How to find the formula of a hydrated salt

A hydrated salt is a salt whose crystals lock a fixed number of water molecules into their structure, the water of crystallisation. Blue copper(II) sulfate crystals are the classic example, written CuSO₄·5H₂O, where the dot means “combined with” and the 5 tells you five water molecules travel with every one formula unit of the salt. The exam question is almost always the same: you are given some masses from a heating experiment and asked to find that number, usually called x. This guide shows the method and the practical details that carry the marks.

The idea behind the question

When you heat a hydrated salt gently, the water of crystallisation is driven off as steam and an anhydrous (water-free) salt is left behind. So the mass you lose on heating is the mass of the water, and the mass that remains is the mass of the anhydrous salt. If you can turn each of those masses into moles, the ratio of moles gives you x directly. It is really the empirical-formula method wearing different clothes: you can see the parent skill on our percentage composition by mass page.

The five-step method

1. Record the mass of the hydrated salt before heating.
2. Heat to constant mass. Heat, cool, reweigh, and repeat until the mass stops changing, that is your proof all the water has gone.
3. Work out the two masses: the anhydrous salt is the final mass; the water lost is the initial mass minus the final mass.
4. Convert each to moles. Moles of anhydrous salt = its mass ÷ its molar mass; moles of water = mass of water ÷ 18.
5. Divide both by the smaller number to get the ratio 1 : x. That x is the number of water molecules in the formula.

You will need the molar mass of the anhydrous salt at step 4, so make sure that skill is solid, practise it at molar mass.

Worked example 1: hydrated copper(II) sulfate

25.0 g of hydrated copper(II) sulfate is heated to constant mass, leaving 16.0 g of white anhydrous copper(II) sulfate. Find the formula. (Ar: Cu = 64, S = 32, O = 16, H = 1)

• Mass of anhydrous CuSO₄ = 16.0 g. Mass of water lost = 25.0 − 16.0 = 9.0 g.
• Molar mass of CuSO₄ = 64 + 32 + (4 × 16) = 160. Moles of CuSO₄ = 16.0 ÷ 160 = 0.1 mol.
• Moles of H₂O = 9.0 ÷ 18 = 0.5 mol.
• Ratio CuSO₄ : H₂O = 0.1 : 0.5 = 1 : 5.
• Formula: CuSO₄·5H₂O.

Worked example 2: finding x for magnesium sulfate

A 4.92 g sample of hydrated magnesium sulfate, MgSO₄·xH₂O, leaves 2.40 g of anhydrous MgSO₄ when heated to constant mass. Find x. (Ar: Mg = 24, S = 32, O = 16, H = 1)

• Mass of water lost = 4.92 − 2.40 = 2.52 g.
• Molar mass of MgSO₄ = 24 + 32 + (4 × 16) = 120. Moles of MgSO₄ = 2.40 ÷ 120 = 0.02 mol.
• Moles of H₂O = 2.52 ÷ 18 = 0.14 mol.
• Ratio MgSO₄ : H₂O = 0.02 : 0.14 = 1 : 7, so x = 7 and the salt is MgSO₄·7H₂O.

The numbers are messier than in the first example, but the five steps do not change. Divide by the smaller value and the ratio appears.

The practical points examiners reward

The calculation is only half the question; Paper 3 and structured questions also test the experiment. Heat the salt gently and keep a lid on the crucible slightly open, so the crystals do not spit out and lower your final mass. Crucially, you must heat to constant mass, heat, cool in a desiccator, weigh, then reheat and weigh again until two readings agree. If you stop too early, some water remains, the water mass looks too small, and your x comes out too low. Cooling in a desiccator matters too, because a warm anhydrous salt will draw moisture back from the air. You can see this and other heating experiments in context on our experiments hub.

The common mistakes

Three errors cause most lost marks. First, dividing the water mass by the wrong number, water is always 18, so use 18, not 16 or 20. Second, using the molar mass of the hydrated salt instead of the anhydrous salt at step 4; the anhydrous salt is what remains, so use its mass and its molar mass together. Third, rounding the ratio carelessly, 1 : 4.9 should round to 1 : 5, but 1 : 5.4 is genuinely closer to 1 : 5 only if the data supports it, so trust clean experimental numbers rather than forcing a value. Hydrated-salt questions sit inside the mole concept, which you can revise in full at the mole concept.

Get these steps clean and this is a very predictable set of marks. If you would like a teacher to check your working and your practical technique together, our online one-to-one lessons with our experienced SPM Chemistry teachers give that kind of feedback, from RM50 an hour with a paid one-hour trial.