Molecular Composition of Gases
Volume-Mass Relationships
of Gases
Measuring and Comparing the
Volumes of Reacting Gases
Early 1800s, Gay-Lussac studied gas
volume relationships with chemical reaction
between H and O
 Observed 2 L H can react with 1 L O to
form 2 L of water vapor at constant
temperature and pressure

Hydrogen gas + oxygen gas  water vapor
2L
1L
2L
2 volumes
1 volume
2 volumes
Reaction shows 2:1:2 relationship between
volumes of reactants and product
 Ratio applies to any proportions (mL, L,
cm3)


Gay-Lussac also noticed ratios by volume
between other reactions of gases
Hydrogen gas + chlorine gas  hydrogen chloride gas
1L
1L
2L
Law of Combining Volumes of
Gases

1808 Gay-Lussac summarized results in
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
Avogadro’s Law

Important point of Dalton’s atomic
theory: atoms are indivisible

Dalton also thought particles of gaseous
elements exist in form of single atoms

Believed one atom of one element always
combines with one atom of another
element to form single particle of product
Gay-Lussac’s results presented problem
for Dalton’s theory
 Ex. Reactions like formation of water

Hydrogen gas + oxygen gas  water vapor
2L
1L
2L

Seems that oxygen involved would have to
divide into two parts

1811 Avogadro found way to explain GayLussac’s simple ratios of combining volumes
without violating Dalton’s idea of indivisible
atoms

Rejected Dalton’s idea that reactant
elements are always in monatomic form
when they combine to form products

Reasoned these molecules could contain
more than 1 atom

Avogadro’s law  equal volumes of
gases at the same temperature and
pressure contain equal numbers of
molecules

At the same temp and pressure, volume
of any given gas varies directly with the
number of molecules
1 mol CO2 at
STP = 22.4 L
1 mol O2 at
STP = 22.4 L
1 mol H2 at
STP = 22.4 L

Consider reaction of H and Cl to produce HCl
According to Avogadro’s law, equal volumes of
H and Cl contain same number of molecules
 b/c he rejected Dalton’s theory that elements
are always monatomic, he concluded H and Cl
components must each consist of 2 or more
atoms joined together

Simplest assumption was that H and Cl
molecules had 2 atoms each
 Leads to following balanced equation:

H2(g) + Cl2(g) →
1 volume
1 volume
1 molecule 1 molecule
2HCl(g)
2 volumes
2 molecules
If the simplest formula for hydrogen chloride,
HCl indicates molecule contains 1 H and 1 Cl
 Then the simplest formulas for hydrogen and
chlorine must be H2 and Cl2


Avogadro’s law also indicates that gas
volume is directly proportional to the
amount of gas, at given temp and
pressure
V = kn
k = constant
 n = amount of gas in moles

Molar Volume of Gases
Remember 1 mol of substance contains
6.022 x 1023
 According to Avogadro’s law, 1 mol of
any gas occupies same volume as 1 mol
of any other gas at same temperature
and pressure, even though masses are
different
 Standard molar volume of gas 
volume occupied by 1 mol of gas at STP
 = 22.4 L


Knowing volume of gas, you can use
1mol/22.4 L as conversion factor

Can find number of moles

Can find mass

Can also use molar volume to find volume if
you have number of moles or mass
Sample Problem

A chemical reaction produces 0.0680
mol of oxygen gas. What volume in
liters is occupied by this gas sample
at STP?
1. Analyze

Given: moles of O2 = 0.0680 mol

Unknown: volume of O2 in liters at STP
2. Plan

moles of O2 → liters of O2 at STP

The standard molar volume can be used
to find the volume of a known molar
amount of a gas at STP
3. Compute
Practice Problems






At STP, what is the volume of 7.08 mol
of nitrogen gas?
159 L N2
A sample of hydrogen gas occupies
14.1 L at STP. How many moles of the
gas are present?
0.629 mol H2
At STP, a sample of neon gas occupies
550. cm3. How many moles of neon gas
does this represent?
0.0246 mol Ne
Sample Problem

A chemical reaction produced 98.0
mL of sulfur dioxide gas, SO2, at STP.
What was the mass (in grams) of the
gas produced?
1. Analyze
Given: volume of SO2 at STP = 98.0 mL
 Unknown: mass of SO2 in grams

2. Plan

liters of SO2 at STP→moles of SO2→grams of SO2
3. Compute

= 0.280 g SO2
Practice Problems






What is the mass of 1.33 × 104 mL of
oxygen gas at STP?
19.0 g O2
What is the volume of 77.0 g of nitrogen
dioxide gas at STP?
37.5 L NO2
At STP, 3 L of chlorine is produced
during a chemical reaction. What is the
mass of this gas?
9 g Cl2
The Ideal Gas Law

A gas sample can be characterized by 4
quantities
Pressure
2. Volume
3. Temperature
4. Number of moles
1.

