CHAPTER 3
STOICHIOMETRY
Questions
23.
Isotope
12
C
13
C
Mass
12 .000 0 u
13.034 u
Abundance
98.89%
1.11%
Average mass = 0.9889 (12.0000) + 0.0111(13.034) = 12.01 u
Note: u is an abbreviation for amu (atomic mass units).
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 u.
Total mass = 10,000 atoms ×
12.01 u
= 1.201 × 105 u
atom
For 1 mole of carbon (6.0221 × 1023 atoms C), the average mass would still be 12.01 u.
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 u
= 7.233 × 1024 u
atom
Total mass in g = 6.0221 × 1023 atoms ×
1g
12.01 u
×
= 12.01 g/mol
atom
6.0221  10 23 u
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.
24.
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 mole 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.
46
CHAPTER 3
25.
STOICHIOMETRY
47
Avogadro’s number of dollars = 6.022 × 1023 dollars/mol dollars
1 mol dollars 
6.022  10 23 dollars
mol dollars
7  10 9 people
= 8.6 × 1013 = 9 × 1013 dollars/person
1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 90 trillion dollars.
26.
Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol
One mol of CO2 contains 6.022  1023 molecules of CO2, 6.022  1023 atoms of C, and 1.204
 1024 atoms of O. We could also break down 1 mol of CO2 into the number of protons and
the number of electrons present (1.325  1025 protons and 1.325  1025 electrons). In order to
determine the number of neutrons present, we would need to know the isotope abundances
for carbon and oxygen.
The mass of 1 mol of CO2 would be 44.01 g. From the molar mass, one mol of CO2 would
contain 12.01 g C and 32.00 g O. We could also break down 1 mol of CO2 into the mass of
protons and mass of electrons present (22.16 g protons and 1.207  102 g electrons). This
assumes no mass loss when the individual particles come together to form the atom. This is
not a great assumption as will be discussed in Chapter 19 on Nuclear Chemistry.
27.
Only in b are the empirical formulas the same for both compounds illustrated. In b, general
formulas of X2Y4 and XY2 are illustrated, and both have XY2 for an empirical formula.
For a, general formulas of X2Y and X2Y2 are illustrated. The empirical formulas for these
two compounds are the same as the molecular formulas. For c, general formulas of XY and
XY2 are illustrated; these general formulas are also the empirical formulas. For d, general
formulas of XY4 and X2Y6 are illustrated. XY4 is also the molecular formula, but X2Y6 has
the empirical formula of XY3.
28.
The molar mass is the mass of 1 mole of the compound. The empirical mass is the mass of 1
mole 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).
29.
The mass percent of a compound is a constant no matter what amount of substance is present.
Compounds always have constant composition.
30.
A balanced equation starts with the correct formulas of the reactants and products. The coefficients necessary to balance the equation 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.
48
CHAPTER 3
STOICHIOMETRY
31.
Only one product is formed in this representation. This product has two Y atoms bonded to
an X. The other substance present in the product mixture is just the excess of one of the
reactants (Y). The best equation has smallest whole numbers. Here, answer c would be this
smallest whole number equation (X + 2 Y  XY2). Answers a and b have incorrect
products listed, and for answer d, an equation only includes the reactants that go to produce
the product; excess reactants are not shown in an equation.
32.
A balanced equation must have the same number and types of atoms on both sides of the
equation, but it also needs to have correct formulas. The illustration has the equation as:
H + O  H2O
Under normal conditions, hydrogen gas and oxygen gas exist as diatomic molecules. So the
first change to make is to change H + O to H2 + O2. To balance this equation, we need one
more oxygen atom on the product side. Trial and error eventually gives the correct balanced
equation of:
2 H2 + O2  2 H2O
This equation uses the smallest whole numbers and has the same number of oxygen atoms
and hydrogen atoms on both sides of the equation (4 H + 2 O atoms). So in your drawing,
there should be two H2 molecules, 1 O2 molecule, and 2 H2O molecules.
33.
The theoretical yield is the stoichiometric amount of product that should form if the limiting
reactant is completely consumed and the reaction has 100% yield.
34.
A reactant is present in excess if there is more of that reactant present than is needed to
combine with the limiting reactant for the process. By definition, the limiting reactant cannot
be present in excess. An excess of any reactant does not affect the theoretical yield for a
process; the theoretical yield is determined by the limiting reactant.
35.
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
molar mass of CB
1 mol A 2 B2
2 mol 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. Moles 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
theoretical mass
calculated in part a. If the actual mass of CB produced is given, then the percent yield can
be determined for the reaction using the percent yield equation.
d. Percent yield =
CHAPTER 3
36.
STOICHIOMETRY
49
One method is to assume each quantity of reactant is limiting, then calculate the amount of
product that could be produced from each reactant. This gives 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 smaller quantity is reached, the limiting reactant runs out, and the reaction
cannot continue.
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 all of it. 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.
Exercises
Atomic Masses and the Mass Spectrometer
37.
Let A = average atomic mass
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 u; from the periodic table, the element is Pb.
Note: u is an abbreviation for amu (atomic mass units).
38.
Average atomic mass = A = 0.0800(45.952632) + 0.0730(46.951764) + 0.7380(47.947947)
+ 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu
This is element Ti (titanium).
39.
Let A = mass of 185Re:
186.207 = 0.6260(186.956) + 0.3740(A), 186.207 − 117.0 = 0.3740(A)
A=
40.
69.2
= 185 u (A = 184.95 u 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 u.
28.09 = 0.9221(27.98) + 0.0470(atomic mass 29Si) + 0.0309(29.97)
Atomic mass 29Si = 29.01 u
The mass of 29Si is actually a little less than 29 u. There are other isotopes of silicon that are
considered when determining the 28.09 u average atomic mass of Si listed in the atomic table.
41.
Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100  x.
50
CHAPTER 3
151.96 =
STOICHIOMETRY
x(150.9196 )  (100  x)(152.9209 )
100
15196 = (150.9196)x + 15292.09  (152.9209)x, 96 = (2.0013)x
x = 48%; 48% 151Eu and 100  48 = 52% 153Eu
42.
If silver is 51.82% 107Ag, then the remainder is 109Ag (48.18%). Determining the atomic
mass (A) of 109Ag:
107.868 =
51.82(106.905)  48.18(A)
100
10786.8 = 5540. + (48.18)A, A = 108.9 u = atomic mass of 109Ag
43.
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
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 masses tell us the two
isotopes are present in about equal abundance. The actual abundance is 50.68% 79Br and
49.32% 81Br.
44.
Because we are not given the relative masses of the isotopes, we need to estimate the masses
of the isotopes. A good estimate is to assume that only the protons and neutrons contribute to
the overall mass of the atom and that the atomic mass of a proton and neutron are each 1.00 u.
So the masses are about: 54Fe, 54.00 u; 56Fe, 56.00 u; 57Fe, 57.00 u; 58Fe, 58.00 u. Using
these masses, the calculated average atomic mass would be:
0.0585(54.00) + 0.9175(56.00) + 0.0212(57.00) + 0.0028(58.00) = 55.91 u
The average atomic mass listed in the periodic table is 55.85 u.
Moles and Molar Masses
45.
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 
46.
500.0 g Fe ×
1 mol Fe
6.022  10
23
atoms Fe

