Gas volumes and molar volumeEdexcel A-Level Chemistry: Revision notes

Equal volumes of different gases, measured at the same temperature and pressure, contain the same number of molecules (Avogadro's law). So the volume of a gas is proportional to its amount in mol. The molar volume is the volume occupied by one mole of gas under stated conditions. At room temperature and pressure (rtp) it is 24.0 dm³ mol⁻¹ (24 000 cm³ mol⁻¹), whatever the gas.

Volume (dm³) = amount (mol) × 24.0, so amount (mol) = volume (dm³) ÷ 24.0. Divide cm³ by 1000 first to get dm³.

When all the reactants and products that matter are gases at the same temperature and pressure, the volume ratio equals the mole ratio in the balanced equation, so no moles or molar volume are needed.

For example, in N₂(g) + 3H₂(g) → 2NH₃(g), 10 cm³ of nitrogen reacts with 30 cm³ of hydrogen to give 20 cm³ of ammonia. Only gases count: liquid water produced in combustion is ignored because its volume is negligible.

Combustion volume questions usually involve an excess of one gas. Work out how much of each gas reacts using the ratio, find the one in excess, then add the leftover gas to any gaseous product.

Worked example: 20 cm³ of methane burns in 100 cm³ of oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O(l). Oxygen used = 40 cm³; CO₂ formed = 20 cm³; oxygen left = 60 cm³. Total gas after cooling = 20 + 60 = 80 cm³. Passing the gases through aqueous sodium hydroxide absorbs the CO₂, leaving 60 cm³.

Use the mole as the bridge between mass and gas volume.

Mass to volume: n = mass ÷ M; use the equation ratio to find n of the gas; V = n × 24.0 dm³. Volume to mass: n(gas) = V ÷ 24.0; use the ratio to find n of the solid; mass = n × M.

Worked example: What volume of hydrogen at rtp forms when 0.486 g of magnesium reacts with excess acid? n(Mg) = 0.486 ÷ 24.3 = 0.0200 mol. Mg + 2HCl → MgCl₂ + H₂ is 1 : 1, so n(H₂) = 0.0200 mol. V = 0.0200 × 24.0 = 0.480 dm³ = 480 cm³.

React a weighed mass of magnesium ribbon (about 0.1 g, balance to 0.001 g) with excess dilute sulfuric or hydrochloric acid. Collect the hydrogen in a gas syringe (or an inverted measuring cylinder over water). Mg + H₂SO₄ → MgSO₄ + H₂.

1. Check the syringe moves freely and the apparatus is airtight.
2. Add the magnesium, bung at once, swirl.
3. Read the volume when the reaction has stopped.
4. n(H₂) = n(Mg) = mass ÷ 24.3; molar volume = V (dm³) ÷ n.

The result should be close to 24.0 dm³ mol⁻¹. A low value suggests gas escaped or magnesium was partly oxidised; a high value may be from the heat of reaction expanding the gas.