100.0 - 10.1 = 89.9% C
47.6 mg cumene
Now solve this empirical formula problem. Out of 100.0 g cumene, we have:
89.9 g C ×
1 mol C
= 7.49 mol C; 10.1 g H ×
12.01 g C
1 mol H
= 10.0 mol H
1.008 g H
10.0
4
= 1.34  , i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C3H4
3
7.49
Empirical formula mass ≈ 3(12) + 4(1) = 40 g/mol
The molecular formula is (C3H4)3 or C9H12 since the molar mass will be between 115 and
125 g/mol (molar mass ≈ 3 × 40 g/mol = 120 g/mol).
80.
First, we will determine composition by mass percent:
1g
12.01 g C
1000 mg
= 4.369 mg C


1000 mg
44.01 g CO 2
g
4.369 mg C
%C =
× 100 = 40.91% C
10.68 mg compound
16.01 mg CO2 ×
4.37 mg H2O ×
%H =
1g
1000 mg

2.016 g H
1000 mg
= 0.489 mg H

18.02 g H 2 O
g
0.489 mg
× 100 = 4.58% H; %O = 100.00 - (40.91 + 4.58) = 54.51% O
10.68 mg
So, in 100.00 g of the compound, we have:
40.91 g C ×
1 mol C
1 mol H
= 3.406 mol C; 4.58 g H ×
= 4.54 mol H
12.01 g C
1.008 g H
CHAPTER 3
STOICHIOMETRY
61
1 mol O
= 3.407 mol O
16.00 g O
4.54
4
Dividing by the smallest number:
= 1.33  ; the empirical formula is C3H4O3.
3.406
3
The empirical formula mass of C3H4O3 is ≈ 3(12) + 4(1) + 3(16) = 88 g.
54.51 g O ×
Because
176.1
= 2.0, the molecular formula is C6H8O6.
88
Balancing Chemical Equations
81.
When balancing reactions, start with elements that appear in only one of the reactants and one
of the products, then go on to balance the remaining elements.
a. C6H12O6(s) + O2(g) → CO2(g) + H2O(g)
Balance C atoms: C6H12O6 + O2 → 6 CO2 + H2O
Balance H atoms: C6H12O6 + O2 → 6 CO2 + 6 H2O
Lastly, balance O atoms: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
b. Fe2S3(s) + HCl(g) → FeCl3(s) + H2S(g)
Balance Fe atoms: Fe2S3 + HCl → 2 FeCl3 + H2S
Balance S atoms: Fe2S3 + HCl → 2 FeCl3 + 3 H2S
There are 6 H and 6 Cl on right, so balance with 6 HCl on left:
Fe2S3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g).
c. CS2(l) + NH3(g) → H2S(g) + NH4SCN(s)
C and S balanced; balance N:
CS2 + 2 NH3 → H2S + NH4SCN
H is also balanced. So: CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s)
82.
One of the most important parts to this problem is writing out correct formulas. If the
formulas are incorrect, then the balanced reaction is incorrect.
a. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq)
c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
62
CHAPTER 3
STOICHIOMETRY
d. Sr(OH)2(aq) + 2 HBr(aq) → 2H2O(l) + SrBr2(aq)
83.
a. 3 Ca(OH)2(aq) + 2 H3PO4(aq) → 6 H2O(l) + Ca3(PO4)2(s)
b. Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l)
c. 2 AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2 HNO3(aq)
84.
a. 2 KO2(s) + 2 H2O(l) → 2 KOH(aq) + O2(g) + H2O2(aq) or
4 KO2(s) + 6 H2O(l) → 4 KOH(aq) + O2(g) + 4 H2O2(aq)
b. Fe2O3(s) + 6 HNO3(aq) → 2 Fe(NO3)3(aq) + 3 H2O(l)
c. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
d. PCl5(l) + 4 H2O(l) → H3PO4(aq) + 5 HCl(g)
e. 2 CaO(s) + 5 C(s) → 2 CaC2(s) + CO2(g)
f.
2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g)
g. FeCO3(s) + H2CO3(aq) → Fe(HCO3)2(aq)
85.
a. The formulas of the reactants and products are C6H6(l) + O2(g) → CO2(g) + H2O(g). To
balance this combustion reaction, notice that all of the carbon in C6H6 has to end up as
carbon in CO2 and all of the hydrogen in C6H6 has to end up as hydrogen in H2O. To
balance C and H, we need 6 CO2 molecules and 3 H2O molecules for every 1 molecule of
C6H6. We do oxygen last. Because we have 15 oxygen atoms in 6 CO2 molecules and 3
H2O molecules, we need 15/2 O2 molecules in order to have 15 oxygen atoms on the
reactant side.
C6H6(l) +
15
2
O2(g) → 6 CO2(g) + 3 H2O(g); Multiply by two to give whole numbers.
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)
b. The formulas of the reactants and products are C4H10(g) + O2(g) → CO2(g) + H2O(g).
C4H10(g) +
13
2
O2(g) → 4 CO2(g) + 5 H2O(g); Multiply by two to give whole numbers.
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
c. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g)
d. 2 Fe(s) +
3
2
e. 2 FeO(s) +
O2(g) → Fe2O3(s); For whole numbers: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
1
2
O2(g) → Fe2O3(s); For whole numbers, multiply by two.
4 FeO(s) + O2(g) → 2 Fe2O3(s)
CHAPTER 3
86.
STOICHIOMETRY
63
a. 16 Cr(s) + 3 S8(s) → 8 Cr2S3(s)
b. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
c. 2 KClO3(s) → 2 KCl(s) + 3 O2(g)
d. 2 Eu(s) + 6 HF(g) → 2 EuF3(s) + 3 H2(g)
87.
a. SiO2(s) + C(s) → Si(s) + CO(g)
Balance oxygen atoms: SiO2 + C → Si + 2 CO
Balance carbon atoms: SiO2(s) + 2 C(s) → Si(s) + 2 CO(g)
b. SiCl4(l) + Mg(s) → Si(s) + MgCl2(s)
Balance Cl atoms: SiCl4 + Mg → Si + 2 MgCl2
Balance Mg atoms: SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s)
c. Na2SiF6(s) + Na(s) → Si(s) + NaF(s)
Balance F atoms:
Na2SiF6 + Na → Si + 6 NaF
Balance Na atoms: Na2SiF6(s) + 4 Na(s) → Si(s) + 6 NaF(s)
88.
CaSiO3(s) + 6 HF(aq) → CaF2(aq) + SiF4(g) + 3 H2O(l)
Reaction Stoichiometry
89.
The stepwise method to solve stoichiometry problems is outlined in the text. Instead of
calculating intermediate answers for each step, we will combine conversion factors into one
calculation. This practice reduces round-off error and saves time.
Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s)
15.0 g Fe ×
90.
2 mol Al 26.98 g Al
1 mol Fe
= 0.269 mol Fe; 0.269 mol Fe ×
= 7.26 g Al

