Ch-5, Homework
ANSWERS TO REVIEW OF CONCEPTS
Section 5.2 (p. 177) 1) (b) < (c) < (a) < (d).
2) It would be easier to drink water with a straw at the foot of Mt. Everest
because the atmospheric pressure is greater there, which helps to push the
water up the straw.
Section 5.3 (p. 183) (a) Volume doubles. (b) Volume increases 1.4 times.
Section 5.4 (p. 186) Greatest volume, (b). Greatest density, (c).
Section 5.6 (p. 202) Blue sphere: 0.43 atm. Green sphere: 1.3 atm. Red sphere: 0.87 atm.
Section 5.7 (p. 210) (c) and (d).
Section 5.8 (p. 213) High pressure and low temperature.
1 atm
 0.739 atm
760 mmHg
5.13
562 mmHg 
5.14
Strategy: Because 1 atm  760 mmHg, the following conversion factor is needed to obtain the
pressure in atmospheres.
1 atm
760 mmHg
For the second conversion, 1 atm  101.325 kPa.
Solution:
? atm  606 mmHg 
? kPa  0.797 atm 
1 atm
 0.797 atm
760 mmHg
101.325 kPa
 80.8 kPa
1 atm
5.32
Strategy: This problem gives the amount, volume, and temperature of CO gas. Is the gas
undergoing a change in any of its properties? What equation should we use to solve for the
pressure? What temperature unit should be used?
Solution: Because no changes in gas properties occur, we can use the ideal gas equation to
calculate the pressure. Rearranging Equation (5.8) of the text, we write:
P 
nRT
V
L  atm 

(6.9 mol)  0.0821
(62  273)K
mol
 K 

P 
 6.2 atm
30.4 L
5.40
In the problem, temperature and pressure are given. If we can determine the moles of CO2, we
can calculate the volume it occupies using the ideal gas equation.
? mol CO2  88.4 g CO2 
1 mol CO2
 2.01 mol CO 2
44.01 g CO2
We now substitute into the ideal gas equation to calculate volume of CO2.
nRT
VCO2 

P
L  atm 

(2.01 mol)  0.0821
(273 K)
mol
 K 

 45.1 L
(1 atm)
Alternatively, we could use the fact that 1 mole of an ideal gas occupies a volume of 22.41 L at
STP. After calculating the moles of CO2, we can use this fact as a conversion factor to convert to
volume of CO2.
? L CO2  2.01 mol CO2 
22.41 L
 45.0 L CO2
1 mol
The slight difference in the results of our two calculations is due to rounding the volume
occupied by 1 mole of an ideal gas to 22.41 L.
5.19
P1  0.970 atm
P2  0.541 atm
V1  725 mL
V2  ?
P1V1  P2V2
V2 
5.36
PV
1 1  (0.970 atm)(725 mL)  1.30  103 mL
P2
0.541 atm
In this problem, the moles of gas and the volume the gas occupies are constant (V1  V2 and n1 
n2). Temperature and pressure change.
PV
PV
1 1
 2 2
n1T1
n2T2
P1
P
 2
T1
T2
The given information is tabulated below.
Initial
conditions
Final Conditions
T1  273 K
T2  (250  273)K  523
K
P1  1.0 atm
P2  ?
The final pressure is given by:
P2 
PT
1 2
T1
P2 
5.35
(1.0 atm)(523 K)
 1.9 atm
273 K
Initial Conditions
Final Conditions
P1  1.2 atm
P2  3.00  103 atm
V1  2.50 L
V2  ?
T1  (25  273)K  298 K
T2  (23  273)K  250 K
PV
PV
1 1
 2 2
T1
T2
V2 
5.48
PV
1 1T2  (1.2 atm)(2.50 L)(250 K)  8.4  102 L
T1P2
(298 K)(3.00  103 atm)
The density can be calculated from the ideal gas equation.
d 
PM
RT
M  1.008 g/mol  79.90 g/mol  80.91 g/mol
T  46  273  319 K
P  733 mmHg 
1 atm
 0.964 atm
760 mmHg
 80.91 g 
(0.964 atm) 

mol  K
 1 mol  
d 
 2.98 g/L
319 K
0.0821 L  atm
Alternatively, we can solve for the density by writing:
density 
mass
volume
Assuming that we have 1 mole of HBr, the mass is 80.91 g. The volume of the gas can be
calculated using the ideal gas equation.
V 
nRT
P
L  atm 

