Molar volume of a gas

The volumes of gases and liquids are often measured in cubic decimetres (dm) in chemistry. A decimetre is equivalent to 10cm, so a cubic decimetre is a volume of space that is 10cm by 10cm by 10cm. Therefore, 1dm³ = 1000cm³.

• Volumes of gases and liquids are measured in cubic centimetres (cm³, or cubic decimetres (dm³) or litres (l)
• 1 mole of any gas at room temperature and pressure occupies 24dm³ (24,000cm³). This value is known as the molar volume
• 1 mole of any gas has the same number of molecules. This means it has the same volume as one mole of any other gas at the same temperature and pressure

You may be asked to calculate the total volume occupied by a given mass of gas. We will work through an example here to show how this is done.

Example

Calculate the volume occupied by 0.02g of H₂ at room temperature and pressure.

1. Calculate the M_r of the gas molecule:

M_r of H₂ = 2×1 = 2gmol^-1

2. Calculate the number of moles of the gas in this mass:
3. Calculate the volume occupied by this number of moles of gas:

As 1 mol of hydrogen can occupy 24,000cm³, 0.01 mol of hydrogen occupies:

0.01 x 24,000cm³ = 240cm³ (0.24dm³)

Working out the volume of gas produced during a reaction

As long as the balanced symbol equation for a reaction is known, and the starting mass of one of the reactants is given, the volume of gas produced during a reaction can be calculated.

Carbon dioxide is produced at room temperature and pressure and an excess of dilute hydrochloric acid is added to 1.00g of calcium carbonate. The equation for this would be:

We can see that 1 mol of CaCO₃ gives 1 mol of CO₂ from the equation to calculate the volume of this gas. To work out the molar mass of CaCO₃, we add up the atomic masses of each element present:

Ca = 40

C = 12

O = 16

Molar mass = 40 + 12 + (16 x 3) = 100g

So 100g of CaCO₃ produces 24dm³ CO₂ at room temperature.

This means that 1g of CaCO₃ produces   x 24dm³ = 0.24dm³.

This tells us that 1g of calcium carbonate produces 0.24dm³ of carbon dioxide.