55.85 g Fe
2 mol Fe
mol Al
0.269 mol Fe ×
1 mol Fe 2 O 3 159.70 g Fe 2 O 3
= 21.5 g Fe2O3

2 mol Fe
mol Fe 2 O 3
0.269 mol Fe ×
1 mol Al2O3 101.96 g Al2O3
= 13.7 g Al2O3

2 mol Fe
mol Al2O3
10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s)
64
CHAPTER 3
52.9 g KClO3 ×
STOICHIOMETRY
1 mol KClO 3
3 mol P4O10
283.88 g P4O10
= 36.8 g P4O10


122.55 g KClO 3 10 mol KClO 3
mol P4O10
3 mol NH4ClO 4 117.49 g NH4ClO 4
1000 g Al
1 mol Al
= 4355 g



kg Al
26.98 g Al
3 mol Al
mol NH4ClO 4
91.
1.000 kg Al ×
92.
a. Ba(OH)2  8H2O(s) + 2 NH4SCN(s) → Ba(SCN)2(s) + 10 H2O(l) + 2 NH3(g)
b. 6.5 g Ba(OH)2  8H2O ×
1 mol Ba(OH) 2  8H 2 O
= 0.0206 mol = 0.021 mol
315.4 g
0.021 mol Ba(OH)2  8H2O ×
2 mol NH4SCN
76.13 g NH4SCN

1 mol Ba(OH) 2  8H 2O
mol NH4SCN
= 3.2 g NH4SCN
93.
1.0 × 104 kg waste ×


1 mol C5 H 7 O 2 N
3.0 g NH 4
1 mol NH 4
1000 g



×


100 kg waste
kg
18.04 g NH 4
55 mol NH 4
113.12 g C5 H 7 O 2 N
= 3.4 × 104 g tissue if all NH4+ converted
mol C5 H 7 O 2 N
Since only 95% of the NH4+ ions react:
mass of tissue = (0.95) (3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue
94.
1.0 × 103 g phosphorite ×
75 g Ca 3 (PO4 ) 2
1 mol Ca 3 (PO4 ) 2


100 g phosphorit e
310.18 g Ca 3 (PO4 ) 2
1 mol P4
2 mol Ca 3 ( PO4 ) 2
95.
a.
1.00 × 102 g C7H6O3 ×
×
123 .88 g P4
= 150 g P4
mol P4
1 mol C 7 H 6 O 3
1 mol C 4 H 6 O 3 102.09 g C 4 H 6 O 3


138.12 g C 7 H 6 O 3 1 mol C 7 H 6 O 3
1 mol C 4 H 6 O 3
= 73.9 g C4H6O3
b.
1.00 × 102 g C7H6O3 ×
1 mol C 7 H 6 O 3
1 mol C 9 H 8 O 4 180.15 g C 9 H 8 O 4


138.12 g C 7 H 6 O 3 1 mol C 7 H 6 O 3
mol C 9 H 8 O 4
= 1.30 × 102 g aspirin
96.
2 LiOH(s) + CO2(g) → Li2CO3(aq) + H2O(l)
The total volume of air exhaled each minute for the 7 astronauts is 7 × 20. = 140 L/min.
CHAPTER 3
STOICHIOMETRY
25,000 g LiOH ×
65
1 mol CO2
44.01 g CO2
1 mol LiOH
100 g air




23.95 g LiOH
2 mol LiOH
mol CO2
4.0 g CO2
1 mL air
1L
1 min
1 hr
= 68 hr = 2.8 days



0.0010 g air
1000 mL
140 L air
60 min
Limiting Reactants and Percent Yield
97.
The product formed in the reaction is NO2; the other species present in the product representtation is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules
react with 3 O2 molecules to form 6 NO2 molecules.
6 NO(g) + 3 O2(g) → 6 NO2(g)
For smallest whole numbers, the balanced reaction is:
2 NO(g) + O2(g) → 2 NO2(g)
98.
In the following table, we have listed three rows of information. The Initial row is the number
of molecules present initially, the Change row is the number of molecules that react to reach
completion, and the Final row is the number of molecules present at completion. To
determine the limiting reactant, let’s calculate how much of one reactant is necessary to react
with the other.
10 molecules O2 ×
4 molecules NH3
= 8 molecules NH3 to react with all of the O2
5 molecules O 2
Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react
with all of the O2, O2 is limiting.
4 NH3(g)
Initial
Change
Final
10 molecules
8 molecules
2 molecules
+
5 O2(g)
10 molecules
10 molecules
0
→
4 NO(g)
0
+8 molecules
8 molecules
+
6 H2O(g)
0
+12 molecules
12 molecules
The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2
+ 8 molecules NO + 12 molecules H2O = 22 molecules total.
99.
1.50 g BaO2 ×
25.0 mL ×
1 mol BaO2
= 8.86 × 10 3 mol BaO2
169 .3 g BaO2
0.0272 g HCl 1 mol HCl
= 1.87 × 10 2 mol HCl

mL
36.46 g HCl
66
CHAPTER 3
STOICHIOMETRY
The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual
ratio is:
1.87  10 2 mol HCl
= 2.11
8.86  10 3 mol BaO2
Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2)
is the limiting reagent.
1 mol H 2O 2 34.02 g H 2O 2
8.86 × 10 3 mol BaO2 ×
= 0.301 g H2O2

mol BaO2
mol H 2O 2
The amount of HCl reacted is:
8.86 × 10 3 mol BaO2 ×
2 mol HCl
mol BaO2
= 1.77 × 10 2 mol HCl
excess mol HCl = 1.87 × 10 2 mol  1.77 × 10 2 mol = 1.0 × 10 3 mol HCl
mass of excess HCl = 1.0 × 10 3 mol HCl ×
100.
36.46 g HCl
= 3.6 × 10 2 g HCl
mol HCl
Ca3(PO4)2 + 3 H2SO4 → 3 CaSO4 + 2 H3PO4
1.0 × 103 g Ca3(PO4)2 ×
1 mol Ca 3 (PO4 ) 2
= 3.2 mol Ca3(PO4)2
310 .18 g Ca 3 (PO4 ) 2
1.0 × 103 g conc. H2SO4 ×
98 g H 2SO 4
1 mol H 2SO 4
= 10. mol H2SO4

