Ch. 11 Molecular
Composition of Gases
11-1 Volume-Mass Relationships of
Gases

Gay-Lussac’s law of combining volumes of
gases-at constant temperature and
pressure, the volumes of gaseous
reactants and products can be expressed
as ratios of small whole numbers
Hydrogen + oxygen ->water vapor
2L
1L
2L
2 volumes
1 volume
2 volumes

Avogadro’s Law


Equal volumes of gases at the same
temperature and pressure contain equal
numbers of molecules (doesn’t matter
which gas) Fig. 11-1
He discovered that some molecules can
have 2 atoms or more (diatomic
molecules)


Avogadro’s law also indicates gas volume
(L) directly proportional to the amount of
a gas (n)
V = kn
Standard molar volume of a gas



Avogadro’s constant = 6.022 X 1023
molecules = 1 mole
Standard molar volume of a gas-the
volume occupied by one mole of a gas at
STP (22.4 L)
Fig. 11-3 1 mole of each gas occupies
22.4 L but different masses
Avogadro’s Law Sample problem
11-1


A chemical reaction produces 0.0680 mol
of oxygen gas. What volume in liters is
occupied by this gas sample at STP?
0.0680 mol X 22.4 L = 1.52 L
1 mol
Avogadro’s Law Practice

A sample of hydrogen gas occupies 14.1 L
at STP. How many moles of the gas are
present?
Converting to grams


Sample problem 11-2
A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2, at STP. What was
the mass in grams of the gas produced?
.098 L X 1 mol X 64.07 g SO2 = 0.280 g
22.4 L
1 mol
Converting to grams practice

What is the volume of 77 g of nitrogen
dioxide gas at STP?
11-2: The Ideal Gas Law



Mathematical relationship among
pressure, volume, temperature, and the
number of moles of a gas.
Combination of Boyle’s, Charles’s, GayLussac’s, and Avogadro’s Laws
PV = nRT


Ideal gas constant (R), is derived by
plugging in all standard values into the
equation:
R = PV = 0.0821
nT
Ideal gas law sample

What is the pressure in atmospheres
exerted by a 0.500 mol sample of nitrogen
gas in a 10 L container at 298 K?
Answer = 1.22 atm
More ideal gas law practice

What is the volume, in liters, of 0.250 mol
of oxygen gas at 20°C and 0.974 atm
pressure?
Answer = 6.17 L
Sample problem 11-5

What mass of chlorine gas, Cl2, in grams,
is contained in a 10 L tank at 27°C and
3.5 atm of pressure?
Answer = 101 g
Finding molar mass or density
PV = mRT
M
M = mRT M = DRT
PV
P

D = MP
RT
Sample Problem

At 28°C and 0.974 atm, 1.00 L of gas has
a mass of 5.16 g. What is the molar mass
of this gas?
11-3 Stoichiometry of Gases

Coefficients can be used as volume ratios:

2CO + O2 -> 2CO2
2 volumes CO
1 volume O2
Sample Problem 11-7 volumevolume
Sample problem 11-8 volume-mass
Sample problem 11-9
11-4 Effusion and Diffusion


Graham’s Law of Effusion-rates of diffusion
and effusion depend on the relative
velocities of gas molecules
Rates of effusion of gases at the same
temperature and pressure are inversely
proportional to the square roots of their
molar masses.
Graham’s Law formula

Rate of effusion A = √MB
Rate of effusion B
√MA
Molar masses can also be replaced by
densities of the gases:

Rate of effusion A = √densityB
Rate of effusion B √denistyA
Graham’s Law Problem



Sample problem 11-10
Compare the rates of effusion of hydrogen
and oxygen at the same temperature and
pressure. (smaller molar mass gas will
diffuse faster-how much faster?)
Smaller molar mass goes on bottom
Diffusion Quicklab pg. 353