Chapter 11
Molecular Composition of Gases
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature
and pressure, the ratio of the volumes of
reacting gases are small whole numbers.
N2
1 volume
+
+
→
3 volumes →
3 H2
V
VNH
H 2 3 32
==
VNH 113
2
2 NH3
2 volumes
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature
and pressure, the ratio of the volumes of
reacting gases are small whole numbers.
Avogadro’s Law
Equal volumes of different gases at the
same temperature and pressure contain
the same number of molecules.
H2 + Cl2 → 2 HCl
hydrogen + chlorine → hydrogen chloride
1 volume 1 volume
2 volumes
1 molecule 1 molecule
2 molecules
1 mol
1 mol
Each molecule of hydrogen and each
molecule of chlorine contains 2 atoms.
2 mol
Mole-Mass-Volume
Relationships
Mole-Mass-Volume Relationships
• Volume of one mole of any gas at STP =
22.4 L.
• 22.4 L at STP is known as the molar
volume of any gas.
Find the mass in grams of 2.80 L of carbon
dioxide?
• (2.8 L CO2)(1 mol CO2/22.4 L CO2)(44 g
CO2 /1 mol CO2 ) = 5.5 g CO2
Density of Gases
grams
m
d=
v
liters
Density of Gases
m
d=
v
depends
on T and P
The molar mass of SO2 is 64.07 g/mol. Determine the
density of SO2 at STP.
1 mole of any gas
occupies 22.4 L at
STP
64.07 g   1 mol 
g

d= 

 = 2.86
L
 mol   22.4 L 
Ideal Gas Equation
nRT
nT
VV=a
PV
= nRT
P
P
atmospheres
nRT
nT
VV=a
PV
= nRT
P
P
liters
nRT
nT
VV=a
PV
= nRT
P
P
moles
nRT
nT
VV=a
PV
= nRT
P
P
Kelvin
nRT
nT
VV=a
PV
= nRT
P
P
Ideal L-atm
Gas
0.0821
mol-K
Constant
nRT
nT
VV=a
PV
= nRT
P
P
A balloon filled with 5.00 moles of helium gas is at a
temperature of 25oC. The atmospheric pressure is
0.987 atm. What is the balloon’s volume?
Step 1. Organize the given information.
Convert temperature to kelvins.
K = oC + 273
K = 25oC + 273 = 298K
A balloon filled with 5.00 moles of helium gas is at a
temperature of 25oC. The atmospheric pressure is .987
atm. What is the balloon’s volume?
Step 2. Write and solve the ideal gas equation
for the unknown.
PV = nRT
nRT
V=
P
Step 3. Substitute the given data into the
equation and calculate.
(5.00 mol)(0.0821 L×atm/mol×K)(298 K)
= 124 L
V=
(0.987 atm)
Determination of Molecular Weights
Using the Ideal Gas Equation
g
molar mass =
mol
M = molar mass
g
mol =
molar mass
g
n = mol =
M
g
RT
PV = nRT PV =
M
gRT
M=
PV
Calculate the molar mass of an unknown gas, if 0.020
g occupies 250 mL at a temperature of 305 K and a
pressure of 0.045 atm.
V = 250 mL = 0.250 L
g = 0.020 g
T = 305 K
P = 0.045 atm
gRT
M=
PV
(0.020 g)(0.082 L × atm/mol × K)(305 K)
g
M=
= 44
(0.045 atm) (0.250 L)
mol
Determination of Density
Using the Ideal Gas Equation
• Density = mass/volume
gRT
M=
PV
D = MP/ RT
The density of a gas was measured at 1.50
atm and 27 ºC and found to be 1.95 g/L.
Calculate the molar mass of the gas.
M =
dRT
P
• (1.95g/L)(0.08206 L atm/K mol)(300K)
1.50 atm
= 32.0 g/mol
Gas Stoichiometry
deals with the quantitative relationships among
reactants and products in a chemical reaction.
• All calculations are done at STP.
• Gases are assumed to behave as ideal
gases.
• A gas not at STP is converted to STP.
Gas Stoichiometry
Primary conversions involved in stoichiometry.
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
• Step 1 Write the balanced equation
2 KClO3  2 KCl + 3 O2
• Step 2 The starting amount is 0.500 mol
KClO3. The conversion is
moles KClO3  moles O2  liters O2
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
2 KClO3  2KCl + 3 O2
• Step 3. Calculate the moles of O2, using the moleratio method.
 3 mol O2 
(0.500 mol KClO3 ) 
= 0.750 mol O2

 2 mol KClO3 
• Step 4. Convert moles of O2 to liters of O2
22.4 L 

(0.750 mol O2 ) 
 = 16.8 L O2
 1 mol 
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
The problem can also be solved in
one continuous calculation.
2 KClO3  2KCl + 3 O2
 3 mol O2   22.4 L 
(0.500 mol KClO3 ) 

 = 16.8 L O2

 2 mol KClO3   1 mol 
What volume of hydrogen, collected at 30.oC and
0.921 atm, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 1 Calculate moles of H2.
grams Al  moles Al  moles H2
 1 mol Al   3 mol H 2 
50.0 g Al 

 = 2.78 mol H 2

 26.98 g Al   2 mol Al 
What volume of hydrogen, collected at 30.oC and
0.921 atm, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 2 Calculate liters of H2.
• Convert oC to K: 30.oC + 273 = 303 K
What volume of hydrogen, collected at 30.oC and 700.
torr, will be formed by reacting 50.0 g of aluminum
with hydrochloric acid?
• Solve the ideal gas equation for V
PV = nRT
nRT
V=
P
(2.78 mol H 2 )(0.0821 L-atm)(303 K)
= 75.1 L H 2
V=
(0.921 atm)(mol-K)
For reacting gases at constant temperature and
pressure: Volume-volume relationships are the same as
mole-mole relationships.
H2(g)
+
Cl2(g)

2HCl(g)
1 mol H2
1 mol Cl2
2 mol HCl
22.4 L
STP
1 volume
22.4 L
STP
1 volume
2 x 22.4 L
STP
2 volumes
What volume of nitrogen will react with 600. mL of
hydrogen to form ammonia? What volume of
ammonia will be formed?
N2(g) + 3H2(g)  2NH3(g)
 1 vol N 2 
600. ml H 2 
= 200. mL N 2

 3 vol H 2 
 2 vol NH 3 
= 400. mL NH 3
600. ml H 2 

 3 vol H 2 
Mass- volume problem
Graham’s Law of Effusion
• Effusion – the process whereby the
molecules of a gas confined in a container
randomly pass through a tiny opening in the
container.
• Rate (like diffusion) depends on the relative
velocities of gas molecules.
• Lighter molecules move faster than heavier
molecules at the same temperature.
Graham’s Law of Effusion
• The rates of effusion of gases at the same
temperature and pressure are inversely
proportional to the square roots of their
molar masses.
• Rate of effusion of A = MB
Rate of effusion of B
MA
Calculate the ratio of the effusion rates of H2
& UF6, a gas used in the enrichment process
to produce fuel for nuclear reactors.
•Rate of effusion of H2
Rate of effusion of UF
•Square root of molar mass of UF6
Square root of molar mass of H2
Square root of 352.02/2.016 =
13.2