100 g conc. H 2SO 4 98.09 g H 2SO 4
The required mole ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The
actual ratio is:
10. mol H 2 SO 4
= 3.1
3.2 mol Ca 3 (PO4 ) 2
This is larger than the required mole ratio, so Ca3(PO4)2 (the denominator), is the limiting
reagent.
101.
3.2 mol Ca3(PO4)2 ×
3 mol CaSO4
136.15 CaSO4
= 1300 g CaSO4 produced

mol Ca 3 (PO4 ) 2
mol CaSO4
3.2 mol Ca3(PO4)2 ×
2 mol H 3 PO4
97.99 g H 3 PO4
= 630 g H3PO4 produced

mol Ca 3 (PO4 ) 2
mol H 3 PO4
An alternative method to solve limiting reagent problems is to assume each reactant is
limiting and calculate how much product could be produced from each reactant. The reactant
that produces the smallest amount of product will run out first and is the limiting reagent.
5.00 × 106 g NH3 ×
1 mol NH3
2 mol HCN
= 2.94 × 105 mol HCN

17.03 g NH3 2 mol NH3
CHAPTER 3
STOICHIOMETRY
67
1 mol O 2
2 mol HCN
= 1.04 × 105 mol HCN

32.00 g O 2
3 mol O 2
5.00 × 106 g O2 ×
5.00 × 106 g CH4 ×
1 mol CH 4
2 mol HCN
= 3.12 × 105 mol HCN

16.04 g CH 4 2 mol CH 4
O2 is limiting because it produces the smallest amount of HCN. Although more product could
be produced from NH3 and CH4, only enough O2 is present to produce 1.04 × 105 mol HCN.
The mass of HCN produced is:
1.04 × 105 mol HCN ×
1 mol O 2
6 mol H 2O 18.02 g H 2O
= 5.63 × 106 g H2O


32.00 g O 2
3 mol O 2
1 mol H 2O
5.00 × 106 g O2 ×
102.
27.03 g HCN
= 2.81 × 106 g HCN
mol HCN
We will use the strategy utilized in the previous problem to solve this limiting reactant
problem.
If C3H6 is limiting:
15.0 g C3H6 ×
1 mol C3H 6
2 mol C3H3 N 53.06 g C3H3 N
= 18.9 g C3H3N


42.08 g C3H 6
2 mol C3H 6
mol C3H3 N
If NH3 is limiting:
5.00 g NH3 ×
1 mol NH3
2 mol C3H3 N 53.06 g C3H3 N
= 15.6 g C3H3N


17.03 g NH3
2 mol NH3
mol C3H3 N
If O2 is limiting:
10.0 g O2 ×
2 mol C3H3 N 53.06 g C3H3 N
1 mol O 2
= 11.1 g C3H3N


32.00 g O2
3 mol O 2
mol C3H3 N
O2 produces the smallest amount of product, thus O2 is limiting and 11.1 g C3H3N can be
produced.
103.
C7H6O3 + C4H6O3 → C9H8O4 + HC2H3O2
1.50 g C7H6O3 ×
1 mol C 7 H 6 O 3
= 1.09 × 10 2 mol C7H6O3
138.12 g C 7 H 6 O 3
2.00 g C4H6O3 ×
1 mol C 4 H 6 O 3
= 1.96 × 10 2 mol C4H6O3
102.09 g C 4 H 6 O 3
C7H6O3 is the limiting reagent because the actual moles of C7H6O3 present are below the
required 1:1 mol ratio. The theoretical yield of aspirin is:
68
CHAPTER 3
1.09 × 10-2 mol C7H6O3 ×
% yield =
104.
STOICHIOMETRY
1 mol C 9 H 8 O 4 180 .15 g C 9 H 8 O 4
= 1.96 g C9H8O4

mol C 7 H 6 O 3
mol C 9 H 8 O 4
1.50 g
× 100 = 76.5%
1.96 g
a. 1142 g C6H5Cl ×
485 g C2HOCl3 ×
1 mol C6 H 5Cl
= 10.1 mol C6H5Cl
112 .55 g C6 H 5Cl
1 mol C2 HOCl3
= 3.29 mol C2HOCl3
147.38 g C2 HOCl3
From the balanced equation, the required mole ratio is
mole ratio present is
2 mol C6 H 5Cl
= 2. The actual
1 mol C 2 HOCl3
10.1 mol C6 H 5Cl
= 3.07. The actual mole ratio is greater than
3.29 mol C 2 HOCl3
the required mole ratio, so the denominator of actual mole ratio (C2HOCl3) is limiting.
3.29 mol C2HOCl3 ×
1 mol C14H9Cl5
354.46 g C14H9Cl5
= 1170 g C14H9Cl5 (DDT)

mol C2 HOCl3
mol C14H 9Cl5
b. C2HOCl3 is limiting and C6H5Cl is in excess.
c. 3.29 mol C2HOCl3 ×
2 mol C6 H5Cl 112 .55 g C6 H5Cl
= 741 g C6H5Cl reacted

mol C2 HOCl3
mol C6 H5Cl
1142 g  741 g = 401 g C6H5Cl in excess
d. % yield =
105.
200.0 g DDT
× 100 = 17.1%
1170 g DDT
2.50 metric tons Cu3FeS3 ×
1000 kg 1000 g 1 mol Cu 3 FeS3
3 mol Cu
63.55 g




metric ton
kg
342.71 g
1 mol Cu 3 FeS3 mol Cu
= 1.39 × 106 g Cu (theoretical)
1.39 × 106 g Cu (theoretical) ×
106.
86.3 g Cu (actual)
= 1.20 × 106 g Cu = 1.20 × 103 kg Cu
100. g Cu (theoretical)
= 1.20 metric tons Cu (actual)
P4(s) + 6 F2(g) → 4 PF3(g); The theoretical yield of PF3 is:
120. g PF3 (actual) ×
100.0 g PF3 (theoretical)
= 154 g PF3 (theoretical)
78.1 g PF3 (actual)
CHAPTER 3
STOICHIOMETRY
154 g PF3 ×
69
1 mol PF3
6 mol F2
38.00 g F2
= 99.8 g F2


