Mass calculations from chemical equations

A balanced equation tells you the ratio in which substances react, counted in moles. Molar mass turns those moles into grams. Putting the two together is called stoichiometry.

The three-step route

1. Grams to moles of the substance you know: divide by its molar mass.
2. Mole ratio: use the numbers in front of the formulas in the balanced equation to find moles of the substance you want.
3. Moles to grams of the substance you want: multiply by its molar mass.

Example 1: heating limestone

CaCO₃ → CaO + CO₂

What mass of quicklime is obtained from 50 g of calcium carbonate?

1. Moles of CaCO₃: 50 ÷ 100.09 = 0.4996 mol.
2. The equation shows 1 mole gives 1 mole of CaO: 0.4996 mol.
3. Mass of CaO: 0.4996 × 56.08 = 28.0 g.

The remaining 22.0 g leaves as carbon dioxide. Mass is conserved.

Example 2: a ratio that is not 1 to 1

2H₂ + O₂ → 2H₂O

What mass of oxygen is needed to burn 4 g of hydrogen?

1. Moles of H₂: 4 ÷ 2.016 = 1.984 mol.
2. 2 moles of hydrogen need 1 mole of oxygen, so oxygen = 1.984 ÷ 2 = 0.992 mol.
3. Mass of O₂: 0.992 × 32.00 = 31.7 g.

Hydrogen needs about eight times its own mass of oxygen.

Example 3: neutralisation

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

What mass of sodium hydroxide neutralises 4.9 g of sulphuric acid? Moles of acid: 4.9 ÷ 98.07 = 0.0500. Moles of NaOH: twice that, 0.0999. Mass: 0.0999 × 40.00 = 4.00 g.

Limiting reagent

When amounts of two reactants are given, one usually runs out first. It is the limiting reagent, and it decides how much product forms. To find it, change both masses to moles and divide each by its number in the equation. The smaller result marks the limiting reagent.

Example. 4 g of hydrogen and 16 g of oxygen are sparked together. Hydrogen: 1.98 mol, divided by 2 gives 0.99. Oxygen: 16 ÷ 32.00 = 0.50 mol, divided by 1 gives 0.50. Oxygen is limiting. Water formed: 2 × 0.50 = 1.00 mol, or 18.0 g. About 2 g of hydrogen is left over.

Percentage yield

The mass worked out from the equation is the theoretical yield. In practice some product is lost, or the reaction does not go to completion.

percentage yield = actual mass obtained ÷ theoretical mass × 100

If the limestone in Example 1 gives 25.2 g of quicklime, the yield is 25.2 ÷ 28.0 × 100 = 90%.

Before you start any such sum

• Balance the equation. An unbalanced equation gives a wrong ratio.
• Write the molar mass under each formula you will use.
• Never compare grams with grams directly. The ratio in an equation is in moles.

Look up the molar masses on the compound pages or work them out with the calculator.

Which values? This site uses the abridged standard atomic weights published by IUPAC (CIAAW), read on 12 November 2026. Your textbook or question paper may round them differently, for example H = 1 and O = 16, so its answer can differ slightly. In an exam, use the values the paper gives.

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