GCSE Chemistry revision

Use of amount of substance in relation to volumes of gases

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

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At room temperature and pressure (RTP), the volume of any gas can be found using the equation: Volume (dm³) = moles × 24 This equation applies to all gases regardless of their type, whether chlorine, oxygen, or water vapour.
The molar volume of a gas at RTP is 24 dm³ per mole.

To find the number of moles from a known volume, rearrange the equation: moles = volume / 24 For example, 60 dm³ of oxygen contains 60 / 24 = 2.5 moles.

When given a mass rather than moles, first calculate moles using: moles = mass / Mr The relative formula mass (Mr) is found by adding together the relative atomic masses of all atoms in the formula.
For example, water (H₂O) has an Mr of 18, so 27 g of water vapour gives 27 / 18 = 1.5 moles.
Once moles are known, multiply by 24 to find the volume: 1.5 × 24 = 36 dm³.

4. Finding Volume from Mass AQA

When given a mass rather than moles, first calculate moles using: moles = mass / Mr The relative formula mass (Mr) is found by adding together the relative atomic masses of all atoms in the formula.
For example, water (H₂O) has an Mr of 18, so 27 g of water vapour gives 27 / 18 = 1.5 moles.
Once moles are known, multiply by 24 to find the volume: 1.5 × 24 = 36 dm³.

When one reactant is in excess, the other reactant is the limiting reagent and determines the amount of product formed.
To find the volume of a gaseous product, first convert the volume of the limiting reagent to moles by dividing by 24.
Use the molar ratio from the balanced equation to find the moles of product, then multiply by 24 to get its volume.

5. Limiting Reagents and Gas Volume Calculations AQA

When one reactant is in excess, the other reactant is the limiting reagent and determines the amount of product formed.
To find the volume of a gaseous product, first convert the volume of the limiting reagent to moles by dividing by 24.
Use the molar ratio from the balanced equation to find the moles of product, then multiply by 24 to get its volume.

The molar ratio is found by comparing the large coefficients (the numbers written in front of chemical formulae) in a balanced equation.
If no number is written in front of a formula, it is assumed to be 1.
For example, in N₂ + 3H₂ → 2NH₃, the molar ratio of N₂ to NH₃ is 1:2.

6. Molar Ratios in Balanced Equations AQA

The molar ratio is found by comparing the large coefficients (the numbers written in front of chemical formulae) in a balanced equation.
If no number is written in front of a formula, it is assumed to be 1.
For example, in N₂ + 3H₂ → 2NH₃, the molar ratio of N₂ to NH₃ is 1:2.

When converting between the volumes of two gases in the same reaction, you can skip calculating moles and simply apply the molar ratio directly to the volumes.
This shortcut works because twice as many moles of a gas occupies twice the volume at the same temperature and pressure.
For example, if 18 dm³ of N₂ reacts with excess H₂, the volume of NH₃ produced is 18 × 2 = 36 dm³ using the 1:2 ratio.

7. Shortcut: Converting Directly Between Gas Volumes AQA

When converting between the volumes of two gases in the same reaction, you can skip calculating moles and simply apply the molar ratio directly to the volumes.
This shortcut works because twice as many moles of a gas occupies twice the volume at the same temperature and pressure.
For example, if 18 dm³ of N₂ reacts with excess H₂, the volume of NH₃ produced is 18 × 2 = 36 dm³ using the 1:2 ratio.

The molar volume of 24 dm³/mol only applies at room temperature and pressure (RTP).
Changing the temperature or pressure alters the volume that a gas occupies, changing the value used in the equation.
For GCSE exams, you only need to use the value of 24 dm³/mol.

8. Conditions: Room Temperature and Pressure (RTP) AQA

The molar volume of 24 dm³/mol only applies at room temperature and pressure (RTP).
Changing the temperature or pressure alters the volume that a gas occupies, changing the value used in the equation.
For GCSE exams, you only need to use the value of 24 dm³/mol.

Volume of gas: Volume (dm³⁾ = moles × 24 Moles from volume: moles = fracvolume (dm³⁾24 Moles from mass: moles = mass (g) / Mr