87.97 g PF3 4 mol PF3
mol F2
99.8 g F2 are needed to produce an actual PF3 yield of 78.1%.
Additional Exercises
107.
0.368 g XeFn
molar mass XeFn =
9.03  10
20
1 mol XeFn
molecules XeFn 
6.022  10 23 molecules
= 245 g/mol
245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6
108.
In one hour, the 1000. kg of wet cereal produced contains 580 kg H2O and 420 kg of cereal.
We want the final product to contain 20.% H2O. Let x = mass of H2O in final product.
x
= 0.20, x = 84 + 0.20 x, x = 105 ≈ 110 kg H2O
420  x
The amount of water to be removed is 580 - 110 = 470 kg/hr.
109.
2 H2(g) + O2(g) → 2 H2O(g)
a. 50 molecules H2 ×
1 molecule O 2
= 25 molecules O2
2 molecules H 2
Stoichiometric mixture. Neither is limiting.
b. 100 molecules H2 ×
1 molecule O 2
= 50 molecules O2;
2 molecules H 2
O2 is limiting since only 40 molecules O2 are present.
c. From b, 50 molecules of O2 will react completely with 100 molecules of H2. We have
100 molecules (an excess) of O2. So, H2 is limiting.
d. 0.50 mol H2 ×
1 mol O 2
= 0.25 mol O2; H2 is limiting because 0.75 mol O2 are present.
2 mol H 2
e. 0.80 mol H2 ×
1 mol O 2
= 0.40 mol O2; H2 is limiting because 0.75 mol O2 are present.
2 mol H 2
f.
1.0 g H2 ×
1 mol H 2
1 mol O 2
= 0.25 mol O2

2.016 g H 2 2 mol H 2
Stoichiometric mixture, neither is limiting.
70
CHAPTER 3
STOICHIOMETRY
1 mol H 2
1 mol O 2 32.00 g O 2
= 39.7 g O2


2.016 g H 2 2 mol H 2
mol O 2
g. 5.00 g H2 ×
H2 is limiting because 56.00 g O2 are present.
0.262 g C7 H5 BiO 4
1 mol C7 H5 BiO 4
1 mol Bi
209.0 g Bi



tablet
362.11 g C7 H5 BiO 4 1 mol C7 H5 BiO 4
mol Bi
110.
2 tablets ×
111.
= 0.302 g Bi consumed
Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; Because 104.14/13.02 = 7.998 ≈ 8,
the molecular formula for styrene is (CH)8 = C8H8.
2.00 g C8H8 ×
112.
1 mol C8 H 8
8 mol H 6.002  10 23 atoms H


= 9.25 × 1022 atoms H
104 .14 g C8 H 8 mol C8 H 8
mol H
41.98 mg CO2 ×
6.45 mg H2O ×
12.01 mg C
11.46 mg
= 11.46 mg C; %C =
× 100 = 57.85% C
44.01 mg CO 2
19.81 mg
2.01 6 mg H
0.772 mg
= 0.722 mg H; %H =
× 100 = 3.64% H
18.02 mg H 2 O
19.81 mg
%O = 100.00 - (57.85 + 3.64) = 38.51% O
Out of 100.00 g terephthalic acid, there are:
57.85 g C ×
1 mol C
1 mol H
= 4.817 mol C; 3.64 g H ×
= 3.61 mol H
12.01 g C
1.008 g H
38.51 g O ×
1 mol O
= 2.407 mol O
16.00 g O
4.817
3.61
2.407
= 2.001;
= 1.50;
= 1.000
2.407
2.407
2.407
The C:H:O mole ratio is 2:1.5:1 or 4:3:2. Empirical formula: C4H3O2
Mass of C4H3O2 ≈ 4(12) + 3(1) + 2(16) = 83
Molar mass =
113.
17.3 g H ×
166
41.5 g
= 166 g/mol;
= 2; Molecular formula: C8H6O4
0.250 mol
83
1 mol H
1 mol C
= 17.2 mol H; 82.7 g C ×
= 6.89 mol C
1.008 g H
12.01 g C
17.2
= 2.50; The empirical formula is C2H5.
6.89
The empirical formula mass is ~29 g, so two times the empirical formula would put the
compound in the correct range of the molar mass. Molecular formula = (C2H5)2 = C4H10
CHAPTER 3
STOICHIOMETRY
2.59 × 1023 atoms H ×
1 molecule C 4 H10
1 mol C 4 H10
= 4.30 × 10 2 mol C4H10

23
10 atoms H
6.022  10 molecules
4.30 × 10 2 mol C4H10 ×
114.
58.12 g
= 2.50 g C4H10
mol C 4 H10
Assuming 100.00 g E3H8:
mol E = 8.73 g H ×
1 mol H 3 mol E
= 3.25 mol E

1.008 g H 8 mol H
xgE
91.27 g E
, x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu

1 mol E 3.25 mol E
115.
Mass of H2O = 0.755 g CuSO4  xH2O - 0.483 g CuSO4 = 0.272 g H2O
0.483 g CuSO4 ×
0.272 g H2O ×
1 mol CuSO 4
= 0.00303 mol CuSO4
159 .62 g CuSO 4
1 mol H 2 O
= 0.0151 mol H2O
18.02 g H 2 O
4.98 mol H 2 O
0.0151 mol H 2O
=
; Compound formula = CuSO4  5 H2O, x = 5
0.00303 g CuSO 4
1 mol CuSO 4
116.
a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:
1 mol C 3 H 3 N 53.06 g C 3 H 3 N
= 33.3 g C3H3N

14.01 g N
1 mol C 3 H 3 N
8.80 g N ×
% C3H3N =
33.3 g C 3 H 3 N
= 33.3% C3H3N
100.00 g polymer
Only butadiene in the polymer reacts with Br2:
0.605 g Br2 ×
% C4H6 =
71
1 mol C4 H 6 54.09 g C4 H 6
1 mol Br2
= 0.205 g C4H6


