CHEMISTRY 161
HW#12 SOLUTIONS
22, 16, 24, 50, 62, 82, 92, 108, 110
3-22
Because gold’s molar mass is relatively high at 197 g/mol, a few ounces do not contain many moles of gold atoms.
3-16
(a) Wolframite, FeWO
4
, contains one atom of Fe per formula unit; so 2.5 mol of wolframite contain 2.5 mol of Fe.
2.5 mol Fe
 
1.5 10 Fe atoms
1 mol
(b) The formula for pyrite, FeS
2
, contains only one iron atom per formula unit as well; so the answer is identical to that calculated in part a.
2.5 mol Fe
 
1.5 10 Fe atoms
1 mol
(c) Magnetite, Fe
3
O
4
, contains three iron atoms in its formula; so 2.5 mol of Fe
3
O
4
contain 2.5

3 = 7.5 mol of iron.
7.5 mol Fe
 
4.5 10 Fe atoms
1 mol
(d) Hematite, Fe
2
O
3
, contains two iron atoms in its formula; so 2.5 mol of hematite contain 2.5

2 = 5.0 mol of iron.
5.0 mol Fe
 
3.0 10 Fe atoms
1 mol
3-24
Because aluminum oxide contains three atoms of O in its formula, 0.55 mol of Al
2
O
3
contains 0.55

3
1.65 mol of O 2– ions. The mass of O 2– ions in the sample, therefore, is
=
1.65 mol O
2– 
16.00 g
1 mol

26.4 g O
2–
3-50
(a) (1) The unbalanced reaction is
CO
2
( g ) + C( s )

CO( g )
Element Left-Hand Side Right-Hand Side
C 1+ 1 = 2 1
O 2 1
Neither the number of C atoms nor the number of O atoms on the reactants and products sides is equal. This reaction is not balanced.
(2) We can start by balancing the O atoms as they appear in only one reactant and product. We can do this by placing a 2 in front of CO on the right-hand side.
CO
2
( g ) + C( s )

2 CO( g )
Element Left-Hand Side Right-Hand Side
C 1+ 1 = 2 2(1) = 2
O 2 2(1) = 2
This also resulted in the O atoms being balanced. The equation is now balanced.
(b) (1) The unbalanced reaction is

K( s ) + H
2
O ( )

KOH( aq ) + H
2
( g )
Element Left-Hand Side Right-Hand Side
K
H
O
1
2
1
1
1 + 2 = 3
1
While the number of K and O atoms on both sides of the equation is equal, the number of H atoms is not. This reaction is not balanced.
(2) We can start by balancing the H atoms. We can do this by placing a coefficient of 1
2 in front of H
2 on the right-hand side.
K( s ) + H
2
O ( )

KOH( aq ) + 1
2
H
2
( g )
Element Left-Hand Side Right-Hand Side
K
H
O
1
2
1
1
1 + 1
2
1
(2) = 2
(3) To eliminate the fractional coefficients we multiply all of the coefficients by 2.
2 K( s ) + 2 H
2
O ( ) 
2 KOH( aq ) + H
2
( g )
Element Left-Hand Side Right-Hand Side
K
H
2(1) = 2
2(2) = 4
2(1) = 2
2(1) + 2( 1
2
(2)) = 4
O 2(1) = 2 2(1) = 2
The equation is now balanced.
(c) (1) The unbalanced reaction is
P
4
( s ) + O
2
( g )

P
2
O
5
( s )
Element Left-Hand Side Right-Hand Side
P 4 2
O 2 5
Neither the number of P atoms nor the number of O atoms on the reactants and products sides is equal. This reaction is not balanced.
(2) We can start by balancing the P atoms as they appear in only one reactant and product. We can do this by placing a 2 in front of P
2
O
5
on the right-hand side.
P
4
( s ) + O
2
( g )

2 P
2
O
5
( s )
Element Left-Hand Side Right-Hand Side
P 4 2(2) = 4
O 2 2(5) = 10
(3) The number of O atoms on the right-hand side is five times the number of that on the left-hand side. To balance the O atoms, therefore, we place 5 as the coefficient before O
2
on the left-hand side of the equation.
P
4
( s ) + 5 O
2
( g )

2 P
2
O
5
( s )
Element Left-Hand Side Right-Hand Side
P 4 2(2) = 4
O 5(2) = 10 2(5) = 10
The equation is now balanced.
3-62
(a) To balance the combustion reaction of charcoal:
(1) The reaction is described by the unbalanced chemical equation
C( s ) + O
2
( g )

