9.4 Density of Gases

2  Molar mass and density of a gas

The molecular weight (molar mass) of any gas is the mass, expressed in grams, of Avogadro’s number of its molecules. This is true regardless of whether the gas is composed of one molecular species or is a mixture.

For a mixture of gases, the molar mass will depend on the molar masses of its components, and on the fractional abundance of each kind of molecule in the mixture. The term “average molecular weight” is often used to describe the molar mass of a gas mixture.

Average molecular weight of a gas

The average molar massof a mixture of gases is just the sum of the mole fractions x_i of each gas, multiplied by the molar mass m_i of that substance:

110-component gas mixture enters Guinness Book of Records in 2010. The U.S. branch of the Linde Corporation, which supplies specialty gases to industry, concocted this one to calibrate instruments that measure atmospheric pollution.

Problem Example 3

Find the average molar mass of dry air whose volume-composition is O₂ (21%), N₂ (78%) and Ar (1%).

Solution: The average molecular weight is the mole-fraction-weighted sum of the molecular weights of its components. The mole fractions, of course, are the same as the volume-fractions (E.V.E.N. principle.)

m = (.21 x 32) + (.78 x 28) + (.01 x 40) = 29

Density of a pure gas

The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar masses of different gases will vary. This means that different gases will have different densities (different masses per unit volume). If we know the molecular weight of a gas, we can calculate its density.

Problem Example 4

Uranium hexafluoride UF₆ gas is used in the isotopic enrichment of natural uranium. Calculate its density at STP.

Solution: The molecular weight of UF₆ is 352.

(352 g mol^–1) ÷ (22.4 L mol^–1) = 15.7 g L^–1

Note: there is no need to look up a “formula” for this calculation; simply combine the molar mass and molar volume in such a way as to make the units come out correctly.

More importantly, if we can measure the density of an unknown gas, we have a convenient means of estimating its molecular weight. This is one of many important examples of how a macroscopic measurement (one made on bulk matter) can yield microscopic information (that is, about molecular-scale objects.)

Gas densities are now measured in industry by electro-mechanical devices such as vibrating reeds which can provide continuous, on-line records at specific locations, as within pipelines.

Determination of the molecular weight of a gas from its density is known as the Dumas method, after the French chemist Jean Dumas (1800-1840) who developed it. One simply measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle’s and Charles’ laws. The weight of the gas divided by its STP volume yields the density of the gas, and the density multiplied by 22.4 mol^–1 gives the molecular weight. Pay careful attention to the examples of gas density calculations shown here and in your textbook. You will be expected to carry out calculations of this kind, converting between molecular weight and gas density.

Problem Example 5

The approximate molar mass of a gas whose measured density is 3.33 g/L at 30°C and 780 torr.

Solution: Find the volume that would be occupied by 1 L of the gas at STP; note that correcting to 273 K will reduce the volume, while correcting to 1 atm (760 torr) will increase it:

The number of moles of gas is n = (.924 L) ÷ (22.4 L mol^–1) = 0.0412 mol

The molecular weight is therefore (3.3 g L^–1) ÷ (.0412 mol L^–1) = 80 g mol^–1

Density of a gas mixture

Gas density measurements can be a useful means of estimating the composition of a mixture of two different gases; this is widely done in industrial chemistry operations in which the compositions of gas streams must be monitored continuously.

Problem Example 6

Find the composition of a mixture of CO₂ (44 g/mol) and methane CH₄ (16 g/mol) that has a STP density of 1.214 g/L.

Solution: The density of a mixture of these two gases will be directly proportional to its composition, varying between that of pure methane and pure CO₂. We begin by finding these two densities:

For CO₂: (44 g/mol) ÷ (22.4 L/mol) = 1.964 g/L

For CH₄: (16 g/mol) ÷ (22.4 L/mol) = 0.714 g/L

If x is the mole fraction of CO₂ and (1–x) is the mole fraction of CH₄, we can write

1.964 x + 0.714 (1–x) = 1.214

(Does this make sense? Notice that if x = 0, the density would be that of pure CH₄, while if it were 1, it would be that of pure CO₂.)

Expanding the above equation and solving for x yields the mole fractions of 0.40 for CO₂ and 0.60 for CH₄.

The mathematical form of the Ideal Gas Law is:

PV = nRT and n = m/MW and d = m/V

Where:

• P – pressure
• V – volume
• n – number of moles
• T – temperature
• m – mass
• d – dendity
• MW – Molecular Weight
• R – ideal gas constant. If the units of P, V, n and T are atm, L, mol and K, respectively, the value of R is 0.0821 L x atm/K x mol or 8.314 J/K x mol.

The density (d) of a gas is defined as

d = m / V

and the moles of a gas is:

n = m / MW

Where m is the mass of the gas, and

M is the molecular weight.

Substituting the definitions to the original Ideal Gas equation, it becomes:

d = P x M / (R x T)

When any three of the four quantities in the equation are known, the forth can be calculated. For example, we’ve known d, P and T, M can be:

M = d x R x T / P

Question: What is the density of an ideal gas with a molecular mass of 50 g/mol at 2 atm and 27 °C?

Figure 9.50 Density determination of Gases

Ref: commons.wikimedia.org/

Solution:

Let’s start with the ideal gas law:

PV = nRT

where
P = pressure
V = volume
n = number of moles of gas
R = gas constant = 0.0821 L·atm/mol·K
T = absolute temperature

We know density ( d ) is mass (m) per unit volume. While the equation has a volume variable, there is no obvious mass variable. The mass can be found in the number of moles of the ideal gas.

The molecular mass ( M ) of the gas is the mass of one mole of the gas. This means n moles of the gas has a mass of nM grams.

m = nM

If we solve this for n we get

n = m/M

Now we

have enough to find the density of the gas. First, solve the ideal gas equation for V.

Substitute n for what we found earlier

Divide both sides by m

Invert the equation

density ( d ) = m/V, so

d= M*P

     R*T

From our question:
M = 50 g/mol
P = 2 atm
T = 27 °C

The first thing we need to do is convert the temperature to absolute temperature. The conversion between Kelvin and Celsius is:

T_K = T_C + 273

T_K = 27 + 273

T_K = 300 K

Another tricky part of ideal gas problems is matching the units on the ideal gas constant R. We’re using liters, atm, and Kelvin so we can use the value

R = 0.0821 L·atm/mol·K

Plug all these values into our equation

d= 

d= 4.06 g/L

Answer: The density of an ideal gas of 50 g/mol at 2 atmospheres and 27 °C is 4.06 g/L