159.80 g Br2
mol Br2
mol C4 H 6
0.205 g
× 100 = 17.1% C4H6
1.20 g
b. If we have 100.0 g of polymer:
33.3 g C3H3N ×
1 mol C 3 H 3 N
= 0.628 mol C3H3N
53.06 g
72
CHAPTER 3
17.1 g C4H6 ×
1 mol C 4 H 6
= 0.316 mol C4H6
54.09 g C 4 H 6
49.6 g C8H8 ×
1 mol C 8 H 8
= 0.476 mol C8H8
104 .14 g C 8 H 8
Dividing by 0.316:
STOICHIOMETRY
0.628
0.316
0.476
= 1.99;
= 1.00;
= 1.51
0.316
0.316
0.316
This is close to a mol ratio of 4:2:3. Thus, there are 4 acrylonitrile to 2 butadiene to 3
styrene molecules in the polymer or (A4B2S3)n.
117.
1.20 g CO2 ×
1 mol C 24 H 30 N 3 O
1 mol CO 2
1 mol C
376.51 g



44.01 g
mol CO 2
24 mol C
mol C 24 H 30 N 3 O
= 0.428 g C24H30N3O
0.428 g C 24 H 30 N 3 O
× 100 = 42.8% C24H30N3O
1.00 g sample
118.
a. CH4(g) + 4 S(s) → CS2(l) + 2 H2S(g) or 2 CH4(g) + S8(s) → 2 CS2(l) + 4 H2S(g)
b. 120. g CH4 ×
1 mol CH 4
1 mol S
= 7.48 mol CH4; 120. g S ×
= 3.74 mol S
16.04 g CH 4
32.07 g S
The required S to CH4 mole ratio is 4:1. The actual S to CH4 mol ratio is:
3.74 mol S
= 0.500
7.48 mol CH 4
This is well below the required ratio so sulfur is the limiting reagent.
The theoretical yield of CS2 is: 3.74 mol S ×
1 mol CS 2 76.15 g CS 2
= 71.2 g CS2

4 mol S
mol CS 2
The same amount of CS2 would be produced using the balanced equation with S8.
119.
126 g B5H9 ×
1 mol
1 mol
= 2.00 mol B5H9; 192 g O2 ×
= 6.00 mol O2
63.12 g
32.00 g
6.00
mol O 2
(actual) =
= 3.00
mol B5 H 9
2.00
The required mol O2 to mol B5H9 ratio is 12/2 = 6. The actual mole ratio is less than the
required mole ratio, thus the numerator (O2) is limiting.
6.00 mol O2 ×
9 mol H 2 O 18.02 g H 2 O
= 81.1 g H2O

12 mol O 2
mol H 2 O
CHAPTER 3
120.
STOICHIOMETRY
25.0 g Ag2O ×
73
1 mol
= 0.108 mol Ag2O
231 .8 g
50.0 g C10H10N4SO2 ×
1 mol
= 0.200 mol C10H10N4SO2
250 .29 g
mol C10 H10 N 4SO 2
0.200
(actual) =
= 1.85
mol Ag2 O
0.108
The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting.
0.200 mol C10H10N4SO2 ×
2 mol AgC10 H 9 N 4SO 2
357.18 g

2 mol C10 H10 N 4SO 2
mol AgC10 H 9 N 4SO 2
= 71.4 g AgC10H9N4SO2 produced
121.
453 g Fe ×
1 mol Fe 2 O 3 159.70 g Fe 2 O 3
1 mol Fe
= 648 g Fe2O3


55.85 g Fe
2 mol Fe
mol Fe 2 O 3
mass %Fe2O3 =
122.
648 g Fe 2 O 3
× 100 = 86.2%
752 g ore
a. Mass of Zn in alloy = 0.0985 g ZnCl2 ×
%Zn =
65.38 g Zn
= 0.0473 g Zn
136 .28 g ZnCl2
0.0473 g Zn
× 100 = 9.34% Zn; %Cu = 100.00  9.34 = 90.66% Cu
0.5065 g brass
b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted
copper could be measured.
123.
Assuming one mol of vitamin A (286.4 g vitamin A):
mol C = 286.4 g vitamin A ×
mol H = 286.4 g vitamin A ×
0.8386 g C
1 mol C
= 20.00 mol C

g vitamin A
12.01 g C
0.1056 g H
1 mol H
= 30.00 mol H

g vitamin A 1.008 g H
Since one mol of vitamin A contains 20 mol C and 30 mol H, the molecular formula of
vitamin A is C20H30E. To determine E, let’s calculate the molar mass of E.
286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol
From the periodic table, E = oxygen and the molecular formula of vitamin A is C20H30O.
Challenge Problems
124.
atoms
atoms
85
87
Rb
= 2.591; Assuming 100 atoms, let x = number of 85Rb atoms
Rb
74
CHAPTER 3
and 100  x = number of
87
STOICHIOMETRY
Rb atoms.
259.1
x
= 2.591, x = 259.1  2.591 x, x =
= 72.15% 85Rb
3.591
100  x
85.4678  61.26
0.7215 (84.9117) + 0.2785 (A) = 85.4678, A =
= 86.92 amu
0.2785
= atomic mass of
125.
87
Rb
First, we will determine composition in mass percent. We assume all the carbon in the 0.213
g CO2 came from 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O
came from the 0.157 g of the compound.
0.213 g CO2 ×
12.01 g C
0.0581 g C
= 0.0581 g C; %C =
× 100 = 37.0% C
44.01 g CO2
0.1571 g compound
0.0310 g H2O ×
3.47  10 3 g
2.016 g H
= 3.47 × 10-3 g H; %H =
= 2.21% H
0.157 g
18.02 g H 2 O
We get %N from the second experiment:
0.0230 g NH3 ×
%N =
14.01 g N
= 1.89 × 10 2 g N
17.03 g NH3
1.89  10 2 g
× 100 = 18.3% N
0.103 g
The mass percent of oxygen is obtained by difference:
%O = 100.00 - (37.0 + 2.21 + 18.3) = 42.5%
So out of 100.00 g of compound, there are:
1 mol C
= 3.08 mol C; 2.21 g H ×
12.01 g C
1 mol N
18.3 g N ×
= 1.31 mol N; 42.5 g O ×
14.01 g N
37.0 g C ×
1 mol H
= 2.19 mol H
1.008 g H
1 mol O
= 2.66 mol O
16.00 g O
The last, and often the hardest part, is to find simple whole number ratios. Divide all mole
values by the smallest number:
3.08
2.19
1.31
2.66
= 2.35;
= 1.67;
= 1.00;
= 2.03
1.31
1.31
1.31
1.31
Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6.
126.
1.0 × 106 kg HNO3 ×
1000 g HNO3
1mol HNO3
= 1.6 × 107 mol HNO3

kg HNO3
63.02 g HNO3
CHAPTER 3
STOICHIOMETRY
75
We need to get the relationship between moles of HNO3 and moles of NH3. We have to use
all 3 equations.
2 mol HNO3
2 mol NO2
4 mol NO 16 mol HNO3



