Percentage of Water in a Hydrated Salt

Introduction

A hydrate's formula tells how much of its mass belongs to water of crystallisation. Heating can sometimes remove that water, allowing an experimental mass loss to be compared with the predicted percentage. The calculation follows percentage composition, but the numerator is the mass of whole H₂O groups, including both hydrogen and oxygen.

Core explanation

For a hydrate written Salt·xH₂O, percentage water by mass is [xMᵣ(H₂O) ÷ Mᵣ(Salt·xH₂O)] × 100%. First calculate the anhydrous salt contribution. Then calculate x times the mass of one water molecule, and add the two for the full hydrate denominator. The dot coefficient multiplies water only; it does not multiply the salt formula.

For CuSO₄·5H₂O with rounded Cu = 64, S = 32, O = 16 and H = 1, anhydrous CuSO₄ contributes 64 + 32 + 4(16) = 160. Five waters contribute 5[2(1) + 16] = 90. Full Mᵣ is 250, so water percentage is 90/250 × 100 = 36%. The remaining 64% is the CuSO₄ contribution. Both percentages refer to the mass of the hydrated crystal, not to the mass of dry salt after heating.

For Na₂CO₃·10H₂O with Na = 23, C = 12, O = 16 and H = 1, the dry salt contributes 2(23) + 12 + 3(16) = 106. Ten waters contribute 180. Full Mᵣ is 286; water percentage is 180/286 × 100 ≈ 62.9%. A hydrate with a larger dot number need not always have a larger water percentage than another hydrate, because the dry salt's mass also changes. Compare ratios, not dot coefficients alone.

The theoretical percentage predicts an ideal mass loss when the only change is complete water removal. If 25.0 g pure CuSO₄·5H₂O is fully dehydrated to CuSO₄ without other loss, expected water loss is 25.0 × 0.36 = 9.0 g and expected residue is 16.0 g in the rounded model. These values also follow from the 90:160 water-to-dry-salt mass ratio.

In a real heating practical, an observed loss may differ. Insufficient heating can leave water behind, reducing the measured loss. Spattering can remove some salt, increasing the apparent loss. A hot crucible weighed before cooling can give unstable readings, and some anhydrous salts can reabsorb moisture while exposed to air. Excessive heating can cause decomposition rather than simple dehydration. A mass change alone therefore needs experimental care before being called exactly the water content.

The calculation is distinct from percentage of oxygen in the hydrate. CuSO₄·5H₂O contains four oxygen atoms in sulfate and five in water, but water mass includes ten hydrogen atoms as well. To find percentage water, use 5Mᵣ(H₂O) = 90 as numerator. To find percentage oxygen, use 9Aᵣ(O) = 144 as numerator and the same full denominator 250, giving 57.6% oxygen in the rounded model. Both answers can be correct for different questions.

If an experimental loss and dry-salt mass are known, their mole amounts may be compared to infer x: n(H₂O) = mass lost/18 and n(salt) = residue mass/M(salt). The ratio n(H₂O):n(salt) can suggest a whole-number hydrate formula. That inverse method assumes loss is water and the residue is the intended anhydrous salt.

Step-by-step reasoning

1. Read the dry salt formula and the dot coefficient x. 2. Calculate Mᵣ(dry salt), xMᵣ(H₂O) and their sum for Mᵣ(hydrate). 3. Divide the water contribution by the full hydrate contribution and multiply by 100. 4. For a sample-mass prediction, multiply the sample mass by the water fraction and check the residue.

Visual explanation

Draw a 250-part bar for CuSO₄·5H₂O, with 160 parts coloured for CuSO₄ and 90 for five waters. A line from the water segment to “90/250 = 36%” shows how the percentage relates to the complete crystal mass.

Real-world analogy

A packaged kit contains a fixed-weight device and five identical batteries. To find what percent of the kit's weight comes from batteries, divide the combined battery weight by the whole kit weight, not by the device weight alone. Hydrate water is the fixed group added to the dry salt.

Real-world example

In a teaching experiment, a student may heat blue copper(II) sulfate pentahydrate and measure the loss in mass after cooling. Comparing the observed percentage loss with the formula's predicted 36% tests whether the simple dehydration model and experimental handling are reasonable.

Why?

Why is the hydrated salt's whole Mᵣ the denominator? A mass percentage asks what part of the original sample's total mass came from water. Dividing by only the anhydrous contribution would give a water-to-dry-salt ratio, a different quantity.

Common misconception

“For CuSO₄·5H₂O, divide 90 by 160 to get percent water.” The result 56.25% compares water mass with dry-salt mass, not with the original hydrated mass. The water percentage of the hydrate is 90/250 × 100 = 36%.

Worked example

Find water percentage in MgSO₄·7H₂O using Mg = 24, S = 32, O = 16 and H = 1. Dry MgSO₄ contributes 24 + 32 + 4(16) = 120. Seven waters contribute 7(18) = 126. Full Mᵣ = 246. Water percentage = 126/246 × 100 ≈ 51.2%. A 12.3 g pure sample would ideally lose 12.3 × 126/246 = 6.30 g water upon complete dehydration without side reactions.

Quick check

1. Which mass belongs in the denominator for water percentage of CuSO₄·5H₂O? Answer: The full hydrate mass contribution, 250 in the rounded example.

Exam focus

Show the dry, water and total formula-mass contributions separately. State “percentage by mass of water of crystallisation” and use the full hydrate denominator. For a heating experiment, distinguish the theoretical formula value from the observed loss and note any relevant experimental limitation.

Advanced insight

An apparent whole-number x inferred from heating depends on a chemical assumption: the mass lost is water and the residue's formula is known. Thermogravimetric measurements can reveal separate mass-loss stages as temperature rises, but assigning them to particular structural waters requires additional evidence. The simple school ratio is the first model, not a full structural determination.

Summary

Water percentage in Salt·xH₂O is xMᵣ(H₂O) divided by the full hydrate Mᵣ, times 100. This predicts an ideal dehydration mass loss if only water leaves. Use the correct numerator and denominator, and interpret experimental differences in light of heating and weighing conditions.

Practice questions

1. Calculate percentage water in CuSO₄·5H₂O using dry contribution 160 and water contribution 90. Answer: 90/(160 + 90) × 100 = 36%. 2. What ideal mass of water is lost from 50.0 g of that pure hydrate? Answer: 50.0 × 0.36 = 18.0 g water in the rounded model. 3. Calculate water percentage in Na₂CO₃·10H₂O with dry contribution 106 and water contribution 180. Answer: 180/(106 + 180) × 100 ≈ 62.9%. 4. Why could an observed mass loss exceed the theoretical water loss? Answer: Spattering or decomposition might remove material besides water, making the apparent loss too large.