The Molar Volume
10:1 Avogadro’s Principle and Molar Volume
What is the relationship between the mass of a gas and its volume? Avogadro’s principle
states that equal volumes of all gases, measured under the same conditions of pressure and
temperature, contain the same number of particles. It has been found experimentally that 1 mole
of any gas at STP contains 6.02 X 1023 particles. This number is called Avogadro’s number of
particles. One mole of any gas has a mass equal to its molecular mass. For example:
1 mole N2 = 28.0 g N2 = 6.02 X 1023 molecules of N2
1 mole CO2 = 44.0 g CO2 = 6.02 X 1023 molecules of CO2
Equal volumes of gases measured at the same temperature and pressure contain the same number
of particles.
A volume of gas which contains one mole of the gas at STP is the molar volume of the
gas. One mole of a gas at STP occupies 22.4 cubic decimeters (dm3). This volume is the same for
all gases at STP. The mass and volume of any gas are related as follows:
1 mole of any gas = molecular mass = 22.4 dm3 of the gas
Although the molar volume does vary slightly with different gases we shall assume we
are dealing with ideal gases in which the volume at STP is always 22.4 dm3.
The relationships among mole, molecular mass, and volume make it possible to calculate
molecular mass, volume, and density under the following conditions.
1. the molecular mass of a gas if the volume and mass of the gas are known.
2. the volume of a specific mass of gas or the mass of a specific volume of gas.
3. the density of the gas from its molecular mass or its molecular formula.
Example 1____________________________________________________________________
A sample of gas has a mass of 1.248 grams and occupies 300.0 cm3 at STP. What is the
molecular mass of this gas?
Solving Process:
The molecular mass of the gas is equal to one mole of the gas. Start with the relationship
between volume and mass of the gas given in the problem and calculate the mass of one mole of
the gas. All units must be divided out except grams per mole.
The conversion ratios are
1.248 grams of gas = 300.0 cm3
22.4 dm3 (STP) = 1 mole gas
1.248 g
22.4 dm3 (STP)
1000 cm3
= 93.2 g/mol
300.0 cm3
molecular
mass
1 mol gas
1 dm3
=
Example 2____________________________________________________________________
How many grams of carbon dioxide, CO2, will occupy a volume of 500 cm3 at STP?
Solving Process:
The conversion ratios are
1 mol CO2 = 22.4 dm3 CO2 (STP)
1 mol CO2 = 44.0 g CO2
500 cm3
1 mol CO2
22.4 dm3
44.0 g CO2
1 mol CO2
1 dm3
1000 cm3
= 0.982 g CO2
Example 3____________________________________________________________________
What is the density of nitrogen dioxide, NO2, at STP in grams per cubic decimeter?
Solving Process:
Calculate the molecular mass of nitrogen dioxide and then convert to grams per cubic
decimeter.
46.0 g NO2
1 mol NO2
= 2.05 g/dm3
1 mol NO2
22.4 dm3 NO2
Example 4____________________________________________________________________
A flask has a volume of 258 cm3. A gas with mass 1.475 grams is introduced into the
flask at a temperature of 300 K and a pressure of 9.86 x 104 Pa. Calculate the molecular mass of
the gas.
Solving Process:
Since 1 mole of a gas occupies 22.4 dm3 when the gas is a STP, the volume must be
adjusted to STP by the kelvin temperature ratio and a pressure ratio. These ratios convert the
original temperature and pressure to the equivalent standard temperature and pressure.
A decrease in temperature causes a volume decrease. An increase in pressure causes a
volume decrease. The ratios are
Temperature ratio
Pressure ratio
273 K
9.86 x 104 Pa
300 K
101.3 x 103 Pa
Correct the volume to STP.
258 cm3
273 K
300 K
9.86 x 104 Pa
101.3 x 103 Pa
= 228 cm3
Calculate the molecular mass using the volume of gas at STP.
1.475
22.4 dm3
1 mol
1000 cm3
1 dm3
= 145 g/mol
Problems
Assume the volumes given are STP unless other conditions are specified.
