# 14.9: Calculating the Molar Mass of a Gas

Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. Small balloons filled with helium are often affordable and available at stores, but large ones are much more expensive (and require a lot more helium).

## Calculating Molar Mass and Density of a Gas

A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.

Example $$\PageIndex{1}$$

A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of $$1.211 \: \text{g}$$ and occupies a volume of $$677 \: \text{mL}$$. The temperature in the laboratory is $$23^\text{o} \text{C}$$ and the air pressure is $$0.987 \: \text{atm}$$. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.

Solution

Step 1: List the known quantities and plan the problem.

Known

• Mass $$= 1.211 \: \text{g}$$
• $$V = 677 \: \text{mL} = 0.677 \: \text{L}$$
• $$T = 23^\text{o} \text{C} = 296 \: \text{K}$$
• $$P = 0.987 \: \text{atm}$$
• $$R = 0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$$

Unknown

• $$n = ? \: \text{mol}$$
• Molar Mass $$= ? \: \text{g/mol}$$

First the ideal gas law will be used to solve for the moles of unknown gas $$\left( n \right)$$. Then the mass of the gas divided by the moles will give the molar mass.

Step 2: Solve.

$n = \frac{PV}{RT} = \frac{0.987 \: \text{atm} \times 0.677 \: \text{L}}{0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 296 \: \text{K}} = 0.0275 \: \text{mol}$

Now divide $$\text{g}$$ by $$\text{mol}$$ to get the molar mass.

$\text{molar mass} = \frac{1.211 \: \text{g}}{0.0275 \: \text{mol}} = 44.0 \: \text{g/mol}$

Since $$\ce{N}$$ has a molar mass of $$14 \: \text{g/mol}$$ and $$\ce{O}$$ has a molar mass of $$16 \: \text{g/mol}$$, the formula $$\ce{N_2O}$$ would produce the correct molar mass.

The $$R$$ value that corresponds to a pressure in $$\text{atm}$$ was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.

## Calculating Density of a Gas

The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas $$\left( \ce{NH_3} \right)$$ at $$0.913 \: \text{atm}$$ and $$20^\text{o} \text{C}$$, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be $$17.04 \: \text{g/mol}$$. Next, assume exactly $$1 \: \text{mol}$$ of ammonia $$\left( n = 1 \right)$$ and calculate the volume that such an amount would occupy at the given temperature and pressure.

$V = \frac{nRT}{P} = \frac{1.00 \: \text{mol} \times 0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 293 \: \text{K}}{0.913 \: \text{atm}} = 26.3 \: \text{L}$

Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.

$\text{Density} = \frac{17.04 \: \text{g}}{26.3 \: \text{L}} = 0.648 \: \text{g/L}$

As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to $$\frac{\left( 170.4 \: \text{g/mol} \right)}{\left( 22.4 \: \text{L/mol} \right)} = 0.761 \: \text{g/L}$$. It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from $$0^\text{o} \text{C}$$ to $$20^\text{o} \text{C}$$) and the decrease in pressure (from $$1 \: \text{atm}$$ to $$0.913 \: \text{atm}$$) would cause the $$\ce{NH_3}$$ molecules to spread out a bit further from one another.

## Summary

• Calculations of molar mass and density of an ideal gas are described.