55.85 g Fe
= 4.64 × 10 20 g Fe
mol 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
47.
1.00 carat ×
48.
5.0 × 1021 atoms C ×
8.3 × 10 3 mol C ×
49.
51
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
50.
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
51.
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
52.
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
53.
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 ×
54.
a. 1.00 g P4O6 ×
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
b. 1.00 g Ca3(PO4)2 ×
1 mol Ca 3 (PO4 ) 2
= 3.22 × 10 3 mol Ca3(PO4)2
310 .18 g
c. 1.00 g Na2HPO4 ×
1 mol Na 2 HPO4
= 7.04 × 10 3 mol Na2HPO4
141 .96 g
52
55.
CHAPTER 3
a. 5.00 mol NH3 ×
17.03 g NH3
= 85.2 g NH3
mol NH3
b. 5.00 mol N2H4 ×
32.05 g N 2 H 4
= 160. g N2H4
mol N 2 H 4
c. 5.00 mol (NH4)2Cr2O7 ×
56.
a. 5.00 mol P4O6 ×
c. 5.00 mol Na2HPO4 ×
310 .18 g
= 1.55 × 103 g Ca3(PO4)2
mol Ca 3 (PO4 ) 2
141 .96 g
= 7.10 × 102 g Na2HPO4
mol Na 2 HPO4
Chemical formulas give atom ratios as well as mole ratios.
a. 5.00 mol NH3 ×
b. 5.00 mol N2H4 ×
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
c. 5.00 mol (NH4)2Cr2O7 ×
58.
59.
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
b. 5.00 mol Ca3(PO4)2 ×
57.
STOICHIOMETRY
a. 5.00 mol P4O6 ×
2 mol N
14.01 g N
= 140. g N

mol ( NH4 ) 2 Cr2O7
mol N
4 mol P
30.97 g P
= 619 g P

mol P4O6
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

17.03 g NH3
mol NH3
= 3.54 × 1022 molecules 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 ×

60.
53
1 mol ( NH4 ) 2 Cr2O7
252.08 g ( NH4 ) 2 Cr2O7
6.022  10 23 formula units ( NH4 ) 2 Cr2O7
= 2.39 × 1021 formula units (NH4)2Cr2O7
mol ( NH4 ) 2 Cr2O7
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 (PO4 ) 2
6.022  10 23 formula units

310 .18 g
mol Ca 3 (PO4 ) 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
61.
Using answers from Exercise 59:
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
62.
Using answers from Exercise 60:
a. 2.74 × 1021 molecules P4O6 ×
63.
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 CCl2F2 = 12.01 + 2(35.45) + 2(19.00) = 120.91 g/mol
5.56 mg CCl2F2 
1g
1 mol
6.022  10 23 molecules


1000 mg
120 .91 g
mol
= 2.77 × 1019 molecules CCl2F2
54
CHAPTER 3
5.56 × 10−3 g CCl2F2 
STOICHIOMETRY
1 mol CCl2 F2
2 mol Cl
35.45 g Cl


120.91 g
1 mol CCl2 F
mol Cl
= 3.26 × 10−3 g = 3.26 mg Cl
64.
The •2H2O is part of the formula of bauxite (they are called waters of hydration). Combining
elements together, the chemical formula for bauxite would be Al2O5H4.
a. Molar mass = 2(26.98) + 5(16.00) + 4(1.008) = 137.99 g/mol
b. 0.58 mol Al2O3•2H2O 
2 mol Al
26.98 g Al
= 31 g Al

mol Al2 O 3  2H 2 O
mol Al
c. 0.58 mol Al2O3•2H2O 
2 mol Al
6.022  10 23 atoms

mol Al 2 O 3  2H 2 O
mol Al
= 7.0 × 1023 atoms Al
d. 2.1 × 1024 formula units Al2O3•2H2O 
1 mol Al2O3  2H 2O
6.022  10
23
formula units

137 .99 g
mol
= 480 g Al2O3•2H2O
65.
a. 150.0 g Fe2O3 ×
b. 10.0 mg NO2 ×
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 ×
66.
a. 20.0 mg C8H10N4O2 ×
1 mol
6.02  10 23 molecules
= 2.5 × 10 8 mol BF3
1g
1 mol
= 1.03 × 10 4 mol C8H10N4O2

1000 mg 194.20 g
b. 2.72 × 1021 molecules C2H5OH ×
1 mol
6.022  10 23 molecules
= 4.52 × 10 3 mol C2H5OH
c. 1.50 g CO2 ×
67.
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
CHAPTER 3
STOICHIOMETRY
5.00 g C2H5O2N 
55
1 mol C2 H5O 2 N
6.022  10 23 molecules C2 H5O2 N

75.07 g C2 H5O 2 N
mol C2 H5O 2 N

1 atom N
= 4.01 × 1022 atoms N
molecule C 2 H5O 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

100.95 g Mg 3 N 2
mol Mg 3 N 2

2 atoms N
= 5.97 × 1022 atoms N
mol Mg 3 N 2
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 
1 mol Ca ( NO3 ) 2
2 mol N
6.022  10 23 atoms N


164 .10 g Ca ( NO3 ) 2
mol Ca ( NO3 ) 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 
68.
4.24 g C6H6 ×
1 mol N 2 O 4
2 mol N
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
5.43 × 10 2 mol C6H6 ×
6.022  10 23 molecules
= 3.27 × 1022 molecules C6H6
mol
Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 atoms total.
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
56
CHAPTER 3
2.71 × 1022 molecules CO2 ×
4.50 × 10 2 mol CO2 ×
1 mol
= 4.50 × 10 2 mol CO2
6.022  10 23 molecules
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 ×
9.27 × 10 3 mol CH3OH ×
69.
32.04 g
= 0.297 g CH3OH
mol
1g
1 mol
= 2.839 × 10 3 mol C6H8O6

1000 mg 176.12 g
2.839 × 10−3 mol 
6.022  10 23 molecules
= 1.710 × 1021 molecules C6H8O6
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 C9H8O4

1000 mg 180.15 g
2.78 × 10 3 mol ×
71.
1 mol
= 9.27 × 10 3 mol CH3OH
6.022  10 23 molecules
Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol
500.0 mg ×
70.
STOICHIOMETRY
a.
2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b. 500.0 g ×
c.
6.022  10 23 molecules
= 1.67 × 1021 molecules C9H8O4
mol
1 mol
= 3.023 mol C2H3Cl3O2
165.39 g
2.0 × 10-2 mol ×
165.39 g
= 3.3 g C2H3Cl3O2
mol
d. 5.0 g C2H3Cl3O2 ×
1 mol
6.022  10 23 molecules 3 atoms Cl


165 .39 g
mol
molecule
= 5.5 × 1022 atoms of chlorine
CHAPTER 3
72.
STOICHIOMETRY
57
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 H3Cl3O2
e.
1.0 g Cl ×
f.
500 molecules ×
1 mol
6.022  10
23
molecules

165 .39 g
= 1.373 × 1019 g C2H3Cl3O2
mol
As we shall see in later chapters, the formula written as (CH3)2N2O tries to tell us something
about how the atoms are attached to each other. For our purposes in this problem, we can
write the formula as C2H6N2O.
a. 2(12.01) + 6(1.008) + 2(14.01) + 1(16.00) = 74.09 g/mol
b. 250 mg 
1g
1 mol
= 3.4 × 10−3 mol

1000 mg 74.09 g
d. 1.0 mol C2H6N2O 
c.
0.050 mol ×
74.09 g
= 3.7 g
mol
6.022  10 23 molecules C2 H 6 N 2O
6 atoms of H

mol C2 H 6 N 2O
molecule C2 H 6 N 2O
= 3.6 × 1024 atoms of hydrogen
e. 1.0 × 106 molecules 
f.
1 mol
6.022  10
23
molecules
1 mol
1 molecule 
6.022  10
23
molecules


