Water Fraction in a Hydrated Salt

Introduction

A hydrate formula predicts how much of its mass belongs to water. Heating can provide a measured mass-loss fraction for comparison. Agreement supports a proposed water count when the residue is the identified anhydrous salt and water is the only material lost.

Core explanation

For salt·xH₂O, theoretical water percentage by mass is [xM(H₂O)]/[M(salt) + xM(H₂O)] × 100%. The numerator counts the water components per formula unit, while the denominator is the full hydrate molar mass. In CuSO₄·5H₂O, M(CuSO₄) ≈ 159.61 g mol⁻¹ and five waters contribute 5 × 18.016 = 90.08 g mol⁻¹. Full M is about 249.69 g mol⁻¹, so theoretical water percentage is 90.08/249.69 × 100 ≈ 36.1%. A 10.0 g ideal sample of this hydrate would contain about 3.61 g formula water and 6.39 g salt component.

The measured mass-loss percentage is (mass before heating − mass of residue)/mass before heating × 100. If 2.50 g of hydrate leaves 1.60 g dry anhydrous residue, loss is 0.90 g or 36.0% of the initial sample. This is consistent with the 36.1% theoretical value at the stated precision. The comparison is a cross-check, not proof from mass alone. An incompletely dried solid, a decomposed salt or external surface moisture can also alter the observed loss.

One can infer a candidate x from a measured water fraction f if the anhydrous salt formula is known: f = xMwater/(Msalt + xMwater), so x = fMsalt/[(1 − f)Mwater]. For f = 0.360 with CuSO₄, x ≈ (0.360 × 159.61)/(0.640 × 18.016) ≈ 4.98, supporting five waters. This algebra is equivalent to dividing moles of water lost by moles of residue. It is useful as a reverse check, but the mole-ratio method keeps the particle meaning clearer.

Different hydrates of the same salt can have very different water fractions. More water units increase both numerator and denominator, but the water fraction rises. A hypothetical CuSO₄·H₂O would have 18.016/[159.61 + 18.016] ≈ 10.1% water, far below pentahydrate's 36.1%. A measured fraction near an intermediate value could mean a mixture of hydration states rather than a new formula with a fractional x, particularly if the material has partially dehydrated before weighing.

External moisture is not necessarily formula water. A damp crystal surface can make initial mass and heating loss too large, inflating inferred x. Conversely, if anhydrous residue reabsorbs water before weighing, apparent loss becomes too small. Heating to constant mass and cooling in a dry environment can reduce these errors. Excess heat can decompose some salts, causing extra mass loss that is not water. A reliable formula assignment requires the residue to be characterized and the heating pathway understood.

The percentages are mass percentages, not the fraction of formula units that are water. CuSO₄·5H₂O contains five water units per one salt unit, but water is not 5/6 of the hydrate's mass; the salt unit is much heavier than one water unit. Count ratios become mass fractions only after multiplying by their separate molar masses.

Step-by-step reasoning

1. Calculate molar mass of the anhydrous salt and water contribution from the proposed x. 2. Divide water contribution by full hydrate molar mass for theoretical water percentage. 3. From heating data, subtract residue mass from initial mass and divide by initial mass. 4. Compare the two percentages at suitable precision and inspect residue identity. 5. If inferring x, use mole ratio or the equivalent fraction equation and check integer closeness.

Visual explanation

Draw a 249.69 g mol⁻¹ stacked bar for CuSO₄·5H₂O, with 159.61 salt and 90.08 water. Mark the water segment as 36.1%. Under it, draw a 2.50 g measured bar splitting into 1.60 g residue and 0.90 g loss, or 36.0%, showing the formula–measurement comparison.

Real-world analogy

A packed product has a heavy core and several light removable inserts. Five inserts out of six counted pieces need not be five-sixths of package mass. Their mass fraction requires insert mass and core mass, just as hydrate water percentage requires both molar-mass contributions.

Real-world example

Washing soda is sodium carbonate decahydrate, Na₂CO₃·10H₂O. The ten waters contribute about 180.16 g mol⁻¹, and Na₂CO₃ contributes about 105.99 g mol⁻¹, giving roughly 63.0% formula water by mass. That high percentage means drying or storage changes can strongly affect the weighed sample's composition.

Why?

Why compare measured mass loss with formula water fraction? If heating removes only the water represented after the dot and leaves the anhydrous salt, the measured lost fraction should equal the formula's predicted water mass share. Agreement is evidence for the composition under those assumptions.

Common misconception

“Five waters per salt unit means 83.3% water by mass.” Five-sixths is a count of formula components. CuSO₄ is much heavier than one H₂O, so the actual water mass share of CuSO₄·5H₂O is about 36.1%.

Worked example

A proposed MgSO₄·7H₂O hydrate uses M(MgSO₄) = 120.37 and M(H₂O) = 18.016 g mol⁻¹. Seven waters contribute 126.112 g mol⁻¹, so full hydrate mass is 246.482 g mol⁻¹. Theoretical water percentage is 126.112/246.482 × 100 = 51.2% to three significant figures. Suppose a 4.00 g sample leaves 1.96 g identified dry MgSO₄. Loss is 2.04 g, or 51.0%, close to the predicted 51.2%. Water moles are 2.04/18.016 = 0.1132, and salt moles are 1.96/120.37 = 0.01628, giving x ≈ 6.96. Both fraction and mole-ratio checks support x = 7 within the measurements' precision. This conclusion still assumes no salt decomposition or external water.

Quick check

1. What mass fraction of a hydrate is water if a 2.50 g sample loses 0.90 g on valid dehydration? Answer: The apparent water fraction is 0.90 divided by 2.50, or 36.0% of the starting mass.

Exam focus

Use full hydrate formula mass as the percentage denominator. Separate water count from water mass share, and compare measured loss only after stating that the residue is anhydrous salt and loss is water. Reverse-check x with a mole ratio if a formula is inferred.

Advanced insight

If a hydrate can lose water in stages, mass-versus-temperature data may reveal plateaus associated with intermediate compositions. A single before-and-after percentage can match more than one physical history, so residue identification and controlled temperature steps improve confidence in the assigned hydrate.

Summary

Theoretical hydrate water percentage is water's molar-mass contribution divided by full hydrate molar mass. A measured heating-loss percentage can support the formula when water alone is lost and the residue is known. Mole-ratio and mass-fraction calculations should agree, while external moisture or decomposition can mislead both.

Practice questions

1. What water mass contribution occurs in one mole of CuSO₄·5H₂O? Answer: About 90.08 g from five moles of formula water. 2. What is its approximate total molar mass using CuSO₄ = 159.61 g mol⁻¹? Answer: About 249.69 g mol⁻¹. 3. What is its approximate water mass percentage? Answer: 36.1% by mass. 4. What loss percentage corresponds to 4.00 g hydrate leaving 1.96 g residue? Answer: 51.0% of the starting sample mass. 5. Why is 5/6 not the water mass fraction in CuSO₄·5H₂O? Answer: Formula-unit counts ignore the very different molar masses of salt and water.