CHAPTER THREE
STOICHIOMETRY
12,15,18,22,28,30,36,40,42,44,46,48,51,56,60,62,67,68,70,
74,82,84,86,90,92,98,104,109,110
Questions
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).
Exercises
Atomic Masses and the Mass Spectrometer
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).
30.
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.
Moles and Molar Masses
40
CHAPTER 3
36.
STOICHIOMETRY
5.0 × 1021 atoms C ×
8.3 × 10 3 mol C ×
37.
41
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
42.
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 ×
44.
a. 5.00 mol P4O6 ×
1 mol Na 2 HPO4
= 7.04 × 10 3 mol Na2HPO4
141 .96 g
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 ×
42
46.
48.
CHAPTER 3
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 P4O6 ×
STOICHIOMETRY
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
51.
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
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 ×
1 mol
= 2.5 × 10 8 mol BF3
23
6.02  10 molecules
CHAPTER 3
54.
STOICHIOMETRY
43
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
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
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 total atoms.
44
CHAPTER 3
3.27 × 1022 molecules C6H6 ×
STOICHIOMETRY
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
6.022  10 23 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 ×
9.27 × 10 3 mol CH3OH ×
1 mol
= 9.27 × 10 3 mol CH3OH
6.022  10 23 molecules
32.04 g
= 0.297 g CH3OH
mol
Percent Composition
60.
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:
CHAPTER 3
STOICHIOMETRY
%O =
62.
45
3(16.00) g O
× 100 = 14.27% O
336 .43 g compound
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)
Empirical and Molecular Formulas
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
68.
d. Molecular formula: C6H12O6; empirical formula = CH2O
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
46
CHAPTER 3
STOICHIOMETRY
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
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
74.
Assuming 100.0 g of compound:
26.7 g P ×
61.2 g Cl ×
1 mol P
30.97 g P
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; 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
Balancing Chemical Equations
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)
CHAPTER 3
STOICHIOMETRY
47
c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
d. Sr(OH)2(aq) + 2 HBr(aq) → 2H2O(l) + SrBr2(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)
86.
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)
90.
10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s)
52.9 g KClO3 ×
92.
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
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
Limiting Reactants and Percent Yield
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
48
CHAPTER 3
STOICHIOMETRY
determine the limiting reactant, let’s calculate how much of one reactant is necessary to react
with the other.
4 molecules NH3
= 8 molecules NH3 to react with all of the O2
5 molecules O 2
10 molecules O2 ×
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.
104.
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 =
200.0 g DDT
× 100 = 17.1%
1170 g DDT
CHAPTER 3
STOICHIOMETRY
49
Additional Exercises
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.
g. 5.00 g H2 ×
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
H2 is limiting because 56.00 g O2 are present.
110.
2 tablets ×
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
= 0.302 g Bi consumed