74.09 g
= 1.2 × 10−16 g
mol
74.09 g
= 1.230 × 10−22 g C2H6N2O
mol
Percent Composition
73.
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
Mass % C =
× 100 = 50.00% C
72.06 g compound
Mass % H =
4.032 g H
× 100 = 5.595% H
72.06 g compound
Mass % O = 100.00  (50.00 + 5.595) = 44.41% O or:
%O=
32.00 g
× 100 = 44.41% O
72.06 g
b. C4H6O2: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00
= 86.09 g/mol
Mass % C =
48.04 g
6.048 g
× 100 = 55.80% C; mass % H =
× 100 = 7.025% H
86.09 g
86.09 g
Mass % O = 100.00  (55.80 + 7.025) = 37.18% O
58
CHAPTER 3
STOICHIOMETRY
c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01
= 53.06 g/mol
74.
Mass % C =
36.03 g
3.024 g
× 100 = 67.90% C; mass % H =
× 100 = 5.699% H
53.06 g
53.06 g
Mass % N =
14.01 g
× 100 = 26.40% N or % N = 100.00 − (67.90 + 5.699)
53.06 g
= 26.40% N
In 1 mole of YBa2Cu3O7, there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu, and 7 moles
of O.
 88.91 g Y 
 137.3 g Ba 
  2 mol Ba 

Molar mass = 1 mol Y 
 mol Y 
 mol Ba 
 63.55 g Cu 
 + 7 mol O
+ 3 mol Cu 
 mol Cu 
 16.00 g O 


 mol O 
Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol
Mass % Y =
Mass % Cu =
75.
88.91 g
274 .6 g
× 100 = 13.35% Y; mass % Ba =
× 100 = 41.22% Ba
666 .2 g
666 .2 g
190.65 g
112 .0 g
× 100 = 28.62% Cu; mass % O =
× 100 = 16.81% O
666 .2 g
666.2 g
14.01 g N
× 100 = 46.68% N
30.01 g NO
NO: Mass % N =
NO2: Mass % N =
14.01 g N
× 100 = 30.45% N
46.01 g NO2
N2O: Mass % N =
2(14.01) g N
× 100 = 63.65% N
44.02 g N 2O
From the calculated mass percents, only NO is 46.7% N by mass, so NO could be this
species. Any other compound having NO as an empirical formula could also be the
compound.
76.
a.
C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol
Mass % C =
96.08 g
8(12.01) g C
× 100 =
× 100 = 49.47% C
194.20 g C8 H10 N 4O 2
194.20 g
b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol
Mass % C =
12(12.01) g C
× 100 = 42.10% C
342 .30 g C12 H 22O11
CHAPTER 3
c.
STOICHIOMETRY
59
C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol
Mass % C =
2(12.01) g C
× 100 = 52.14% C
46.07 g C 2 H 5OH
The order from lowest to highest mass percentage of carbon is:
sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH)
77.
There are 0.390 g Cu for every 100.000 g of fungal laccase. Assuming 100.00 g fungal
laccase:
1 mol Cu
1 mol fungal laccase
Mol fungal laccase = 0.390 g Cu ×
= 1.53 × 10 3 mol

63.55 g Cu
4 mol Cu
x g fungal laccase
100 .000 g
=
, x = molar mass = 6.54 × 104 g/mol
3
mol fungal laccase 1.53  10 mol
78.
There are 0.347 g Fe for every 100.000 g hemoglobin (Hb). Assuming 100.000 g
hemoglobin:
Mol Hb = 0.347 g Fe 
1 mol Fe
1 mol Hb
= 1.55 × 10−3 mol Hb

55.85 g Fe 4 mol Fe
x g Hb
100 .000 g Hb
=
, x = molar mass = 6.45 × 104 g/mol
mol Hb 1.55  10 3 mol Hb
Empirical and Molecular Formulas
79.
 12.01 g C 
 1.008 g H 
 + 2 mol H 

a. Molar mass of CH2O = 1 mol C 
 mol C 
 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
× 100 = 40.00%;
180.16 g C 6 H12 O 6
% O = 100.00  (40.00 + 6.714) = 53.29%
%H=
12.(1.008) g
× 100 = 6.714%
180 .16 g
60
CHAPTER 3
STOICHIOMETRY
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%
80.
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.
81.
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
82.
a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol
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
d. SN: 32.07 + 14.01 = 46.08 g/mol;
83.
184.32 g
= 4.000; molecular formula: S4N4
46.08 g
Out of 100.00 g of compound, there are:
48.64 g C ×
1 mol C
1 mol H
= 4.050 mol C; 8.16 g H ×
= 8.10 mol H
12.01 g C
1.008 g H
% O = 100.00 – 48.64 – 8.16 = 43.20%; 43.20 g O ×
Dividing each mole value by the smallest number:
1 mol O
= 2.700 mol O
16.00 g O
CHAPTER 3
STOICHIOMETRY
61
4.050
8.10
2.700
= 1.500;
= 3.00;
= 1.000
2.700
2.700
2.700
Because a whole number ratio is required, the C : H : O ratio is 1.5 : 3 : 1 or 3 : 6 : 2. So the
empirical formula is C3H6O2.
84.
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 mole value by the smallest number:
5.302
9.72
0.8838
= 6.000;
= 11.0;
= 1.000
0.8837
0.8837
0.8837
The empirical formula for nylon-6 is C6H11NO
85.
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 mole 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
0.401 g Hg ×
1 mol Hg
1 mol O
= 2.00 × 103 mol Hg; 0.016 g O ×
= 1.0 × 103 mol O
200 .6 g Hg
16.00 g O
The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O.
86.
1.121 g N ×
1 mol N
14.01 g N
= 8.001 × 10 2 mol N; 0.161 g H ×
0.480 g C ×
1 mol C
12.01 g C
= 4.00 × 10 2 mol C; 0.640 g O ×
Dividing all mole values by the smallest number:
1 mol H
1.008 g H
1 mol O
16.00 g O
= 1.60 × 10 1 mol H
= 4.00 × 10 2 mol O
62
CHAPTER 3
8.001  10 2
4.00  10 2
= 2.00;
1.60  10 1
4.00  10 2
= 4.00;
4.00  10 2
4.00  10 2
STOICHIOMETRY
= 1.00
The empirical formula is N2H4CO.
87.
Out of 100.0 g, there are:
69.6 g S ×
1 mol S
32.07 g S
= 2.17 mol S; 30.4 g N ×
1 mol N
14.01 g N
= 2.17 mol N
The empirical formula is SN because the mole values are in a 1 : 1 mole ratio.
The empirical formula mass of SN is ~ 46 g/mol. Because 184/46 = 4.0, the molecular
formula is S4N4.
88.
Assuming 100.0 g of compound:
26.7 g P ×
1 mol P
30.97 g P
61.2 g Cl ×
1 mol Cl
35.45 g Cl
= 0.862 mol P; 12.1 g N ×
1 mol N
14.01 g N
= 0.864 mol N
= 1.73 mol Cl
1.73
= 2.01; the empirical formula is PNCl2.
0.862
The empirical formula mass is  31.0 + 14.0 + 2(35.5) = 116 g/mol.
Molar mass
580
= 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10.

