Homework, chapter 7: 1, 5, 15, 26, 33
1. Determine the molar masses of these compounds:
a) KBr
molar mass K: 39.10, Br: 79.90; molar mass KBr = 119.80
b) Na2SO4
Na: 22.99, S: 32.07, O: 16.00; molar mass Na2SO4 =142.05
c) Pb(NO3)2
331.2
d) C2H5OH
46.07
e) HC2H3O2
60.05
f) Fe3O4
231.6
g) C12H22O11
342.3
h) Al2(SO4)3
342.2
i) (NH4)2HPO4
132.1
5. Calculate the number of grams in each of the following:
 197.0 g Au 
a) 0.550 mol Au
0.550 mol Au x 
 = 108 g Au
 1 mol Au 
 18.02 g H2O 
b) 15.8 mol H2O
15.8 mol H2O x  1 mol H O  = 285 g H2O


2
c) 12.5 mol Cl2
d) 3.15 mol NH4NO3.
886.3 g Cl2
252 g NH4NO3.
15. Exactly 1 mol of carbon disulfide contains:
a) How many carbon disulfide molecules?
b) How many carbon atoms?
c) How many sulfur atoms?
e) How many total atoms of all kinds?
6.022 x 1023 molecules CS2.
6.022 x 1023 atoms C.
1.204 x 1024 atoms S.
1.807 x 1024 atoms total.
26. A sample of ethylene chloride was analyzed to contain 6.00 g of C, 1.00 g of H, and
17.75 g of Cl. Calculate the percent composition of ethylene chloride.
6.00 g C x 
1mol C atoms 
1mol C atoms 
= 0.500 mol C atoms; 1.00 g H x 
12.01 g C 
 1.008 g H  = 1 mol H atoms
1mol Cl atoms 
17.75 g Cl x 
 35.45 g Cl  = 0.501 mol Cl atoms.
Empirical formula CH2Cl; molar mass: 49.48
12.01
2.016
35.45
%C =  49.48  x 100 = 24.3%; %H =  49.48  x 100 = 4.07%; %Cl =  49.48  x 100 = 71.6%

33. A sample of tin having a mass of 3.996 g was oxidized and found to have combined
with 1.077 g of oxygen. Calculate the empirical formula of this oxide of tin.
1mol Sn atoms 
3.996 g Sn x 
 118.7 g Sn  = 0.03367 mol Sn atoms;
1mol O atoms 
1.077 g O x 
 16.00 g O  = 0.06731 mol O atoms
divide with smaller #  Sn = 1, O = 2
Empirical formula: SnO2.
Homework, chapter 9: 3, 7, 13, 15, 23, 29
3.Calculate the number of grams in these quantities:
a) 2.55 mol Fe(OH)3; molar mass Fe(OH)3 = 72.86
 72.86 g Fe(OH)3 
2.55 mol Fe(OH)3 x  1 mol Fe(OH)  = 186 g Fe(OH)3

3 
b) 125 kg CaCO3;
c) 10.5 mol NH3;
d) 72 millimol HCl
1.25 x 105 g CaCO3.
179 g NH3
2.6 g HCl
 3.19 g Br2 
e) 500 mL of liquid Br2 (d = 3.119 g/mL); 500 mL Br2 x  1 mL Br  = 1.56 kg

2 
7. Given the equation for the combustion of isopropyl alcohol:
2 C3H7OH + 9 O2  6 CO2 + 8 H2O
set up the mole ratio of
6
mol
CO
2


 2 mol C3H7OH 
a) CO2 to C3H7OH: 
b) C3H7OH to O2: 


9 mol O2
 2 mol C3H7OH 


 9 mol O2 
c) O2 to CO2: 

 6 mol CO2 
 5 mol CO2 
e) CO2 to H2O: 

 8 mol H2O 
 8 mol H2O 
d) H2O to C3H7OH: 

 2 mol C3H2OH 
 8 mol H2O 
f) H2O to O2: 

 9 mol O2 
13. How many grams of sodium hydroxide can be produced from 500. g calcium
hydroxide according to this equation?
Ca(OH)2 + Na2CO3  2 NaOH + CaCO3 (molar mass CaCO3: 100.1, NaOH: 40.00)
 1 mol CaCO3   2 mol NaOH   40.00g NaOH 
500. g CaCO3 x  100.1 g CaCO  x  1 mol CaCO  x  1 mol NaOH  = 400. g NaOH


3 
3 
15. In a blast furnace, iron(III) oxide reacts with coke (carbon) to produce molten iron
and carbon monoxide: Fe2O3 + C  2 Fe + 3 CO; How many kg of Fe would be formed
from 125 kg Fe2O3?
 1 mol Fe2O3   2 mol Fe   55.85g Fe 
1.25 x 103 g Fe2O3 x 
x
x
 = 87.4 kg Fe
 159.7 g Fe2O3   1 mol Fe2O3   1 mol Fe 
23. In the following equations, determine which reactant is the limiting reactant and
which is in excess. The amounts used are given below each reactant. Show evidence for
your answers.
a) KOH + HNO3  KNO3 + H2O
16.0 g 12.0 g
 1 mol KOH   1 mol KNO3 
16.0 g KOH x  56.11 g KOH  x  1 mol KOH  = 0.285 mol KNO3;

 

 1 mol HNO3   1 mol KNO3 
12.0 g HNO3 x  63.02 g HNO  x  1 mol HNO  = 0.190 mol KNO3.

3 
3
The limiting reactant is HNO3 and KOH is in excess.
The limiting reactant is HNO3 and KOH is in excess.
b) 2 NaOH + H2SO4 Na2SO4 + H2O
10.0 g 10.0 g
 1 mol NaOH   1 mol H2O 
10.0 g NaOH x  40.00 g NaOH  x  2 mol NaOH  = 0.125mol H2O;

 

 1 mol H2SO4   1 mol H2O 
10.0 g H2SO4 x  98.09 g H SO  x  1 mol H SO  = 0.102 mol H2O;

2
4 
2
4
The limiting reactant is H2SO4 and NaOH is in excess.
29. Aluminum reacts with bromine to form aluminum bromide: 2 Al + 3 Br2  2 AlBr3;
If 25.0 g of Al and 100. g of Br2 are reacted, and 64.2 g of AlBr3 product are recovered,
what is the percent yield for the reaction?
 1 mol Al   2 mol Al   266.7g AlBr3 
25.0 g Al x  26.98 g Al  x  2 mol AlBr  x  1 mol AlBr = 247 g AlBr3;

 
3 
3 
1
mol
Br
3
mol
Br
266.7g
AlBr
2  
3

 
100. g Br2 x  159.8 g Br  x  2 mol AlBr  x  1 mol AlBr = 250. g AlBr3

2 
3 
3 
The limiting reactant is AlBr3, so 247 g AlBr3 should be obtained for 100% yield.
Hence the percent yield for the reaction is:
 64.2 g AlBr3 
 247 g AlBr  x 100 = 26.0%

3 