(1 mol)  0.0821
(319 K)
mol  K 

V 
 27.2 L
0.964 atm
Now, we can calculate the density of HBr gas.
density 
5.49
mass
80.91 g

 2.97 g/L
volume
27.2 L
METHOD 1:
The empirical formula can be calculated from mass percent data. The molar mass can be
calculated using the ideal gas equation. The molecular formula can then be determined.
To calculate the empirical formula, assume 100 g of substance.
64.9 g C 
1 mol C
 5.40 mol C
12.01 g C
13.5 g H 
1 mol H
 13.4 mol H
1.008 g H
21.6 g O 
1 mol O
 1.35 mol O
16.00 g O
This gives the formula C5.40H13.4O1.35. Dividing by 1.35 gives the empirical formula, C4H10O.
To calculate the molar mass, first calculate the number of moles of gas using the ideal gas
equation.

1 atm 
 750 mmHg 
 (1.00 L)
760 mmHg 
PV

n 

 0.0306 mol
L  atm 
RT

0.0821
(120

273)K

mol  K 

Solving for the molar mass:
M 
mass (in g)
2.30 g

 75.2 g/mol
mol
0.0306 mol
The empirical mass is 74.0 g/mol which is essentially the same as the molar
mass. In this case, the molecular formula is the same as the empirical formula, C4H10O.
METHOD 2:
First calculate the molar mass using the ideal gas equation.

1 atm 
 750 mmHg 
 (1.00 L)
760
mmHg 
PV

n 

 0.0306 mol
L  atm 
RT

 0.0821 mol  K  (120  273)K


Solving for the molar mass:
M 
mass (in g)
2.30 g

 75.2 g/mol
mol
0.0306 mol
Next, multiply the mass % (converted to a decimal) of each element by the molar mass to
convert to grams of each element. Then, use the molar mass to convert to moles of each
element.
nC  (0.649)  (75.2 g) 
1 mol C
 4.06 mol C
12.01 g C
nH  (0.135)  (75.2 g) 
1 mol H
 10.07 mol H
1.008 g H
nO  (0.216)  (75.2 g) 
1 mol O
 1.02 mol O
16.00 g O
Since we used the molar mass to calculate the moles of each element present in the compound,
this method directly gives the molecular formula. The formula is C4H10O.
5.26
This is a gas stoichiometry problem that requires knowledge of Avogadro’s law to solve.
Avogadro’s law states that the volume of a gas is directly proportional to the number of moles
of gas at constant temperature and pressure.
The volume ratio, 1 vol. Cl2 : 3 vol. F2 : 2 vol. product, can be written as a mole ratio,
1 mol Cl2 : 3 mol F2 : 2 mol product.
Attempt to write a balanced chemical equation. The subscript of F in the product will be three
times the Cl subscript, because there are three times as many F atoms reacted as Cl atoms.
1Cl2(g)  3F2(g) 
 2ClxF3x(g)
Balance the equation. The x must equal one so that there are two Cl atoms on each side of the
equation. If
x  1, the subscript on F is 3.
Cl2(g)  3F2(g) 
 2ClF3(g)
The formula of the product is ClF3.
5.62
The balanced equation is:
C2H5OH(l)  3O2(g) 
 2CO2(g)  3H2O(l)
The moles of O2 needed to react with 227 g ethanol are:
227 g C2 H5OH 
1 mol C2 H5OH
3 mol O2

