Relative formula mass — AQA GCSE Chemistry
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Relative formula mass explained
Percentage by mass shows the contribution of one element to the total mass of a compound.
Read the full explanation
First calculate the relative formula mass, Mᵣ, by summing all relative atomic masses in the formula. Then find the total mass of the element of interest by multiplying its relative atomic mass by the number of its atoms in the formula. Divide that element's total mass by Mᵣ and multiply by 100. For example, in CO₂, Mᵣ = 12 + (2 × 16) = 44; carbon contributes 12 ÷ 44 × 100 = 27.3% and oxygen contributes 32 ÷ 44 × 100 = 72.7%. The percentages should sum to 100%, which is a useful check.
Your focus
- Calculate the percentage by mass of an element in a compound.
- Use relative atomic masses and relative formula mass correctly.
- Check percentage composition answers by summing to 100%.
Relative formula mass exam tips
Marking Points
- Calculate Mᵣ by summing the relative atomic masses of all atoms in the formula.
- Find the total mass of the chosen element using its relative atomic mass and atom count.
- Divide the element's total mass by Mᵣ and multiply by 100 to obtain percentage by mass.
- Worked example: in CO₂, carbon is 12 ÷ 44 × 100 = 27.3% and oxygen is 32 ÷ 44 × 100 = 72.7%.
- The percentages of all elements in a compound should sum to 100%.
- Give answers to an appropriate number of significant figures, typically three.
Examiner Tips
- 💡Write down Mᵣ first and label it clearly before starting the percentage calculation.
- 💡Show the numerator as the element's total mass and the denominator as Mᵣ, then multiply by 100.
- 💡Check that the percentages for all elements add up to 100% as a final verification.
- 💡State the element and the compound in your answer, for example 'percentage by mass of carbon in CO₂'.
Common Mistakes
- Dividing by the element's mass instead of Mᵣ; the correction is to divide the element's total mass by the compound's Mᵣ.
- Forgetting to multiply the relative atomic mass by the number of atoms of that element; the correction is to include the atom count, for example 2 × 16 for oxygen in CO₂.
- Rounding intermediate values too early, causing a final answer that is slightly wrong; the correction is to keep full values during the calculation and round only at the end.