Empirical formula mass
116
89.
Assuming 100.00 g of compound:
47.08 g C ×
46.33 g Cl ×
1 mol C
1 mol H
= 3.920 mol C; 6.59 g H ×
= 6.54 mol H
12.01 g C
1.008 g H
1 mol Cl
= 1.307 mol Cl
35.45 g Cl
Dividing all mole values by 1.307 gives:
3.920
6.54
1.307
= 2.999;
= 5.00;
= 1.000
1.307
1.307
1.307
The empirical formula is C3H5Cl.
The empirical formula mass is 3(12.01) + 5(1.008) + 1(35.45) = 76.52 g/mol.
CHAPTER 3
STOICHIOMETRY
63
Molar mass
153
=
= 2.00 ; the molecular formula is (C3H5Cl)2 = C6H10Cl2.
Empirical formula mass
76.52
90.
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 ×
1 mol C
12.01 g C
= 3.446 mol C; 3.47 g H ×
55.14 g O ×
1 mol O
16.00 g O
= 3.446 mol O
1 mol H
1.008 g H
= 3.44 mol H
All are the same mole 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
Molar mass
116
= 4.00; molecular formula = (CHO)4 = C4H4O4

Empirical mass
29.02
91.
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 fructose combusted, the masses of C and H are:
mass C = 2.20 g CO2 ×
1 mol CO 2
44.01 g CO 2
mass H = 0.900 g H2O ×

1 mol C
12.01 g C
= 0.600 g C

mol CO 2
mol C
1 mol H 2 O
2 mol H
1.008 g H
= 0.101 g H


18.02 g H 2 O
mol H 2 O
mol H
Mass O = 1.50 g fructose  0.600 g C  0.101 g H = 0.799 g O
So, in 1.50 g of the fructose, we have:
0.600 g C ×
1 mol C
1 mol H
= 0.0500 mol C; 0.101 g H ×
= 0.100 mol H
12.01 g C
1.008 g H
0.799 g O ×
1 mol O
= 0.0499 mol O
16.00 g O
Dividing by the smallest number:
92.
0.100
= 2.00; the empirical formula is CH2O.
0.0499
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 the carbon in
33.5 mg CO2 came from the 35.0 mg of compound and all the hydrogen in 41.1 mg H2O
came from the 35.0 mg of compound.
64
CHAPTER 3
3.35 × 10 2 g CO2 ×
Mass % C =

1 mol C
12.01 g C
= 9.14 × 10 3 g C

mol CO2
mol C
9.14  10 3 g C
3.50  10 2 g compound
4.11 × 10 2 g H2O ×
Mass % H =
1 mol CO2
44.01 g CO2
STOICHIOMETRY
× 100 = 26.1% C
1 mol H 2O
2 mol H
1.008 g H
= 4.60 × 10 3 g H


18.02 g H 2O
mol H 2O
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:
Mass % 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
1 mol H
= 2.17 mol C; 13.1 g H ×
= 13.0 mol H
12.01 g C
1.008 g H
60.8 g N ×
1 mol N
= 4.34 mol N
14.01 g N
Dividing all mole values by 2.17 gives:
2.17
13.0
4.34
= 1.00;
= 5.99;
= 2.00
2.17
2.17
2.17
The empirical formula is CH6N2.
93.
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 ×
Mass % H =
1g
1000 mg

2.016 g H
1000 mg
= 4.79 mg H

18.02 g H 2O
g
4.79 mg H
× 100 = 10.1% H; mass % C = 100.0  10.1 = 89.9% C
47.6 mg cumene
Now solve the empirical formula problem. Out of 100.0 g cumene, we have:
89.9 g C ×
1 mol C
1 mol H
= 7.49 mol C; 10.1 g H ×
= 10.0 mol H
12.01 g C
1.008 g H
CHAPTER 3
STOICHIOMETRY
65
10.0
4
= 1.34  ; the mole H to mole C ratio is 4 : 3. The empirical formula is C3H4.
3
7.49
Empirical formula mass  3(12) + 4(1) = 40 g/mol.
The molecular formula must be (C3H4)3 or C9H12 because the molar mass of this formula will
be between 115 and 125 g/mol (molar mass  3 × 40 g/mol = 120 g/mol).
94.
There are several ways to do this problem. We will determine composition by mass percent:
16.01 mg CO2 ×
%C=

12.01 g C
1000 mg
= 4.369 mg C

44.01 g CO2
g
4.369 mg C
× 100 = 40.91% C
10.68 mg compound
4.37 mg H2O ×
%H=
1g
1000 mg
1g
1000 mg

2.016 g H
1000 mg
= 0.489 mg H

18.02 g H 2O
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
54.51 g O ×
1 mol O
= 3.407 mol O
16.00 g O
Dividing by the smallest number:
4.54
4
= 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/mol.
Because
176.1
= 2.0, the molecular formula is C6H8O6.
88
Balancing Chemical Equations
95.
When balancing reactions, start with elements that appear in only one of the reactants and one
of the products, and 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)
66
CHAPTER 3
STOICHIOMETRY
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. CS2(l) + 2 NH3(g)  H2S(g) + NH4SCN(s)
96.
An important part 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)
d. Sr(OH)2(aq) + 2 HBr(aq)  2H2O(l) + SrBr2(aq)
MnO2
catalyst
97.
2 H2O2(aq)
2 H2O(l) + O2(g)
98.
Fe3O4(s) + 4 H2(g)  3 Fe(s) + 4 H2O(g)
Fe3O4(s) + 4 CO(g)  3 Fe(s) + 4 CO2(g)
99.
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)
100.
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)
CHAPTER 3
STOICHIOMETRY
67
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)
101.
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.
15
C6H6(l) +
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)
102.
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)
103.
a. SiO2(s) + C(s)  Si(s) + CO(g); Si is balanced.
Balance oxygen atoms: SiO2 + C  Si + 2 CO
Balance carbon atoms: SiO2(s) + 2 C(s)  Si(s) + 2 CO(g)
68
CHAPTER 3
STOICHIOMETRY
b. SiCl4(l) + Mg(s)  Si(s) + MgCl2(s); Si is balanced.
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); Si is balanced.
Balance F atoms:
Na2SiF6 + Na  Si + 6 NaF
Balance Na atoms: Na2SiF6(s) + 4 Na(s)  Si(s) + 6 NaF(s)
104.
CaSiO3(s) + 6 HF(aq)  CaF2(aq) + SiF4(g) + 3 H2O(l)
Reaction Stoichiometry
105.
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 ×
106.
1 mol Fe
2 mol Al 26.98 g Al
= 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 2O3 159.70 g Fe 2O3
= 21.5 g Fe2O3

2 mol Fe
mol Fe 2O3
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)
52.9 g KClO3 ×
107.
1.000 kg Al ×
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



kg Al
26.98 g Al
3 mol Al
mol NH4ClO 4
= 4355 g = 4.355 kg NH4ClO4
108.
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
CHAPTER 3
STOICHIOMETRY
0.021 mol Ba(OH)2•8H2O ×
69
2 mol NH 4 SCN
76.13 g NH 4 SCN

1 mol Ba(OH)2  8H 2 O
mol NH 4 SCN
= 3.2 g NH4SCN
109.
a. 1.0 × 102 mg NaHCO3 
1 mol NaHCO3
1 mol C6 H8O7
1g


1000 mg 84.01 g NaHCO3 3 mol NaHCO3

b. 0.10 g NaHCO3 
192.12 g C6 H8O7
= 0.076 g or 76 mg C6H8O7
mol C6 H8O7
1 mol NaHCO3
3 mol CO2
44.01 g CO2


84.01 g NaHCO3 3 mol NaHCO3
mol CO2
= 0.052 g or 52 mg CO2
110.
a. 1.00 × 102 g C7H6O3 ×
1 mol C7 H6O3
1 mol C4 H 6O3 102.09 g C4 H6O3


138.12 g C7 H 6O3
1 mol C7 H 6O3
1 mol C4 H 6O3
= 73.9 g C4H6O3
b. 1.00 × 102 g C7H6O3 ×
1 mol C7 H 6O3
1 mol C9 H8O 4
180.15 g C9 H8O4