3 mol NO2
2 mol NO
4 mol NH3
24 mol NH3
Thus, we can produce 16 mol HNO3 for every 24 mol NH3 we begin with:
1.6 × 107 mol HNO3 ×
24 mol NH3
17.03 g NH3
= 4.1 × 108 g or 4.1 × 105 kg

16 mol HNO3
mol NH3
This is an oversimplified answer. In practice, the NO produced in the third step is recycled
back continuously into the process in the second step. If this is taken into consideration, then
the conversion factor between mol NH3 and mol HNO3 turns out to be 1:1, i.e., 1 mol of NH3
produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer
of 2.7 × 105 kg NH3 reacted.
127.
The two relevant equations are:
4 FeO(s) + O2(g) → 2 Fe2O3(s) and 4 Fe3O4(s) + O2(g) → 6 Fe2O3(s)
Let x = mass FeO, so 5.430 - x = mass Fe3O4. The mol of each are:
moles FeO =
x
5.430  x
and moles Fe3O4 =
71.85
231 .55
Thus, moles Fe2O3 is:
 5.430  x 6 moles Fe 2O3 
 x
2 moles Fe 2O3 


  


4 moles FeO 
4 moles Fe3O 4 
 71.85
 231.35
and mass Fe2O3 is:
 x
2   5.430  x 6  
 
   = 5.779 g
159.70 g/mol 
4
 71.85 4   231 .35
Solving: x = 2.10 g; Thus, the mixture is
128.
2.10 g
× 100 = 38.7% FeO by mass.
5.430 g
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
30.07 g/mol
44.09 g/mol
Let x = mass C2H6, so 9.780  x = mass C3H8. Use the balanced reaction to set up an
equation for the mol of O2 required.
x
7 9.780  x 5
 
 = 1.120 mol O2
30.07 2
44.09
1
76
CHAPTER 3
Solving: x = 3.7 g C2H6 ;
129.
STOICHIOMETRY
3.7 g
× 100 = 38% C2H6 by mass
9.780 g
The two relevant equations are:
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
Let x = mass Mg, so 10.00  x = mass Zn.
From the balanced equations, moles H2 = moles Zn + moles Mg.
mol H2 = 0.5171 g H2 ×
Thus, 0.2565 =
1 mol H 2
= 0.2565 mol H2
2.016 g H 2
x
10.00  x
+
; Solving, x = 4.008 g Mg.
24.31
65.38
4.008 g
× 100 = 40.08% Mg
10.00 g
130.
Let M = unknown element
mass %M =
mass M
2.077
× 100 = 56.01% M
 100 
total mass compound
3.708
100.00 – 56.01 = 43.99% O
Assuming 100.00 g compound:
43.99 g O ×
1 mol O
= 2.749 mol O
16.00 g O
56.01 g M
= 20.37 g/mol.
2.749 mol M
This is too low for the molar mass. We must have fewer moles of M than mol O present in
If MO is the formula of the oxide, then M has a molar mass =
the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which
to try. Let’s assume an MO2 formula. Then the molar mass of M is:
56.01 g M
= 40.75 g/mol
1 mol M
2.749 mol O 
2 mol O
This is close to calcium but calcium forms an oxide having the CaO formula, not CaO2.
If MO3 is assumed to be the formula then the molar mass of M calculates to be 61.12 g/mol
which is too large. Therefore, the mol O to mol M ratio must be between 2 and 3. Some
reasonable possibilities are 2.25, 2.33, 2.5, 2.67, 2.75 (these are reasonable since they will
lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 to 1 gives a molar
mass of:
CHAPTER 3
STOICHIOMETRY
77
56.01 g M
= 50.94 g/mol.
1 mol M
2.749 mol O 
2.5 mol O
This is the molar mass of vanadium and V2O5 is a reasonable formula for an oxide of
vanadium. The other choices for the O:M mol ratios between 2 and 3 do not give as
reasonable results. Therefore, M is vanadium and the formula is V2O5.
131.
We know water is a product, so one of the elements in the compound is hydrogen.
XaHb + O2 → H2O + ?
To balance the H atoms, the mole ratio between XaHb : H2O =
mol compound =
2
.
b
1.39 g
1.21 g
= 0.0224 mol; mol H2O =
= 0.0671 mol
62.09 g / mol
18.02 g / mol
2
0.0224
and b = 6; XaH6 has a molar mass of 62.09 g/mol.

b
0.0671
62.09 = a × molar mass of X + 6 × 1.008, a × molar mass of X = 56.04
Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), Li (a = 8).
N fits the data best so N4H6 is the formula.
132.
The balanced equation is: 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g)
2
.
x
1 mol Sc
moles Sc = 2.25 g Sc ×
= 0.0500 mol Sc
44.96 g Sc
The mol ratio of Sc : H2 =
mol H2 = 0.1502 g H2 ×
1 mol H 2
= 0.07450 mol H2
2.016 g H 2
2
0.0500
=
, x = 3; The formula is ScCl3.
0.07450
x
133.
Total mass of copper used:
10,000 boards ×
(8.0 cm  16.0 cm  0.060 cm) 8.96 g
= 6.9 × 105 g Cu

board
cm3
Amount of Cu removed = 0.80 × 6.9 × 105 g = 5.5 × 105 g Cu
5.5 × 105 g Cu ×
1 mol Cu( NH3 ) 4 Cl 2 202.59 g Cu( NH3 ) 4 Cl 2
1 mol Cu


63.55 g Cu
mol Cu
mol Cu( NH3 ) 4 Cl 2
78
CHAPTER 3
STOICHIOMETRY
= 1.8 × 106 g Cu(NH3)4Cl2
5.5 × 105 g Cu ×
134.
4 mol NH3 17.03 g NH3
1 mol Cu
= 5.9 × 105 g NH3


63.55 g Cu
mol Cu
mol NH3
a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for
every 4 mol of C6H5O3N reacted. The actual yield is 3 moles of acetaminophen
compared to a theoretical yield of 4 moles of acetaminophen. Solving for percent yield
by mass (where M = molar mass acetaminophen):
3 mol  M
% yield =
× 100 = 75%
4 mol  M
b. The product of the percent yields of the individual steps must equal the overall yield,
75%.
(0.87) (0.98) (x) = 0.75, x = 0.88; Step III has a % yield = 88%.
135.
10.00 g XCl2 + excess Cl2 → 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4
contains 2.55 g Cl and 10.00 g XCl2. From mol ratios, 10.00 g XCl2 must also contain 2.55 g
Cl; mass X in XCl2 = 10.00  2.55 = 7.45 g X.
2.55 g Cl ×
1 mol Cl 1 mol XCl 2
1 mol X
= 3.60 × 10 2 mol X


