Use of amount of substance in relation to volumes of gases (4.3.5.0) — AQA GCSE Chemistry Revision Notes

Revision notes for AQA GCSE Chemistry specification point 4.3.5.0, Use of amount of substance in relation to volumes of gases.

Amount of Substance and Volumes of Gases

Equal amounts in moles of ANY gas occupy the SAME VOLUME under the same conditions of temperature and pressure. It does not matter which gas it is.

 

The number you must remember

At ROOM TEMPERATURE AND PRESSURE (20°C and 1 atmosphere), one mole of any gas occupies 24 dm³, which is the same as 24 000 cm³.

volume of gas (dm³) = moles × 24

Rearranged: moles = volume (dm³) ÷ 24

 

Going from mass to volume

Convert the mass to MOLES first, then multiply by 24.

Example: what volume does 8.0 g of methane, CH4 (Mr = 16), occupy at room temperature and pressure?

  • moles = mass ÷ Mr = 8.0 ÷ 16 = 0.50 mol
  • volume = 0.50 × 24 = 12 dm³

 

Gas volumes straight from a balanced equation

Because equal moles means equal volumes, the BALANCING NUMBERS of the gases in an equation are also the ratio of their VOLUMES. You do not need to convert to moles at all.

Example: N2 + 3H2 → 2NH3. If 60 cm³ of hydrogen reacts completely, what volume of ammonia is made?

  • The ratio of H2 to NH3 is 3 : 2
  • volume of NH3 = 60 × (2 ÷ 3) = 40 cm³

 

Changing the subject of the equation

You may be asked to rearrange rather than substitute. The three forms you need are:

  • volume = moles × 24
  • moles = volume ÷ 24
  • mass = moles × Mr

 

EXAM TIP: the volume ratio shortcut only works for substances that are GASES. If the equation also contains solids or solutions, you cannot use their balancing numbers as volumes.

WATCH THE UNITS: 24 dm³ is per mole. If the question gives you cm³, either work in cm³ using 24 000, or divide by 1000 first to get dm³. Do not mix the two.

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