Number of moles present always affects
at least one of the other 3 quantities

Collision rate per unit area of container
wall depends on number of moles

Increase moles, increase collision rate,
increase pressure
Pressure, volume, temperature, moles
are all interrelated
 A mathematical relationship exists to
describe behavior of gas for any
combination of these conditions


Ideal gas law  mathematical
relationship among pressure, volume,
temperature, and number of moles of a
gas
Derivation of Ideal Gas Law
Derived by combining the other gas laws
 Boyle’s law: at constant temp, the
volume of a given mass of gas is
inversely proportional to the pressure.


Charles’s law: At constant pressure,
volume of given mass of gas is directly
proportional to Kelvin temperature
VαT

Avogadro’s law: at constant temp and
pressure, volume of given mass of gas is
directly proportional to the number of moles
Vαn
Volume is proportional to pressure, temp and
moles in each equation
 Combine the 3:

Can change proportion to equality by adding
constant, this time R

This equation says the volume of a gas
varies directly with the number of moles
of gas and its Kelvin temperature

Volume also varies inversely with
pressure

Ideal gas law combines Boyle’s, Charles’s,
Gay-Lussac’s, and Avogadro’s laws

Ex: PV = nRT  n and T are constant, nRT
is constant b/c R is also constant

This makes PV = constant which is Boyle’s
law
The Ideal Gas Constant


Ideal gas constant  R
Value depends on units for volume, pressure,
temp
Unit of R
Value of
R
Unit of P Unit of V Unit of T Unit of n
62.4
mmHg
L
K
Mol
0.0821
Atm
L
K
Mol
8.314
Pa
m3
K
Mol
8.314
kPa
L
K
Mol
Sample Problem

What is the pressure in atmospheres
exerted by a 0.500 mol sample of
nitrogen gas in a 10.0 L container at
298 K?
1. Analyze
Given:
 V of N2 = 10.0 L
 n of N2 = 0.500 mol
 T of N2 = 298 K

Unknown:
 P of N2 in atm

2. Plan
n,V,T → P
 The gas sample undergoes no change
in conditions
 Therefore, the ideal gas law can be
rearranged and used to find the
pressure as follows

3. Compute
= 1.22 atm
Practice Problems
What pressure, in atmospheres, is
exerted by 0.325 mol of hydrogen gas in
a 4.08 L container at 35°C?
 2.01 atm

A gas sample occupies 8.77 L at
20°C.What is the pressure, in
atmospheres, given that there are 1.45
mol of gas in the sample?
 3.98 atm







What is the volume, in liters, of 0.250 mol of
oxygen gas at 20.0°C and 0.974 atm
pressure?
6.17 L O2
A sample that contains 4.38 mol of a gas at
250 K has a pressure of 0.857 atm. What is
the volume?
105 L
How many liters are occupied by 0.909 mol
of nitrogen at 125°C and 0.901 atm
pressure?
33.0 L N2
What mass of chlorine gas, Cl2, in grams,
is contained in a 10.0 L tank at 27°C and
3.50 atm of pressure?
 101 g Cl2
 How many grams of carbon dioxide gas
are there in a 45.1 L container at 34°C and
1.04 atm?
 81.9 g CO2
 What is the mass, in grams, of oxygen
gas in a 12.5 L container at 45°C and 7.22
atm?
 111 g O2

A sample of carbon dioxide with a
mass of 0.30 g was placed in a 250
mL container at 400. K. What is the
pressure exerted by the gas?
 0.90 atm

Finding Molar Mass or Density
from Ideal Gas Law
If P, V, T and mass are known you can
calculate number of moles (n) in sample
 Can calculate molar mass (g/mol)

Equation shows relationship between
density, P, T, molar mass
 Mass divided by molar mass gives moles
 Substitute m/M for n in equation PV=nRT

Density (D) = mass (m) per unit volume (V)
 D = m/V

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?
1. Analyze
Given:
 P of gas = 0.974 atm
 V of gas = 1.00 L
 T of gas = 28°C + 273 = 301 K
 m of gas = 5.16 g

Unknown:
 M of gas in g/mol

2. Plan
P, V, T, m → M
 You can use the rearranged ideal gas law
provided earlier to find the answer

3. Compute

= 131 g/mol
Practice Problems
What is the molar mass of a gas if
0.427 g of the gas occupies a volume
of 125 mL at 20.0°C and 0.980 atm?
 83.8 g/mol
 What is the density of a sample of
ammonia gas, NH3, if the pressure is
0.928 atm and the temperature is
63.0°C?
 0.572 g/L NH3

The density of a gas was found to be
2.0 g/L at 1.50 atm and 27°C. What is
the molar mass of the gas?
 33 g/mol
 What is the density of argon gas,Ar,
at a pressure of 551 torr and a
temperature of 25°C?
 1.18 g/L Ar

Stoichiometry of Gases

You can apply gas laws to calculate
stoichiometry of reactions involving gases

Coefficients in balanced equations represent
mole AND volume ratios
Volume-Volume Calculations
Propane, C3H8, is a gas that is sometimes
used as a fuel for cooking and heating. The
complete combustion of propane occurs
according to the following equation.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
 (a) What will be the volume, in liters, of
oxygen required for the complete
combustion of 0.350 L of propane? (b) What
will be the volume of carbon dioxide
produced in the reaction? Assume that all
volume measurements are made at the same
temperature and pressure.