35.45 g Cl 2 mol Cl
mol XCl 2
So, 3.60 × 10 2 mol X has a mass equal to 7.45 g X. The molar mass of X is:
7.45 g X
= 207 g/mol X; Atomic mass = 207 amu so X is Pb.
3.60  10 2 mol X
136.
4.000 g M2S3 → 3.723 g MO2
There must be twice as many mol of MO2 as mol of M2S3 in order to balance M in the
reaction. Setting up an equation for 2 mol MO2 = mol M2S3 where A = molar mass M:


4.000 g
3.723 g
8.000
3.723
 
2
,

2 A  96.21
A  32.00
 2 A  3(32.07)  A  2(16.00)
8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 g/mol; atomic mass
= 184 amu
137.
Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2
anions. The simplest compound of the two elements is Al2O3. Similarly, we would expect the
formula of any group 6A element with Al to be Al2X3. Assuming this, out of 100.00 g of
compound there are 18.56 g Al and 81.44 g of the unknown element, X. Let’s use this
information to determine the molar mass of X which will allow us to identify X from the
periodic table.
CHAPTER 3
STOICHIOMETRY
18.56 g Al ×
1 mol Al
3 mol X
= 1.032 mol X

26.98 g Al 2 mol Al
81.44 g of X must contain 1.032 mol of X.
81.44 g X
= 78.91 g/mol X.
1.032 mol X
From the periodic table, the unknown element is selenium and the formula is Al2Se3.
The molar mass of X =
138.
NaCl(aq) + Ag+(aq) → AgCl(s); KCl(aq) + Ag+(aq) → AgCl(s)
8.5904 g AgCl ×
1 mol AgCl
1 mol Cl 

= 5.991 × 10 2 mol Cl
143 .4 g AgCl 1 mol AgCl
The molar masses of NaCl and KCl are 58.44 and 74.55 g/mol, respectively.
Let x = mass NaCl and y = mass KCl:
x + y = 4.000 g;
x + y = 5.991 × 10 2 total mol Cl or 1.276 x + y = 4.466
58.44
74.55
Solving using simultaneous equations:
1.276 x + y = 4.466
x  y = -4.000
0.276 x
= 0.466,
%NaCl =
139.
x = 1.69 g NaCl and y = 2.31 g KCl
1.69 g
× 100 = 42.3% NaCl; %KCl = 57.7%
4.00 g
The balanced equations are:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g) → 4 NO2(g)
+ 6 H2O(g)
Let 4x = number of mol of NO formed, and let 4y = number of mol of NO2 formed. Then:
4x NH3 + 5x O2 → 4x NO + 6x H2O and 4y NH3 + 7y O2 → 4y NO2 + 6y H2O
All the NH3 reacted, so 4x + 4y = 2.00. 10.00  6.75 = 3.25 mol O2 reacted, so 5x + 7y
= 3.25.
Solving by the method of simultaneous equations:
20 x + 28 y = 13.0
20 x  20 y = 10.0
8 y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12
mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed
79
80
140.
CHAPTER 3
STOICHIOMETRY
CxHyOz + oxygen → x CO2 + y/2 H2O
2.20 g CO 2 
mass %C in aspirin =
1 mol C
1 mol CO 2
12.01 g C


44.01 g CO 2 mol CO 2
mol C
= 60.0% C
1.00 g aspirin
1 mol H 2 O 2 mol H 1.008 g H


18.02 g H 2 O mol H 2 O
mol H
mass %H in aspirin =
= 4.48% H
1.00 g aspirin
mass %O = 100.00  (60.0 + 4.48) = 35.5% O
0.400 g H 2 O 
Assuming 100.00 g aspirin:
60.0 g C ×
1 mol C
1 mol H
= 5.00 mol C; 4.48 g H ×
= 4.44 mol H
12.01 g C
1.008 g H
1 mol O
= 2.22 mol O
16.00 g O
5.00
4.44
Dividing by the smallest number:
= 2.25;
= 2.00
2.22
2.22
35.5 g O ×
Empirical formula = (C2.25 H2.00O)4 = C9H8O4
Empirical mass 9(12) + 8(1) + 4(16) = 180 g/mol; This is in the 170 – 190 g/mol range so
the molecular formula is also C9H8O4.
Balance the aspirin synthesis reaction to determine the formula for salicylic acid.
CaHbOc + C4H6O3 → C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3
Integrative Problems
141.
1 mol Fe
6.022  10 23 atoms Fe

= 113 atoms Fe
55.85 g Fe
mol Fe
a. 1.05 × 10 20 g Fe ×
b. The total number of platinum atoms is 14 × 20 = 280 atoms (exact number). The mass of
these atoms is:
280 atoms Pt ×
1 mol Pt
195.1 g Pt
= 9.071 × 10 20 g Pt

23
mol Pt
6.022  10 atoms Pt
c. 9.071 × 10 20 g Ru ×
142.
1 mol Ru
6.022  10 23 atoms Ru

= 540.3 = 540 atoms Ru
101 .1 g Ru
mol Ru
Assuming 100.00 g of tetrodotoxin:
CHAPTER 3
STOICHIOMETRY
41.38 g C ×
5.37 g H ×
81
1 mol C
1 mol N
= 3.445 mol C; 13.16 g N ×
= 0.9393 mol N
12.01 g C
14.01 g N
1 mol H
1 mol O
= 5.33 mol H; 40.09 g O ×
= 2.506 mol O
1.008 g H
16.00 g O
Divide by the smallest number:
3.445
5.33
2.506
= 3.668;
= 5.67;
= 2.668
0.9393
0.9393
0.9393
To get whole numbers for each element, multiply through by 3.
empirical formula = (C3.668H5.67NO2.668)3 = C11H17N3O8
The mass of the empirical formula is 319.3 g/mol.
molar mass tetrodotoxin =
1.59  10 21 g
= 319 g/mol
1 mol
3 molecules 
6.022  10 23 molecules
Because the empirical mass and molar mass are the same, the molecular formula is the same
as the empirical formula, C11H17N3O8.
165 lb ×
1 kg
10. μg 1  10 6 g
1 mol
6.022  10 23 molecules




