OCR OCR 2019-style practice: Magnesium burns in oxygen to form magnesium oxide: 2Mg + O_2 → 2MgO. Calculate the mass of magnesium oxide formed when 6.0 g of magnesium burns completely. Relative atomic masses: Mg = 24, O = 16.

A C3.1 reacting-mass calculation. Reward: moles of Mg = 6.024 = 0.25 mol (1 mark). The mole ratio of Mg to MgO is 2:2, so moles of MgO = 0.25 mol (1 mark). The relative formula mass of MgO = 24 + 16 = 40 (1 mark). Mass of MgO = 0.25 × 40 = 10\ g (1 mark). Markers credit converting mass to moles, using the correct mole ratio, the M_r of MgO, and the final mass. A common error is forgetting that the ratio is 2:2 (which is 1:1) and using the wrong relative formula mass.

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What is the mole, and how do we use it to calculate reacting masses?

Relative formula mass, the mole and the Avogadro constant, calculating moles from mass, using mole ratios to find reacting masses, limiting reactants, and percentage yield.

A focused answer to OCR Gateway GCSE Chemistry A topic C3.1 on the mole and reacting masses, covering relative formula mass, the mole and the Avogadro constant, calculating moles from mass, using mole ratios for reacting masses, limiting reactants, and percentage yield.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Relative formula mass and the mole
Calculating moles from mass
Reacting masses
Limiting reactants (Higher)
Percentage yield

What this dot point is asking

OCR wants you to calculate the relative formula mass of a compound, understand the mole and the Avogadro constant, calculate moles from mass, use mole ratios from balanced equations to find reacting masses, identify the limiting reactant (Higher), and calculate percentage yield. This is the heart of quantitative chemistry and a major part of the maths assessment.

Relative formula mass and the mole

For example, the $M_r$ of water ( $\text{H}_2\text{O}$ ) is $(2 \times 1) + 16 = 18$ , so one mole of water has a mass of $18$ g. The $M_r$ of calcium carbonate ( $\text{CaCO}_3$ ) is $40 + 12 + (3 \times 16) = 100$ .

Calculating moles from mass

Reacting masses

To find the mass of a product or reactant, use the three-step method, which works for any balanced equation:

Convert the known mass to moles ( $\text{moles} = \dfrac{\text{mass}}{M_r}$ ).
Use the mole ratio from the balanced equation to find the moles of the other substance.
Convert that back to mass ( $\text{mass} = \text{moles} \times M_r$ ).

The same steps work in grams, kilograms or tonnes, because the mole ratio is just a number.

Limiting reactants (Higher)

To find the limiting reactant, work out the moles of each reactant and compare them with the mole ratio in the equation. The reactant that provides fewer moles relative to the ratio is limiting, which is not always the one with the smaller mass.

Percentage yield

Reacting mass with a limiting reactant

$8.0$ g of methane ( $\text{CH}_4$ , $M_r = 16$ ) is burned in $40$ g of oxygen ( $\text{O}_2$ , $M_r = 32$ ): $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ . Find the mass of carbon dioxide produced ( $M_r$ of $\text{CO}_2 = 44$ ).

step 1: Find the moles of each reactant

Moles of $\text{CH}_4 = \dfrac{8.0}{16} = 0.50$ mol. Moles of $\text{O}_2 = \dfrac{40}{32} = 1.25$ mol.

step 2: Identify the limiting reactant

The ratio of $\text{CH}_4$ to $\text{O}_2$ is $1:2$ , so $0.50$ mol of methane needs $1.0$ mol of oxygen. There is $1.25$ mol of oxygen, which is more than enough, so oxygen is in excess and methane is limiting.

step 3: Use the limiting reactant in the ratio

The ratio of $\text{CH}_4$ to $\text{CO}_2$ is $1:1$ , so moles of $\text{CO}_2 = 0.50$ mol.

step 4: Convert to mass

Mass of $\text{CO}_2 = 0.50 \times 44 = 22\ \text{g}$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksMagnesium burns in oxygen to form magnesium oxide:

2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

. Calculate the mass of magnesium oxide formed when 6.0 g of magnesium burns completely. Relative atomic masses: Mg = 24, O = 16.

Show worked answer →

A C3.1 reacting-mass calculation. Reward: moles of Mg $= \dfrac{6.0}{24} = 0.25$ mol (1 mark). The mole ratio of Mg to MgO is $2:2$ , so moles of MgO $= 0.25$ mol (1 mark). The relative formula mass of MgO $= 24 + 16 = 40$ (1 mark). Mass of MgO $= 0.25 \times 40 = 10\ \text{g}$ (1 mark). Markers credit converting mass to moles, using the correct mole ratio, the $M_r$ of MgO, and the final mass. A common error is forgetting that the ratio is $2:2$ (which is $1:1$ ) and using the wrong relative formula mass.

OCR 20225 marksNitrogen reacts with hydrogen to make ammonia:

\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3

. A reaction uses 28 g of nitrogen with excess hydrogen and produces 27 g of ammonia. Calculate the theoretical mass of ammonia and the percentage yield. Relative atomic masses: N = 14, H = 1.

Show worked answer →

A Higher tier two-stage calculation. Reward: moles of $\text{N}_2 = \dfrac{28}{28} = 1.0$ mol (1 mark). The ratio of $\text{N}_2$ to $\text{NH}_3$ is $1:2$ , so moles of $\text{NH}_3 = 2.0$ mol (1 mark). The $M_r$ of $\text{NH}_3 = 14 + 3 = 17$ , so the theoretical mass $= 2.0 \times 17 = 34\ \text{g}$ (1 mark). Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}} \times 100 = \dfrac{27}{34} \times 100 = 79.4\%$ (1 mark for the method, 1 mark for the answer, about 79%). Markers reward the theoretical mass via moles and the ratio, and the percentage yield formula with the correct substitution. A common slip is to use mass directly in the yield without first finding the theoretical mass.

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Sources & how we know this

OCR GCSE (9-1) Gateway Science Suite Chemistry A (J248) specification — OCR (2016)