138.12 g C7 H 6O3
1 mol C7 H 6O3
mol C9 H8O4
= 1.30 × 102 g aspirin
111.
1.0 × 104 kg waste ×
1 mol C5H 7 O 2 N
3.0 kg NH4 
1 mol NH4 
1000 g




100 kg waste
kg
18.04 g NH4
55 mol NH4 
×
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
Because 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
112.
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
×
113.
1.0 ton CuO 
1 mol P4
2 mol Ca 3 ( PO4 ) 2
×
123 .88 g P4
= 150 g P4
mol P4
907 kg 1000 g
1 mol CuO
1 mol C
12.01 g C 100. g coke





ton
kg
79.55 g CuO 2 mol CuO
mol C
95 g C
= 7.2 × 104 g or 72 kg coke
114.
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.
70
CHAPTER 3
25,000 g LiOH ×
STOICHIOMETRY
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
1h
= 68 h = 2.8 days



0.0010 g air
1000 mL
140 L air
60 min
Limiting Reactants and Percent Yield
115.
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)
116.
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. Now use the 10 molecules of O2 and the molecule
relationships given in the balanced equation to determine the number of molecules of each
product formed, then complete the table.
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.
117.
a. The strategy we will generally use to solve limiting reactant problems is to assume each
reactant is limiting, and then calculate the quantity of product each reactant could
produce if it were limiting. The reactant that produces the smallest quantity of product
is the limiting reactant (runs out first) and therefore determines the mass of product that
can be produced.
Assuming N2 is limiting:
1.00 × 103 g N2 
2 mol NH3
17.03 g NH3
1 mol N 2
= 1.22 × 103 g NH3


28.02 g N 2
mol N 2
mol NH3
CHAPTER 3
STOICHIOMETRY
71
Assuming H2 is limiting:
2 mol NH3
17.03 g NH3
1 mol H 2
= 2.82 × 103 g NH3


2.016 g H 2
3 mol H 2
mol NH3
5.00 × 102 g H2 
Because N2 produces the smaller mass of product (1220 g vs. 2820 g NH3), N2 is limiting
and 1220 g NH3 can be produced. As soon as 1220 g of NH3 is produced, all of the N2 has
run out. Even though we have enough H2 to produce more product, there is no more N2
present as soon as 1220 g of NH3 have been produced.
b. 1.00  103 g N2 
1 mol N 2
3 mol H 2 2.016 g H 2
= 216 g H2 reacted


28.02 g N 2
mol N 2
mol H 2
Excess H2 = 500. g H2 initially – 216 g H2 reacted = 284 g H2 in excess (unreacted)
118.
Ca3(PO4)2 + 3 H2SO4  3 CaSO4 + 2 H3PO4
Assuming Ca3(PO4)2 is limiting:
1.0 × 103 g Ca3(PO4)2 ×
1 mol Ca 3 (PO4 ) 2
3 mol CaSO4
136.15 g CaSO4


310.18 g Ca 3 (PO4 ) 2 mol Ca 3 (PO4 ) 2
mol CaSO4
= 1300 g CaSO4
Assuming concentrated H2SO4 reagent is limiting:
1.0 × 103 g conc. H2SO4 ×
98 g H 2SO 4
1 mol H 2SO 4

100 g conc. H 2SO 4 98.09 g H 2SO 4

3 mol CaSO4 136.15 g CaSO4
= 1400 g CaSO4

3 mol H 2SO 4
mol CaSO4
Because Ca3(PO4)2 produces the smaller quantity of product, Ca3(PO4)2 is limiting and
1300 g CaSO4 can be produced.
1.0 × 103 g Ca3(PO4)2 
1 mol Ca 3 (PO4 ) 2
2 mol H3PO4
97.99 g H3PO4


310.18 g Ca 3 (PO4 ) 2
mol Ca 3 (PO4 ) 2
mol H3PO4
= 630 g H3PO4 produced
119.
Assuming BaO2 is limiting:
1.50 g BaO2 ×
1 mol BaO2
1 mol H 2O2 34.02 g H 2O2
= 0.301 g H2O2


169.3 g BaO2
mol BaO2
mol H 2O2
Assuming HCl is limiting:
25.0 mL ×
0.0272 g HCl 1 mol HCl 1 mol H 2 O 2 34.02 g H 2 O 2
= 0.317 g H2O2



mL
36.46 g HCl 2 mol HCl
mol H 2 O 2
72
CHAPTER 3
STOICHIOMETRY
BaO2 produces the smaller amount of H2O2, so it is limiting and a mass of 0.301 g of H2O2
can be produced.
Initial mol HCl present: 25.0 mL 
0.0272 g HCl
1 mol HCl
= 1.87 × 102 mol HCl

mL
36.46 g HCl
The amount of HCl reacted:
1.50 g BaO2 ×
1 mol BaO2
2 mol HCl
= 1.77 × 10 2 mol HCl

169.3 g BaO2
mol BaO2
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 ×
120.
36.46 g HCl
= 3.6 × 10 2 g HCl unreacted
mol HCl
Assuming Ag2O is limiting:
25.0 g Ag2O ×
2 mol AgC10H9 N 4SO 2 357.18 g AgC10H9 N 4SO 2
1 mol Ag2O


231.8 g Ag2O
mol Ag2O
mol AgC10H9 N 4SO 2
= 77.0 g AgC10H9N4SO2
Assuming C10H10N4SO2 is limiting:
50.0 g C10H10N4SO2 ×
1 mol C10H10N 4SO 2
2 mol AgC10H9 N 4SO 2

250.29 g C10H10N 4SO 2
2 mol C10H10N 4SO 2

357.18 g AgC10H9 N 4SO 2
= 71.4 g AgC10H9N4SO2
mol AgC10H9 N 4SO 2
Because C10H10N4SO2 produces the smaller amount of product, it is limiting and 71.4 g of
silver sulfadiazine can be produced.
121.
To solve limiting-reagent problems, we will generally assume each reactant is limiting and
then 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 ×
5.00 × 106 g O2 ×
5.00 × 106 g CH4 ×
1 mol NH3
2 mol HCN
= 2.94 × 105 mol HCN

17.03 g NH3
2 mol NH3
1 mol O 2
2 mol HCN
= 1.04 × 105 mol HCN

32.00 g O 2
3 mol O 2
1 mol CH 4
2 mol HCN
= 3.12 × 105 mol HCN

16.04 g CH 4
2 mol CH 4
CHAPTER 3
STOICHIOMETRY
73
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 ×
5.00 × 106 g O2 ×
122.
27.03 g HCN
= 2.81 × 106 g HCN
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
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.
123.
C2H6(g) + Cl2(g)  C2H5Cl(g) + HCl(g)
If C2H6 is limiting:
300. g C2H6 ×
1 mol C2 H 6
1 mol C2 H5Cl 64.51 g C2 H5Cl
= 644 g C2H5Cl


30.07 g C2 H 6
mol C2 H 6
mol C2 H5Cl
If Cl2 is limiting:
650. g Cl2 ×
1 mol C2 H5Cl 64.51 g C2 H5Cl
1 mol Cl2
= 591 g C2H5Cl


70.90 g Cl2
mol Cl2
mol C2 H5Cl
Cl2 is limiting because it produces the smaller quantity of product. Hence, the theoretical
yield for this reaction is 591 g C2H5Cl. The percent yield is:
percent yield =
124.
490 . g
actual
× 100 =
× 100 = 82.9%
591 g
theoretical
a. 1142 g C6H5Cl ×
1 mol C6 H5Cl
1 mol C14H9Cl5 354.46 g C14H9Cl5