1. Analyze
Given:
 balanced chemical equation
 V of propane = 0.350 L

Unknown:
 a. V of O2 in L;
 b. V of CO2 in L

2. Plan

a. V of C3H8 → V of O2;

b. V of C3H8 → V of CO2

All volumes are to be compared at the
same temperature and pressure

Therefore, volume ratios can be used
like mole ratios to find the unknowns.
3. Compute

= 0.175 L O2

= 1.05 L CO2
Practice Problem

Assuming all volume measurements
are made at the same temperature
and pressure, what volume of
hydrogen gas is needed to react
completely with 4.55 L of oxygen gas
to produce water vapor?

9.10 L H2

What volume of oxygen gas is
needed to react completely with 0.626
L of carbon monoxide gas, CO, to
form gaseous carbon dioxide?
Assume all volume measurements
are made at the same temperature
and pressure.

0.313 L O2
Volume-Mass and Mass-Volume
Calculations





gas volume A → moles A → moles B → mass
B
or
mass A → moles A → moles B → gas volume
B
You must know the conditions under which
both the known and unknown gas volumes
have been measured
The ideal gas law is useful for calculating
values at standard and nonstandard conditions
SAMPLE PROBLEM
Calcium carbonate, CaCO3, also known
as limestone, can be heated to produce
calcium oxide (lime), an industrial
chemical with a wide variety of uses.
The balanced equation for the reaction
follows.
Δ
CaCO3(s) → CaO(s) + CO2(g)
 How many grams of calcium carbonate
must be decomposed to produce 5.00 L
of carbon dioxide gas at STP?

1. Analyze
Given:
 balanced chemical equation
 desired volume of CO2 produced at STP
= 5.00 L

Unknown:
 mass of CaCO3 in grams

2. Plan





The known volume is given at STP
This tells us the pressure and temperature
The ideal gas law can be used to find the
moles of CO2
The mole ratios from the balanced
equation can then be used to calculate the
moles of CaCO3 needed
(Note that volume ratios do not apply here
because calcium carbonate is a solid)
3. Compute

= 0.223 mol CO2
Practice Problem

What mass of sulfur must be used to
produce 12.61 L of gaseous sulfur
dioxide at STP according to the
following equation?
S8(s) + 8O2(g) → 8SO2(g)

18.0 g S8

How many grams of water can be
produced from the complete reaction
of 3.44 L of oxygen gas, at STP, with
hydrogen gas?

5.53 g H2O

Tungsten, W, a metal used in light-bulb
filaments, is produced industrially by the
reaction of tungsten oxide with hydrogen.
WO3(s) + 3H2(g) →W(s) + 3H2O(l)


How many liters of hydrogen gas at 35°C
and 0.980 atm are needed to react
completely with 875 g of tungsten oxide?
292 L H2




What volume of chlorine gas at 38°C and
1.63 atm is needed to react completely with
10.4 g of sodium to form NaCl?
3.54 L Cl2
How many liters of gaseous carbon
monoxide at 27°C and 0.247 atm can be
produced from the burning of 65.5 g of
carbon according to the following
equation?
2C(s) + O2(g) → 2CO(g)
544 L CO
Effusion and
Diffusion
Graham’s Law of Effusion

Rates of effusion and diffusion depend
on relative velocities of gas molecules

Velocity varies inversely with mass
(lighter molecules move faster)
Average kinetic energy ½ mv2
 For two gases, A and B, at same temp:

½ MAvA2 = ½ MBvB2
MA and MB = molar masses of A and B
 Multiply by 2

MAvA2 = MBvB2
MAvA2 = MBvB2

Suppose you wanted to compare the
velocities of the two gases

You would first rearrange the equation
above to give the velocities as a ratio

Take square root of each side
This equation shows that velocities of two gases
are inversely proportional to the square roots of
their molar masses

b/c rates of effusion are directly
proportional to molecular velocities, can
write
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
Application of Graham’s Law
Graham’s experiments dealt with
densities of gases
 Density varies directly with molar mass
 So…square roots of molar masses from
equation can be replaced with square
roots of densities

Sample Problem

Compare the rates of effusion of
hydrogen and oxygen at the same
temperature and pressure.
1. Analyze
Given:
 identities of two gases, H2 and O2

Unknown:
 relative rates of effusion

2. Plan

molar mass ratio → ratio of rates of
effusion

The ratio of the rates of effusion of two
gases at the same temperature and
pressure can be found from Graham’s
law
3. Compute

Hydrogen effuses 3.98 times faster than
oxygen.
Practice Problems
A sample of hydrogen effuses through a
porous container about 9 times faster
than an unknown gas. Estimate the
molar mass of the unknown gas.
 160 g/mol
 Compare the rate of effusion of carbon
dioxide with that of hydrogen chloride
at the same temperature and pressure.
 CO2 will effuse about 0.9 times as fast as
HCl


If a molecule of neon gas travels at
an average of 400 m/s at a given
temperature, estimate the average
speed of a molecule of butane gas,
C4H10, at the same temperature.

about 235 m/s