1. Calculate the number of moles contained in each of the following gas volumes.
a. 50 000 cm3 H2
d. 15 000 cm3 NH3
3
b. 1000 cm N2
e. 2500 cm3 O2
c. 6500 cm3 SO2
f. 2000 cm3 CO2
2. Calculate the mass of each of the following volumes of gas.
a. 20.0 m3 CH4
d. 0.300 m3 N2O
b. 1500.0 cm3 Cl2
e. 3.0 m3 N2
3
c. 70.0 cm SO3
f. 3500.0 cm3 H2S
3. Calculate the volume in dm3 of each of the following:
a. 4.0 moles Br2
e. 2.50 moles NH3
b. 200.0 grams H2S
f. 50.0 grams NO2
c. 25.5 grams SO2
g. 7.00 moles O2
d. 600.0 grams Cl2
h. 10.0 grams HCl
4. Calculate the density in g/dm3 for each of the following:
a. hydrogen sulfide, H2S
c. ammonia, NH
b. methane, CH4
d. chlorine, Cl2
5. A carbon dioxide sample has a mass of 5.148 grams and occupies 2600.0 cm3 at STP. What
is the density (in grams per liter) of the carbon dioxide?
6. A sample of hydrogen chloride has a mass of 2.459 grams and occupies 1500.0 cm3. What is
the density?
10:2: Molar Volume and Gases Collected Over Water
We can apply what we know about molar volume to laboratory situations involving gases
collected over water.
Example 5____________________________________________________________________
A reaction produces 200 cm3 of oxygen which is collected over water and measures at
22ºC and 9.90 X 104 Pa. How many grams of the gas are produced? Assume a water vapor
pressure of 2.6 kPa at 22 ºC.
Solving Process:
Convert the volume of oxygen produced to the volume of dry oxygen at STP, because 1
mole of gas occupies 22.4 dm3 only when dry and at STP. Once the number of moles of oxygen
has been calculated, the grams of oxygen can be determined using the conversion ration, 1 mole
O2 = 32.0 grams O2.
The numbers will be easier to work with if the pressures are all in kilopascals. Since the
pressure increases, a decrease in gas volume will occur and the pressure ratio (allowing for water
vapor pressure correction) is
99.0 – 2.6 kPa
or
96.4 kPa
101.3 kPa
101.3 kPa
The temperature decrease causes a decrease in gas volume. The kelvin temperature ratio is
273 K
295 K
Correct the volume to STP.
200 cm3
96.4 kPa
273 K
= 176 cm3 O2
101.3 kPa
295 K
Calculate the grams of O2 produced.
176 cm3
1 mol
32.0 g O2
1 dm3
= 0.251 grams O2
3
3
22.4 dm
1 mol O2
1000 cm
10:3: Idea Gas Equation
The ideal gas equation combines the four physical variables (pressure, volume,
temperature, and number of particles) for gases into one equation. Remember that an ideal gas is
composed of point masses with no mutual attraction. All real gases deviate somewhat from the
gas laws since the molecules of real gases occupy volume and attract one another.
The ideal gas equation is PV = nRT, where P is the pressure in kilopascals, V is the
volume in cubic decimeters and T is the temperature in kelvin. The n represents the number of
moles of a gas. With these units, the value of the constant R is 8.31 dm3 · kPa / mol · K. There
are other values of R depending upon the units used to derive R.
We can use the ideal gas equation to determine the molecular mass (M) of a gas. The
number of moles (n) of any species is equal to its mass (m) divided by the molecular mass (M).
Thus, the ideal gas equation can also be written as
PV = nRT
or
PV = m RT
M
or
M = mRT
PV
Example 6____________________________________________________________________
Use the data from Example 4 to calculate the molecular mass of the gas using the ideal
gas equation.
Solving Process:
M = mRT
PV
1.475 g
9.86 X 104 Pa
8.31 dm3 · kPa
mol · K
300 K
258 cm3
1000 Pa
1 kPa
1000 cm3
1 dm3
= 145 g/mol
Remember that the units of volume, pressure, temperature, and quantity of gas must be
consistent with the value of R.
Example 7____________________________________________________________________
In an experiment, an oxygen generator is heated gently and the oxygen formed displaces
water in a flask. The displaced water is collected and its temperature and volume are determined
and assumed to be that of oxygen gas produced. Calculate the molar volume of the oxygen gas
from the data given.
1. mass of generator before heating
23.784 g
2. pressure
100.0 kPa
3. mass of generator after heating
23.505 g
4. temperature of oxygen
21.0 ºC
5. vapor pressure of water
2.5 kPa
6. pressure of dry oxygen (100.0 – 2.5 = 97.5)
97.5 kPa
7. volume of oxygen collected
225 cm3
Solving Process:
Determine the volume of dry oxygen gas at STP.
225 cm3 O2
97.5 kPa
273 K
101.3 kPa
294 K
Find the experimental molar volume.
Molar volume = molecular mass O2
32.0 g O2
201 cm3 O2
(23.784 – 23.505) g O2
= 201cm3 O2
volume O2
mass O2
1 dm3
1000 cm3
= 23.1 dm3 O2
Problems
7. What is the molecular mass of sulfur dioxide if 300.0 cm3 of the gas has a mass of 0.855
grams?