112.55 g C6 H5Cl
2 mol C6 H5Cl
mol C14H9Cl5
= 1798 C14H9Cl5
74
CHAPTER 3
485 g C2HOCl3 ×
STOICHIOMETRY
1 mol C2 HOCl3
1 mol C14H9Cl5 354.46 g C14H9Cl5


147.38 g C2 HOCl3
mol C2 HOCl3
mol C14H9Cl5
= 1170 g C14H9Cl5
From the masses of product calculated, C2HOCl3 is limiting and 1170 g C14H9Cl5 can be
produced.
b. C2HOCl3 is limiting, and C6H5Cl is in excess.
c. 485 g C2HOCl3 ×
1 mol C2 HOCl3
2 mol C6 H5Cl 112.55 g C6 H5Cl


147.38 g C2 HOCl3
mol C2 HOCl3
mol C6 H5Cl
= 741 g C6H5Cl reacted
1142 g  741 g = 401 g C6H5Cl in excess
d. Percent yield =
125.
200.0 g DDT
× 100 = 17.1%
1170 g DDT
2.50 metric tons Cu3FeS3 
1 mol Cu 3FeS3
1000 kg
1000 g
3 mol Cu



metric ton
kg
342.71 g
1 mol Cu 3FeS3

1.39 × 106 g Cu (theoretical) ×
126.
63.55 g
= 1.39 × 106 g Cu (theoretical)
mol Cu
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) 
154 g PF3 
100.0 g PF3 (theoretical)
= 154 g PF3 (theoretical)
78.1 g PF3 (actual)
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 is needed to actually produce 120. g of PF3 if the percent yield is 78.1%.
Additional Exercises
127.
12
C21H6: 2(12.000000) + 6(1.007825) = 30.046950 u
12
C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 u
14
N16O: 1(14.003074) + 1(15.994915) = 29.997989 u
The peak results from 12C1H216O.
CHAPTER 3
128.
STOICHIOMETRY
75
We would see the peaks corresponding to:
10
B35Cl3 [mass  10 + 3(35) = 115 u], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119),
10
B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122)
We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120,
121, and 122.
129.
0.368 g XeFn
Molar mass XeFn =
9.03  10
20
molecules XeFn 
1 mol XeFn
= 245 g/mol
6.022  10 23 molecules
245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6
130.
a.
14 mol C ×
12.01 g
1.008 g
14.01 g
+ 18 mol H ×
+ 2 mol N ×
mol C
mol H
mol N
+ 5 mol O ×
1 mol C14H18N 2O5
= 3.40 × 102 mol C14H18N2O5
294.30 g C14H18N 2O5
b. 10.0 g C14H18N2O5 ×
c.
1.56 mol ×
d. 5.0 mg ×
16.00 g
= 294.30 g
mol O
294.3 g
= 459 g C14H18N2O5
mol
1g
1 mol
6.022  10 23 molecules


1000 mg 294 .30 g
mol
= 1.0 × 1019 molecules C14H18N2O5
e. The chemical formula tells us that 1 molecule of C14H18N2O5 contains 2 atoms of N. If
we have 1 mole of C14H18N2O5 molecules, then 2 moles of N atoms are present.
1.2 g C14H18N2O5 ×
1 mol C14H18N 2O5
2 mol N

294.30 g C14H18N 2O5 mol C14H18N 2O5

f.
1.0 × 109 molecules ×
g. 1 molecule 
131.
1 mol
6.022  10
1 mol
6.022  10
23
atoms
23

atoms
6.022  10 23 atoms N
= 4.9 × 1021 atoms N
mol N

294 .30 g
= 4.9 × 1013 g
mol
294 .30 g
= 4.887 × 1022 g C14H18N2O5
mol
Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol
Mass % C =
20(12.01) g C
× 100 = 71.40% C
336 .43 g compound
76
CHAPTER 3
Mass % H =
Mass % F =
STOICHIOMETRY
29(1.008) g H
× 100 = 8.689% H
336 .43 g compound
19.00 g F
× 100 = 5.648% F
336 .43 g compound
Mass % O = 100.00  (71.40 + 8.689 + 5.648) = 14.26% O or:
%O=
132.
3(16.00) g O
× 100 = 14.27% O
336 .43 g compound
In 1 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/h.
133.
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 mole 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.
134.
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
1.008 g H
= 6.85 mol 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
Because a whole number ratio is required, the empirical formula is C3H5O2.
CHAPTER 3
STOICHIOMETRY
77
Empirical formula mass: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol
146.1
Molar mass
=
= 1.999; molecular formula = (C3H5O2)2 = C6H10O4
Empirical formula mass 73.07
135.
There are many valid methods to solve this problem. We will assume 100.00 g of compound,
and then determine from the information in the problem how many moles of compound
equals 100.00 g of compound. From this information, we can determine the mass of one
mole of compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:
1 mol Co
1 mol cyanocobalamin
mol cyanocobalamin = 4.34 g Co ×

58.93 g Co
mol Co
= 7.36 × 10 2 mol cyanocobalamin
x g cyanocobalamin
100 .00 g
=
, x = molar mass = 1.36 × 103 g/mol
2
1 mol cyanocobalamin
7.36  10 mol
136.
2 tablets ×
0.262 g C7 H5 BiO 4
1 mol C7 H5BiO 4
1 mol Bi
209.0 g Bi



tablet
362.11 g C7 H5BiO 4 1 mol C7 H5BiO 4
mol Bi
= 0.302 g Bi consumed
137.
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 ×
138.
1 mol C 8 H 8
8 mol H
6.022  10 23 atoms H
= 9.25 × 1022 atoms H


104.14 g C 8 H 8
mol C 8 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
78
CHAPTER 3
STOICHIOMETRY
The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2.
Mass of C4H3O2  4(12) + 3(1) + 2(16) = 83g/mol.
Molar mass =
139.
17.3 g H ×
41.5 g
166
= 166 g/mol;
= 2.0; the molecular formula is 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/mol, so two times the empirical formula would put the
compound in the correct range of the molar mass. Molecular formula = (C2H5)2 = C4H10.
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 ×
140.
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 u

1 mol E
3.25 mol E
Note: From the periodic table, element E is silicon, Si.
141.
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•5H2O, x = 5
0.00303 g CuSO 4
1 mol CuSO 4
142.
a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:
8.80 g N ×
1 mol C3H3 N 53.06 g C3H 3 N
= 33.3 g C3H3N

14.01 g N
1 mol C3H 3 N
CHAPTER 3
STOICHIOMETRY
% C3H3N =
79
33.3 g C3H3 N
= 33.3% C3H3N
100.00 g polymer
Only butadiene in the polymer reacts with Br2:
0.605 g Br2 ×
% C4H6 =
1 mol C4 H 6
54.09 g C4 H 6
1 mol Br2
= 0.205 g C4H6


159.8 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 C3H 3 N
= 0.628 mol C3H3N
53.06 g
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 C8 H8
= 0.476 mol C8H8
104.14 g C8 H8
Dividing by 0.316:
0.628
0.316
0.476
= 1.99;
= 1.00;
= 1.51
0.316
0.316
0.316
This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3
styrene molecules in the polymer, or (A4B2S3)n.
143.
1.20 g CO2 ×
1 mol C24H30N3O
1 mol CO2
1 mol C
376.51 g



44.01 g
mol CO2
24 mol C
mol C24H30N3O
= 0.428 g C24H30N3O
0.428 g C 24 H 30 N 3 O
× 100 = 42.8% C24H30N3O (LSD)
1.00 g sample
144.
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 ×
120. g S ×
1 mol CH 4
1 mol CS2
76.15 g CS2
= 570. g CS2


