How to calculate mass of product from a balanced equation

Once you can balance an equation, the next skill SPM asks for is turning that equation into a number: given the mass of one substance, work out the mass of another. It sounds like a big calculation, but it is really the same five steps every single time. The equation is a recipe, and the mole is the measuring cup that lets you read the recipe in grams. Learn the five steps, and this becomes one of the most reliable sources of marks in Paper 2.

Why you cannot just compare masses

Students often try to shortcut the problem by comparing masses directly, “10 g reacted, so 10 g is produced.” That almost never works, because a balanced equation counts particles, not grams. Two magnesium atoms react with one oxygen molecule; the equation tells you the ratio of moles, not the ratio of masses. So you must convert into moles, use the ratio, then convert back into grams. That round trip is the whole method.

The five-step method

Use this exact routine every time:

1. Write the balanced equation. Nothing works if the equation is wrong, so balance it first and check every atom.
2. Find the moles of the substance you are given. Use moles = mass ÷ molar mass (Mr).
3. Use the mole ratio from the equation to find the moles of the substance you want. This is the step everyone rushes, read the big numbers in front of the formulae.
4. Convert those moles back to mass. Use mass = moles × Mr of the substance you want.
5. State the answer with a unit, rounded sensibly.

You can drill this pattern on our mass of product or reactant page, and practise reading ratios correctly at mole ratio in equations.

Worked example 1: burning magnesium

Calculate the mass of magnesium oxide formed when 4.8 g of magnesium burns completely in oxygen. (Ar: Mg = 24, O = 16)

• Step 1: 2Mg + O₂ → 2MgO.
• Step 2: moles of Mg = 4.8 ÷ 24 = 0.2 mol.
• Step 3: the ratio Mg : MgO is 2 : 2, which is 1 : 1, so moles of MgO = 0.2 mol.
• Step 4: Mr of MgO = 24 + 16 = 40. Mass = 0.2 × 40 = 8.0 g.

Notice the mass went up from 4.8 g to 8.0 g, the extra 3.2 g is the oxygen that combined with the magnesium. That is the whole point of converting through moles: masses do not simply carry across.

Worked example 2: when the ratio is not 1 : 1

The ratio in the first example was friendly. Most exam questions are not, so here is one that tests step 3 properly.

Calculate the mass of aluminium oxide formed when 5.4 g of aluminium reacts completely with oxygen. (Ar: Al = 27, O = 16)

• Step 1: 4Al + 3O₂ → 2Al₂O₃.
• Step 2: moles of Al = 5.4 ÷ 27 = 0.2 mol.
• Step 3: the ratio Al : Al₂O₃ is 4 : 2, which simplifies to 2 : 1. So moles of Al₂O₃ = 0.2 ÷ 2 = 0.1 mol.
• Step 4: Mr of Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102. Mass = 0.1 × 102 = 10.2 g.

If you had skipped the ratio and assumed 1 : 1, you would have doubled the answer. Step 3 is where the marks are won and lost.

When the product is a gas

Sometimes the product is a gas and the question asks for a volume, not a mass. The method is identical up to step 3; only the last step changes. Instead of multiplying the moles by Mr, multiply them by the molar volume, 24 dm³ mol⁻¹ at room conditions. For example, if a reaction produces 0.1 mol of carbon dioxide, its volume at room conditions is 0.1 × 24 = 2.4 dm³. Everything before that stays the same, because moles are still the bridge.

The mistakes that cost marks

Three slips account for most lost marks. First, an unbalanced equation, every later step inherits the error, so check it before you calculate. Second, reading the ratio upside down or ignoring it; always write the two numbers that matter (given : wanted) before dividing. Third, using the wrong molar mass, count every atom in the formula, including subscripts, so Al₂O₃ is 102 and not 43. A quick habit that catches most errors is to sanity-check the size of your answer, as we did when the magnesium mass rose rather than fell.

Work through five of these in a row and the five-step pattern becomes automatic. You can check any answer with our mole calculator. If you would like a teacher to watch you set out the ratio step and catch a habit before the exam does, our online one-to-one lessons with our experienced SPM Chemistry teachers give exactly that feedback, from RM50 an hour with a paid one-hour trial.