2.2046 lb
kg
μg
319 .3 g
1 mol
= 1.4 × 1018 molecules tetrodotoxin is the LD50 dosage
143.
0.105 g
molar mass X2 =
8.92  10
20
1 mol
molecules 
6.022  10 23 molecules
= 70.9 g/mol
The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine.
Assuming 100.00 g of MX3 compound:
54.47 g Cl ×
1 mol
= 1.537 mol Cl
35.45 g
1.537 mol Cl ×
molar mass M =
1 mol M
= 0.5123 mol M
3 mol Cl
45.53 g M
= 88.87 g/mol M
0.5123 mol M
82
CHAPTER 3
STOICHIOMETRY
M is the element yttrium (Y) and the name of YCl3 is yttrium(III) chloride.
The balanced equation is: 2 Y + 3 Cl2 → 2 YCl3
Assuming Cl2 is limiting:
1.00 g Cl2 ×
2 mol YCl3
195.26 g YCl3
1 mol Cl2
= 1.84 g YCl3


70.90 g Cl2
3 mol Cl2
1 mol YCl3
Assuming Y is limiting:
1.00 g Y ×
2 mol YCl3
195.26 g YCl3
1 mol Y
= 2.20 g YCl3


88.91 g Y
2 mol Y
1 mol YCl3
Cl2 is the limiting reagent and the theoretical yield is 1.84 g YCl3.
144.
2 As + 4 AsI3 → 3 As2I4
Volume of As cube = (3.00 cm)3 = 27.0 cm3
27.0 cm3 ×
5.72 g As
1 mol As
= 2.06 mol As

3
74.92 g As
cm
1.01 × 1024 molecules AsI3 ×
1 mol AsI3
= 1.68 mol AsI3
6.022  10 23 molecules AsI3
From the balanced equation, we need twice the number of moles of AsI3 as As to react.
Because the mole of AsI3 present are less than the mole of As present, AsI3 is limiting.
1.68 mol AsI3 ×
0.756 =
3 mol As2 I 4 657.44 g As2 I 4
= 828 g As2I4

4 mol AsI3
2 mol As2 I 4
actual yield
, actual yield = 0.756 × 828 g = 626 g As2I4
828 g
Marathon Problems
145.
To solve the limiting reagent problem, we must determine the formulas of all the compounds
so we can get a balanced reaction.
a. 40 million trillion = 40 × 106 × 1012 = 4.000 × 1019 (assuming 4 S.F.)
4.000 × 1019 molecules A ×
Molar mass of A =
1 mol A
= 6.642 × 10-5 mol A
6.0221  10 23 molecules A
4.26  10 3 g A
= 64.1 g/mol
6.642  10 5 mol A
CHAPTER 3
STOICHIOMETRY
83
Mass of carbon in one mol of A is:
64.1 g A ×
37.5 g C
= 24.0 g carbon = 2 mol carbon in substance A
100 .0 g A
The remainder of the molar mass (64.1 g - 24.0 g = 40.1 g) is due to the alkaline earth
metal. From the periodic table, calcium has a molar mass of 40.08 g/mol. The formula
of substance A is CaC2.
b. 5.36 g H + 42.5 g O = 47.9 g; Substance B only contains H and O. Determining the
empirical formula of B:
5.36 g H ×
5.32
1 mol H
= 5.32 mol H;
= 2.00
2.66
1.008 g H
42.5 g O ×
2.66
1 mol O
= 2.66 mol O;
= 1.00
2.66
16.00 g O
Empirical formula = H2O; The molecular formula of substance B could be H2O, H4O2,
H6O3, etc. The most reasonable choice is water (H2O) for substance B.
c. Substance C + O2 → CO2 + H2O; Substance C must contain carbon and hydrogen, and
may contain oxygen. Determining the mass of carbon and hydrogen in substance C:
1 mol CO2
1 mol C
12.01 g C
= 9.22 g carbon


44.01 g CO2
mol CO2
mol C
1 mol H 2O
2 mol H
1.008 g H
6.92 g H2O ×
= 0.774 g hydrogen


18.02 g H 2O mol H 2O
mol H
33.8 g CO2 ×
9.22 g carbon + 0.774 g hydrogen = 9.99 g; Substance C initially weighed 10.0 g, so
there is no oxygen present in substance C. Determining the empirical formula for substance C:
9.22 g ×
1 mol C
= 0.768 mol carbon
12.01 g C
0.774 g H ×
1 mol H
= 0.768 mol hydrogen
1.008 g H
mol C/mol H = 1.00; The empirical formula is CH which has an empirical formula mass
≈ 13. The mass spectrum data indicates a molar mass of 26 g/mol, thus the molecular
formula for substance C is C2H2.
d. Substance D is Ca(OH)2.
Now we can answer the question. The balanced equation is:
CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(aq)
84
CHAPTER 3
45.0 g CaC2 ×
STOICHIOMETRY
1 mol CaC2
= 0.702 mol CaC2
64.10 g CaC2
1 mol H 2 O
= 1.28 mol H2O
18.02 g H 2 O
mol H 2O
1.28
= 1.82

mol CaC2
0.702
23.0 g H2O ×
Because the actual mole ratio present is smaller than the required 2:1 mole ratio from the
balanced equation, H2O is limiting.
1.28 mol H2O ×
146.
a. i.
1 mol C2 H 2
26.04 g C2 H 2
= 16.7 g C2H2 = mass of product C

2 mol H 2O
mol C2 H 2
If the molar mass of A is greater than the molar mass of B, then we cannot determine
the limiting reactant, because, while we have a smaller number of moles of A, we
also need fewer moles of A (from the balanced reaction).
ii. If the molar mass of B is greater than the molar mass of A, then B is the limiting
reactant because we have a smaller number of moles of B and we need more B (from
the balanced reaction).
b. A + 5 B → 3 CO2 + 4 H2O
To conserve mass : 44.01 + 5(B) = 3(44.01) + 4(18.02); solving: B = 32.0 g/mol
Because it is diatomic, the best choice for B is O2.
c. We can solve this without mass percent data simply by balancing the equation:
A + 5 O2 → 3 CO2 + 4 H2O
A must be C3H8. This is also the empirical formula.
Note:
3(12.01)
× 100 = 81.71%. So this checks.
3(12.01)  8(1.008)