8. A sample of hydrogen iodide has a mass of 2.28 grams and occupies 400.0 cm3 at STP. What
is the molecular mass of this compound?
9. If 0.179 grams of methane occupy 0.250 dm3, what is the molecular mass of methane if the
volume is given at standard conditions?
10. The density of s sample of phosphorous trifluoride, PF3 is 3.90 g/dm3. What is the molecular
mass of this gas?
11. What is the molecular mass of nitrous oxide, N2O, if the density of the gas is 1.978 g/dm3?
12. From the volume, temperature, and pressure, calculate the number of moles and the mass in
grams for each gas listed:
a. 750 cm3 O2 at 27 ºC and 99.0 kPa
b. 3.00 dm3 CO2 at -10 ºC and 103.0 kPa
c. 2000.0 cm3 NH3 at 10 ºC and 105.0 kPa
d. 5.00 dm3 SO2 at 21 ºC and 100.0 kPa
13. Calculate the volume each gas will occupy under the conditions listed:
a. 3.00 moles H2 at 24 ºC and 100.5 kPa
b. 150.0 grams Cl2 at -10 ºC and 98.5 kPa
c. 5.00 moles CH4 at 27 ºC and 97.2 kPa
d. 200.0 grams NH3 at 12 ºC and 104.5 kPa
10:4: Gas Concentration
The concentration of a gas can be expressed in many ways – percent by volume, parts per
million, and milligrams per cubic meter. Percent by volume is the volume of the given
compound contained in 100 volumes of air. For example, air contains about 20% oxygen on a
volume basis.
Parts per million (ppm) has been useful for very low concentrations commonly
encountered in pollution work. Parts per million (ppm) expresses how many parts of a very
small component are in a million parts of the whole. On a mass basis, ppm is the number of
milligrams of component per one kilogram of the whole, since 1000 milligrams = 1 grams and
1000 grams – 1 kilogram. For example, if 1 kilogram of fish contains 4 milligrams of mercury,
the concentration of mercury could be expressed as 4 parts per million.
Extremely small components of the whole can be expressed as parts per billion (ppb). A
microgram (μg) is one billionth of a kilogram, thus ppb is the number of micrograms per
kilogram or micrograms per cubic decimeter of solution.
Some comparisons of ppm and ppb are
1 ppm is one penny in $10 000
and
1 ppb is one penny in $10 000 000
1 ppm is 1 mm in 1 km
and
1 ppb is 1 mm in 1000 km
Chapter Review Problems
14. What volume will be occupied by 5.70 grams of oxygen which is collected over water at 98.0
kPa and 27 ºC? The vapor pressure of water at this temperature is 3.6 kPa.
15. If 0.850 grams of hydrogen are produced by a reaction, what volume in cubic decimeters will
the gas occupy at the following conditions?
a. standard temperature and pressure
b. 20 ºC and 1.0 X 105 Pa
c. collected over water at 20 ºC and 1.0 X 105 Pa, dry
16. How many moles and grams of nitrogen are produced if 250.0 cm3 is collected over water at
20 ºC and 9.87 X 104 Pa?
17. A gas sample occupies a volume of 300.0 cm3 when measured over water at 20 ºC and 1.00
X 105 Pa. What is the molecular mass of this gas if the sample has a mass of 0.700 gramss?
18. A gas sample has a mass of 1.25 grams and occupies 125 cm3 when collected over water at
27 ºC and 96.4 kPa. What is the molecular mass of this gas sample?
19. Calculate the density of oxygen under the following conditions:
a. standard temperature and pressure
b. 27 ºC and 98.6 kPa
20. A sample of gas has a density of 1.25 g/dm3 at 30 ºC and 1.02 X 105 Pa. What is the
molecular mass of this gas?
21. What is the molecular mass of a gas if 5.75 grams of the gas occupy a volume of 3.50 dm3?
The pressure was recorded as 9.525 X 104 Pa and the temperature is 50 ºC.
22. Determine the molecular mass of a gas if its density is 2.70 g/dm3 at a pressure of 132.0 kPa
and temperature of 30 ºC.
23. At a pressure of 98.0 kPa and temperature of 31 ºC, ammonia gas occupies 18.9 dm3.
Calculate the grams of ammonia in this volume.
24. Air is a mixture of gases but it can be assumed that the approximate composition of air is
21% O2 and 79% N2 by volume. Determine the mass of the molar volume of air at STP.
Note that the answer is the apparent molecular mass of air.