How do we use balanced equations to calculate reacting masses?
Using the mole and balanced symbol equations to calculate reacting masses, the conservation of mass, and finding empirical formulae from mass or percentage data.
A CCEA GCSE Chemistry answer on reacting-mass calculations, covering the conservation of mass, how to use the mole and a balanced equation to calculate the mass of a product or reactant, and how to find an empirical formula from masses or percentages.
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What this dot point is asking
CCEA wants you to use the conservation of mass and a balanced equation to calculate the mass of a reactant or product, working through moles, and to find an empirical formula from masses or percentage composition.
Conservation of mass
Apparent mass changes usually mean a gas was lost or gained: when a metal burns the product gains mass by combining with oxygen, while a fizzing reaction that releases gas appears to lose mass.
The reacting-mass method
The ratio in step 2 comes straight from the big numbers in the balanced equation. If the equation says , then 2 moles of magnesium make 2 moles of magnesium oxide, a 1:1 ratio.
Empirical formulae
To find it from masses or percentages: divide each element's mass (or percentage) by its to get moles, then divide all the results by the smallest to get the simplest ratio. If the answers are not whole numbers, multiply through to clear them.
Worked example
Examples in context
- Example 1. Scaling up a reaction
- A factory making lime heats limestone () and uses reacting-mass calculations to work out how much lime each tonne of limestone yields. The mole method lets them plan raw-material orders and predict output exactly.
- Example 2. Identifying an unknown
- When chemists analyse a new compound, they measure the mass of each element and calculate the empirical formula, which is often the first clue to its identity. The same percentage-to-moles method used in the worked example is a routine analytical step.
- Example 3. Checking a burning reaction
- When a strip of magnesium burns, weighing it before and after shows a mass increase, which seems to break conservation of mass until you remember the magnesium has combined with oxygen from the air. Adding the mass of oxygen that joined accounts for the increase exactly, confirming that mass is conserved once every reactant and product, including gases, is included.
Working with the ratio carefully
The equation ratio is where most marks are won or lost. It is the ratio of the big numbers in front of the formulae, not the ratio of the masses. For , one mole of nitrogen reacts with three moles of hydrogen to give two moles of ammonia, a 1:3:2 ratio in moles. To turn that into masses you must convert each through its own relative formula mass. Keeping moles and masses separate, and only using the ratio on the moles, is the habit that makes reacting-mass questions reliable, and it is exactly the skill extended into titration and gas-volume calculations in Unit 2.
Try this
Q1. State what is meant by the conservation of mass. [1 mark]
- Cue. The total mass of products equals the total mass of reactants.
Q2. A compound has moles of C and H in the ratio 1 : 3. Write its empirical formula. [1 mark]
- Cue. .
Exam-style practice questions
Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
CCEA 20194 marksCalculate the mass of magnesium oxide formed when 6.0 g of magnesium burns in oxygen. 2Mg + O2 gives 2MgO. (Ar: Mg = 24, O = 16)Show worked answer →
Markers want moles of magnesium, then the ratio, then the mass.
First find moles of magnesium: of Mg is 24, so
The equation shows a 2:2 (so 1:1) ratio of Mg to MgO, so moles of MgO formed = 0.25 mol.
Relative formula mass of MgO is , so
Markers reward 0.25 mol Mg, the 1:1 ratio giving 0.25 mol MgO, and the answer 10 g.
CCEA 20214 marksA compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula. (Ar: C = 12, H = 1)Show worked answer →
The marks are for moles of each element, then the simplest ratio.
Find moles of each element:
Divide both by the smaller number (0.20):
So the empirical formula is .
Markers reward moles of C and H, dividing by the smaller to get the 1:4 ratio, and the formula CH4.
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Sources & how we know this
- CCEA GCSE Chemistry specification (1110) — CCEA (2017)