16.04 g CH 4
mol CH 4
mol CS2
1 mol CS2
76.15 g CS2
1 mol S
= 71.2 g CS2


32.07 g S
4 mol S
mol CS2
Because S produces the smaller quantity of CS2, sulfur is the limiting reactant and 71.2 g
CS2 can be produced. The same amount of CS2 would be produced using the balanced
equation with S8.
80
145.
CHAPTER 3
126 g B5H9 ×
192 g O2 ×
STOICHIOMETRY
1 mol B5H9
9 mol H 2O 18.02 g H 2O
= 162 g H2O


63.12 g B5H9 2 mol B5H9
mol H 2O
1 mol O2
9 mol H 2 O 18.02 g H 2 O
= 81.1 g H2O


32.00 g O2
12 mol O2
mol H 2 O
Because O2 produces the smallest quantity of product, O2 is limiting and 81.1g H2O can be
produced.
146.
2 NaNO3(s)  2 NaNO2(s) + O2(g); the amount of NaNO3 in the impure sample is:
0.2864 g NaNO2 
2 mol NaNO3
85.00 g NaNO3
1 mol NaNO2


69.00 g NaNO2
2 mol NaNO2
mol NaNO3
= 0.3528 g NaNO3
0.3528 g NaNO3
Mass percent NaNO3 =
× 100 = 83.40%
0.4230 g sample
147.
453 g Fe ×
1 mol Fe 2O3 159.70 g Fe 2O3
1 mol Fe
= 648 g Fe2O3


55.85 g Fe
2 mol Fe
mol Fe 2O3
Mass percent Fe2O3 =
148.
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.
149.
Assuming 1 mole 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
Because 1 mole 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.
CHAPTER 3
150.
STOICHIOMETRY
81
a. At 40.0 g of Na added, Cl2 and Na both run out at the same time (both are limiting reactants). Past 40.0 g of Na added, Cl2 is limiting, and because the amount of Cl2 present in
each experiment was the same quantity, no more NaCl can be produced. Before 40.0 g of
Na added, Na was limiting. As more Na was added (up to 40.0 g Na), more NaCl was
produced.
b. 20.0 g Na 
1 mol Na
2 mol NaCl 58.44 g NaCl
= 50.8 g NaCl


22.99 g Na
2 mol Na
mol NaCl
c. At 40.0 g Na added, both Cl2 and Na are present in stoichiometric amounts.
40.0 g Na 
1 mol Cl2 70.90 g Cl2
1 mol Na
= 61.7 g Cl2


22.99 g Na 2 mol Na
mol Cl2
61.7 g Cl2 was present at 40.0 g Na added, and from the problem, the same 61.7 g Cl2 was
present in each experiment.
d. At 50.0 g Na added, Cl2 is limiting:
61.7 g Cl2 
e. 20.0 g Na 
1 mol Cl2
2 mol NaCl 58.44 g NaCl
= 101.7 g = 102 g NaCl


70.90 g Cl2
1 mol Cl2
mol NaCl
1 mol Cl2 70.90 g Cl2
1 mol Na
= 30.8 g Cl2 reacted


22.99 g Na 2 mol Na
mol Cl2
Excess Cl2 = 61.7 g Cl2 initially – 30.8 g Cl2 reacted = 30.9 g Cl2 in excess
Note: We know that 40.0 g Na is the point where Na and the 61.7 g of Cl2 run out at the
same time. So if 20.0 g of Na are reacted, one-half of the Cl2 that was present at 40.0 g
Na reacted will be in excess. The previous calculation confirms this.
For 50.0 g Na reacted, Cl2 is limiting and 40.0 g Na will react as determined previously.
Excess Na = 50.0 g Na initially – 40.0 g Na reacted = 10.0 g Na in excess.
151.
X2Z: 40.0% X and 60.0% Z by mass;
40.0/A x
(40.0)A z
mol X
or Az = 3Ax,
2

mol Z
60.0/A z
(60.0)A x
where A = molar mass.
For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax.
Mass percent X =
Ax
× 100 = 14.3% X; % Z = 100.0  14.3 = 85.7% Z
7A x
ChemWork Problems
The answers to the problems 152-159 (or a variation to these problems) are found in OWL. These
problems are also assignable in OWL.
82
CHAPTER 3
STOICHIOMETRY
Challenge Problems
160.
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)
Ga (0.40)
69
0.36
0.24
71
0.24
0.16
Ga (0.60)
Ga (0.40)
288
161.
71
290
292
The volume of a gas is proportional to the number of molecules of gas. Thus the formulas
are:
I: NH3 ;
II: N2H4;
III: HN3
The mass ratios are:
I:
4.634 g N
82.25 g N
6.949 g N
41.7 g N
=
; II:
; III:
17.75 g H
gH
gH
gH
If we set the atomic mass of H equal to 1.008, then the atomic mass, A, for nitrogen is:
I: 14.01;
II: 14.01;
For example, for compound I:
85
162.
87
Rb atoms
= 2.591;
Rb atoms
III. 14.0
A
4.634
=
, A = 14.01
3(1.008)
1
if we had exactly 100 atoms, x = number of 85Rb atoms, and
100  x = number of 87Rb atoms.
CHAPTER 3
STOICHIOMETRY
83
x
259 .1
= 2.591, x = 259.1  (2.591)x, x =
= 72.15; 72.15% 85Rb
100  x
3.591
0.7215(84.9117) + 0.2785(A) = 85.4678, A =
163.
85.4678  61.26
= 86.92 u
0.2785
First, we will determine composition in mass percent. We assume that all the carbon in the
0.213 g CO2 came from the 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
0.157 g compound
0.0310 g H2O ×
2.016 g H
3.47  10 3 g
= 3.47 × 103 g H; % H =
 100 = 2.21% H
0.157 g
18.02 g H 2O
We get the mass percent of 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% O
So, out of 100.00 g of compound, there are:
37.0 g C ×
1 mol C
1 mol H
= 3.08 mol C; 2.21 g H ×
= 2.19 mol H
12.01 g C
1.008 g H
18.3 g N ×
1 mol N
1 mol O
= 1.31 mol N; 42.5 g O ×
= 2.66 mol O
14.01 g N
16.00 g O
Lastly, and often the hardest part, we need 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.
84
164.
CHAPTER 3
1.0 × 106 kg HNO3 ×
STOICHIOMETRY
1000 g HNO3
1mol HNO3
= 1.6 × 107 mol HNO3

kg HNO3
63.02 g HNO3
We need to get the relationship between moles of HNO3 and moles of NH3. We have to use
all three equations.
2 mol HNO3
16 mol HNO3
2 mol NO2
4 mol NO



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 NH3

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; that is, 1 mole of
NH3 produces 1 mole of HNO3. Taking into consideration that NO is recycled back gives an
answer of 2.7 × 105 kg NH3 reacted.
165.
Fe(s) +
1
2
O2 (g)  FeO(s) ; 2 Fe(s) +
20.00 g Fe 
3
2
O 2 (g)  Fe 2O3 (s)
1 mol Fe
= 0.3581 mol
55.85 g
(11.20  3.24) g O 2 
1 mol O 2
= 0.2488 mol O2 consumed (1 extra sig. fig.)
32.00 g
Let’s assume x moles of Fe reacts to form x moles of FeO. Then 0.3581 – x, the remaining
moles of Fe, reacts to form Fe2O3. Balancing the two equations in terms of x:
1
x O 2  x FeO
2
3
(0.3581  x) mol Fe 
2
x Fe +
 0.3581  x 
 0.3581  x 

 mol O 2  
 mol Fe 2O3
2
2




Setting up an equation for total moles of O2 consumed:
1
2
x 
3
4
(0.3581  x)  0.2488 mol O2 , x  0.0791  0.079 mol FeO
0.079 mol FeO 
71.85 g FeO
= 5.7 g FeO produced
mol
Mol Fe2O3 produced =
0.140 mol Fe2O3 
0.3581  0.079
= 0.140 mol Fe2O3
2
159 .70 g Fe 2O3
= 22.4 g Fe2O3 produced
mol
CHAPTER 3
166.
STOICHIOMETRY
85
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 equation to set up a
mathematical expression for the moles of O2 required.
x
7
9.780  x 5