CO
2
( g )
Element Left-Hand Side
C 1
O 2
Right-Hand Side
1
2

This equation is already balanced.
To balance the combustion reaction of propane:
(1) The reaction is described by the unbalanced chemical equation
C
3
H
8
( g ) + O
2
( g )

CO
2
( g ) + H
2
O ( )
Element Left-Hand Side
C 3
H
O
8
2
Right-Hand Side
1
2
2 + 1 = 3
(2) To balance the carbon atoms we place 3 as the coefficient in front of CO
2
.
C
3
H
8
( g ) + O
2
( g )

3 CO
2
( g ) + H
2
O ( )
Element Left-Hand Side
C 3
H
O
8
2
Right-Hand Side
3

1 = 3
2
3(2) + 1 = 7
(3) To balance the hydrogen atoms we place 4 as the coefficient in front of H
2
O.
C
3
H
8
( g ) + O
2
( g )

3 CO
2
( g ) + 4 H
2
O ( )
Element Left-Hand Side
C 3
H
O
8
2
Right-Hand Side
3

1 = 3
4(2) = 8
3(2) + 4(1) = 10
Now we see that the reactants side has 2 oxygen atoms while the products side has 10 oxygen atoms.
We can balance them by placing 5 as the coefficient in front of O
2
.
C
3
H
8
( g ) + 5 O
2
( g )

3 CO
2
( g ) + 4 H
2
O ( )
Element Left-Hand Side
C 3
H
O
8
5(2) = 10
Right-Hand Side
3

1 = 3
4(2) = 8
3(2) + 4(1) = 10
The equation is now balanced.
(b) For the combustion of charcoal
500.0 g C

1 mol C
12.01 g

1 mol CO
2

44.01 g CO
1 mol C 1 mol
2

1832 g CO
2
For the combustion of propane
500.0 g C H
3 8

1 mol C H
8 
3 mol CO
2
44.10 g 1 mol C H

44.01 g CO
2
1 mol

1497 g CO
2
3-82
(a) The moles of P and S in the compound are found by
27.87 g P

1 mol
30.97

0.8999 mol P
72.13 g S

1 mol
32.07

2.249 mol S
Dividing each by 0.8999 gives a ratio of 1 P to 2.5 S. Multiplying this ratio by 2 to give whole numbers gives the empirical formula of P
2
S
5
+ (5
. The molar mass of the empirical formula is (2

30.97)

32.07) =
222.29 g/mol. This is half that of the given molecular molar mass, so the molecular formula is P
4
S
10
.
(b) Knowing that the formula for iron(II) sulfide is FeS and that the phosphorus–sulfur product is
P
4
S
10
, we can write the balanced equation:
4 Fe
2
P( s ) + 18 FeS
2
( s )

26 FeS( s ) + P
4
S
10
( s )

3-92
0.1783 g CO
2

1 mol CO
2

1 mol C
44.01 g 1 mol CO
2
  –3
4.051 10 mol C
 
12.011 g C
1 mol


1 mol H O

2 mol H
18.02 g 1 mol H O


1.008 g H
1 mol
 2
 

 2
 2
Mass of O present = 0.100 g – 0.05687 g = 0.043 g O
Moles of O in compound = 0.043 g O

1 mol
15.999 g

2.7 10 mol O
Dividing the moles of C, H, and O by the lowest molar amount (2.7

10
–3
mol) gives a ratio of 1.5 C :
3 H : 1 O. Multiplying this through by 2 to obtain whole-number ratios, we get an empirical formula of C
3
H
6
O
2
.
3-108
The theoretical yield of NaHCO
3
is
10.0 g NaCl

1 mol NaCl
58.44 g

1 mol NaHCO
3

84.01 g NaHCO
1 mol NaCl 1 mol
3

14.4 g NaHCO
3
The percent yield for this reaction is
4.2 g
% yield =
14.4 g

100

29%
3-110
(a) The molar amount of each ion present in seawater is
19.4 g Cl
– 
1 mol Cl
–
35.453 g

0.547 mol Cl
–
10.8 g Na +

1 mol Na +
22.990 g

0.470 mol Na +
1.29 g Mg 2+ 
1 mol Mg
2+
24.305 g

0.0531 mol Mg 2+
(b) To form sodium chloride (NaCl), we would need 0.470 mol of chloride ions. To form 0.0531 mol of magnesium chloride (MgCl
0.576 mol of Cl
–
2
), we would need 0.0531

2 = 0.106 mol Cl
–
. Adding these gives
needed to form the chloride salts. We have only 0.547 mol of Cl
–
, so we do not have sufficient chloride ion to form the salts.