 14.8 mol O2
46.07 g C2 H5OH 1 mol C2 H5OH
14.8 moles of O2 correspond to a volume of:
VO2 
nO2 RT
P
L  atm 

(14.8 mol O2 )  0.0821
(35  273 K)
mol  K 


 3.60  102 L O2


1 atm
 790 mmHg 

760
mmHg 

Since air is 21.0 percent O2 by volume, we can write:
 100% air 
 100% air 
2
3
Vair  VO2 
  (3.60  10 L O2 ) 
  1.71  10 L air
21%
O
21%
O
2 
2 


5.100 The reaction is: HCO3(aq)  H(aq) 
 H2O(l)  CO2(g)
The mass of HCO3 reacted is:
3.29 g tablet 
32.5% HCO3
 1.07 g HCO3
100% tablet
mol CO2 produced  1.07 g HCO3 
VCO2 
5.67
nCO2 RT
P
1 mol HCO3
61.02 g
HCO3

1 mol CO2
1 mol HCO3
 0.0175 mol CO2
L  atm 

(0.0175 mol CO2 )  0.0821
(37  273)K
mol  K 


 0.445 L  445 mL
(1.00 atm)
First, we calculate the mole fraction of each component of the mixture. Then, we can calculate
the partial pressure of each component using the equation, Pi  iPT.
The number of moles of the combined gases is:
n  nCH4  nC2H6  nC3H8  0.31 mol  0.25 mol  0.29 mol  0.85 mol
 CH4 
0.31 mol
 0.36
0.85 mol
 C2 H 6 
0.25 mol
 0.29
0.85 mol
 C3H8 
0.29 mol
 0.34
0.85 mol
The partial pressures are:
PCH4   CH 4  Ptotal  0.36  1.50 atm  0.54 atm
PC2H6   C2H6  Ptotal  0.29  1.50 atm  0.44 atm
PC3H8   C3H8  Ptotal  0.34  1.50 atm  0.51 atm
5.72
Strategy: To solve for moles of H2 generated, we must first calculate the partial pressure of H2
in the mixture. What gas law do we need? How do we convert from moles of H2 to amount of
Zn reacted?
Solution: Dalton’s law of partial pressure states that
PTotal  P1  P2  P3  . . .  Pn
In this case,
PTotal  PH2 + PH2O
PH2  PTotal  PH2O
 1 atm 
PH2  0.980 atm  (23.8 mmHg) 
  0.949 atm
 760 mmHg 
5.88
The rate of effusion is the number of molecules passing through a porous barrier in a given time.
The molar mass of CH4 is 16.04 g/mol. Using Equation (5.17) of the text, we find the molar mass of
Ni(CO)x.
r1

r2
3.3

1.0
10.89 
M2
M1
M Ni(CO) x
16.04 g/mol
M Ni(CO) x
16.04 g/mol
MNi(CO) x  174.7 g/mol
To find the value of x, we first subtract the molar mass of Ni from 174.7 g/mol.
174.7 g  58.69 g  116.0 g
116.0 g is the mass of CO in 1 mole of the compound. The mass of 1 mole of CO is 28.01 g.
116.0 g
 4.141  4
28.01 g
This calculation indicates that there are 4 moles of CO in 1 mole of the compound. The value of
x is 4.
Now that we know the pressure of H2 gas, we can calculate the moles of H2. Then, using the
mole ratio from the balanced equation, we can calculate moles of Zn.
nH2 
nH2 
PH2 V
RT
(0.949 atm)(7.80 L)
mol  K

 0.303 mol H 2
(25  273) K
0.0821 L  atm
Using the mole ratio from the balanced equation and the molar mass of zinc, we can now
calculate the grams of zinc consumed in the reaction.
? g Zn  0.303 mol H 2 
1 mol Zn 65.39 g Zn

 19.8 g Zn
1 mol H2
1 mol Zn