 = 1.120 mol O2
30.07 2
44.09
1
Solving: x = 3.7 g C2H6;
167.
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 produced =
moles Zn reacted + moles Mg reacted.
Mol H2 = 0.5171 g H2 ×
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
168.
a N2H4 + b NH3 + (10.00  4.062) O2  c NO2 + d H2O
Setting up four equations to solve for the four unknowns:
2a + b = c
(N mol balance)
2c + d = 2(10.00  4.062)
(O mol balance)
4a + 3b = 2d
(H mol balance)
a(32.05) + b(17.03) = 61.00
(mass balance)
Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4.
1.1 mol N 2 H 4  32.05 g / mol N 2 H 4
 100 = 58% N2H4
61.00 g
169.
We know that water is a product, so one of the elements in the compound is hydrogen.
XaHb + O2  H2O + ?
86
CHAPTER 3
To balance the H atoms, the mole ratio between XaHb and H2O =
Mol compound =
STOICHIOMETRY
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
, 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), and Li (a = 8). N fits
the data best, so N4H6 is the most likely formula.
170.
The balanced equation is 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g)
The mole ratio of Sc : H2 =
Mol Sc = 2.25 g Sc ×
2
.
x
1 mol Sc
= 0.0500 mol Sc
44.96 g Sc
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.
x
0.07450
171.
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

3
board
cm
Amount of Cu to be recovered = 0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu.
5.5 × 105 g Cu ×
1mol Cu(NH3 ) 4 Cl2 202.59 g Cu(NH3 ) 4 Cl2
1mol Cu


63.55 g Cu
mol Cu
mol Cu(NH3 ) 4 Cl2
= 1.8 × 106 g Cu(NH3)4Cl2
5.5 × 105 g Cu ×
172.
4 mol NH3 17.03 g NH3
1mol 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 mol of acetaminophen compared
to a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass
(where M = molar mass acetaminophen):
percent yield =
3 mol  M
× 100 = 75%
4 mol  M
CHAPTER 3
STOICHIOMETRY
87
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 percent yield of 88%.
173.
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 the mole 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 XCl 2
1 mol Cl
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 u, so X is Pb.
3.60  10 2 mol X
174.
4.000 g M2S3  3.723 g MO2
There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the
reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M:


4.000 g
3.723 g
8.000
3.723
2
,


A  32.00
 2A  3(32.07)  A  2(16.00) 2A  96.21
(8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A = 102.2, A = 184 g/mol; atomic mass
= 184 u
Note: From the periodic table, M is tungsten, W.
175.
Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2
anions. The simplest compound between 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.
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.
Molar mass 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.
88
176.
CHAPTER 3
STOICHIOMETRY
Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g.
Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol
Mol KCl =
x
y
; mol KNO3 =
74.55
101.11
Knowing that the mixture is 43.2% K, then in the 100.0 g mixture, an expression for the mass
of K is:
y 
 x
39.10 

 = 43.2
 74.55 101.11 
We have two equations and two unknowns:
(0.5245)x + (0.3867)y = 43.2
x +
y = 100.0
Solving, x = 32.9 g KCl;
177.
32.9 g
 100 = 32.9% KCl
100.0 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 moles of NO formed, and let 4y = number of moles 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:
20x + 28y = 13.0
20x  20y = 10.0
8y = 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
178.
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
CHAPTER 3
STOICHIOMETRY
89
0.400 g H 2 O 
Mass % H in 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
= 4.48% H
1.00 g aspirin
Mass % O = 100.00  (60.0 + 4.48) = 35.5% 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
35.5 g O ×
1 mol O
= 2.22 mol O
16.00 g O
Dividing by the smallest number:
5.00
4.44
= 2.25;
= 2.00
2.22
2.22
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
179.
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
6.022  10
c. 9.071 × 10 20 g Ru ×
180.
23
atoms Pt

195 .1 g Pt
= 9.071 × 10 20 g Pt
mol Pt
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:
41.38 g C ×
5.37 g H ×
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
90
CHAPTER 3
STOICHIOMETRY
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
181.
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 (= MCl3) compound:
54.47 g Cl ×
1 mol
= 1.537 mol Cl
35.45 g
1.537 mol Cl ×
Molar mass of M =
1 mol M
= 0.5123 mol M
3 mol Cl
45.53 g M
= 88.87 g/mol M
0.5123 mol M
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
CHAPTER 3
STOICHIOMETRY
91
Assuming Y is limiting:
2 mol YCl3
195.26 g YCl3
1 mol Y
= 2.20 g YCl3


88.91 g Y
2 mol Y
1 mol YCl3
1.00 g Y ×
Because Cl2, when it all reacts, produces the smaller amount of product, Cl2 is the limiting
reagent, and the theoretical yield is 1.84 g YCl3.
182.
2 As + 4 AsI3  3 As2I4
Volume of As cube = (3.00 cm)3 = 27.0 cm3
27.0 cm3 ×
5.72 g As
cm
3

3 mol As 2 I 4
657.44 g As 2 I 4
1 mol As
= 2030 g As2I4


74.92 g As
2 mol As
mol As 2 I 4
1.01 × 1024 molecules AsI3 ×
1 mol AsI3
6.022  10
23
molecules AsI3


3 mol As 2 I 4
4 mol AsI3
657.44 g As 2 I 4
 827 g As 2 I 4
mol As 2 I 4
Because the reactant AsI3 produces the smaller quantity of product, then AsI3 is the limiting
reactant and 827 g As2I4 is the theoretical yield.
0.756 =
actual yield
, actual yield = 0.756 × 827 g = 625 g As2I4
827 g
Marathon Problems
183.
Let M = unknown element; mass O in oxide = 3.708 g – 2.077 g = 1.631 g O
In 3.708 g of compound:
1.631 g O ×
1 mol O
= 0.1019 g mol O
16.00 g O
If MO is the formula of the oxide, then M has a molar mass of
2.077 g M
= 20.38 g/mol.
0.1019 mol M
This is too low for the molar mass. We must have fewer moles of M than moles O present in
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:
2.077 g M
= 40.77 g/mol
1 mol M
0.1019 mol O 
2 mol O
This is close to calcium, but calcium forms an oxide having the CaO formula, not CaO2.
92
CHAPTER 3
STOICHIOMETRY
If MO3 is assumed to be the formula, then the molar mass of M calculates to be 61.10 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, and 2.75 (these are reasonable because they
will lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 : 1 gives a molar
mass of:
2.077 g M
= 50.96 g/mol
1 mol M
0.1019 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 mole ratios between 2 and 3 do not give as
reasonable results. Therefore, M is most likely vanadium, and the formula is V 2O5.
184.
a. i.
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 fewer 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 fewer 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 B 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 (which has a similar molar mass to CO2). This is also the empirical
formula.
Note:
3(12.01)
× 100 = 81.71% C. So this checks.
3(12.01)  8(1.008)