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Gravimetric analysis: learning from mass changes

Key idea: Heat, cool and weigh a hydrated salt repeatedly to constant mass. If only water is lost, convert water loss and remaining salt mass to moles to find the water-to-salt ratio.

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  • GCE A-Level H2 Chemistry 2027
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A balance measures mass, not the amount of a particular substance. Gravimetric analysis uses a known chemical change to connect a measured mass to the amount of substance. The crucial question is: what exactly has been gained or lost?

Here you will use heating to remove water from a hydrated salt. You need the mass-to-mole relationship, n = m/M, with molar mass M in g mol⁻¹, and a balanced description of the change:

MgSO₄ · xH₂O(s) → MgSO₄(s) + xH₂O(g)

The formula calculation assumes that only water leaves and that the anhydrous salt remains stable. Another volatile substance, lost solid or decomposition would break that model.

Check your understanding 1 Predict a loss of solid

Suppose dry salt spits out during heating, but all lost mass is counted as water. What happens to the calculated water-to-salt ratio?

Calculated ratio
Show answer

Both effects raise the ratio: the apparent water amount grows, while the remaining salt amount falls. Prevent solid loss with the method’s controlled heating and lid position.

Why heat, cool and weigh again?

Record the mass of a clean, dry crucible and lid before adding the sample. Use the same crucible and lid for every later weighing, so their mass can be subtracted consistently.

Under the supervised method, heat the sample, allow it to cool, and weigh it. Then repeat the heating–cooling–weighing cycle. Constant mass means that successive cooled masses agree within the tolerance specified for the balance and method. It suggests that further heating is no longer removing a measurable amount of material. A steady reading alone cannot tell you whether the sample is pure or whether all the water has gone.

  • A falling mass suggests that more volatile material is still leaving.
  • Weighing while hot can give unreliable balance readings. Cool as directed; a desiccator, where specified, limits uptake of atmospheric moisture.
  • Strong heating or spitting can lose solid as well as water. Heat at the rate specified and use the lid as directed, allowing vapour to escape.
  • A dehydrated salt may absorb moisture again. Keep the cooling and weighing conditions consistent.
Heating requires supervised laboratory practice

Hot crucibles can look cold. Use suitable tongs and a heat-resistant surface, and follow the instructed heating and cooling method. The salt and any gases released determine the additional precautions. Do not improvise a heating procedure from the calculation below.

Worked example: find the water-to-salt ratio

Worked example 1

Determine the formula of hydrated magnesium sulfate

Problem

Use these illustrative data to find x in MgSO₄ · xH₂O. Assume that only water is lost. Use molar masses 120.4 g mol⁻¹ for MgSO₄ and 18.0 g mol⁻¹ for water.

WeighingMass / g
Empty crucible and lid25.00
Crucible, lid and hydrated salt27.46
After first heating and cooling26.21
After second heating and cooling26.20
After third heating and cooling26.20
Show full solution

  1. Identify the final solid

    Method

    Use the final constant mass and subtract the container.

    Reason

    The balance reading includes both the salt and the crucible with its lid.

    Working

    Answer

    m(MgSO₄) = 26.20-25.00 = 1.20 g

  2. Identify the lost water

    Method

    Subtract the final mass from the mass before heating.

    Reason

    The container is unchanged, so this difference is the mass lost from the sample.

    Working

    Answer

    m(H₂O) = 27.46-26.20 = 1.26 g

  3. Compare amounts

    Method

    Convert each mass to moles before forming the ratio.

    Reason

    x counts water molecules per formula unit, so it is an amount ratio, not a mass ratio.

    Working

    Answer

    n(H₂O) = 1.26/18.0 = 0.0700 mol
    n(MgSO₄) = 1.20/120.4 = 0.00997 mol

    The water-to-salt amount ratio is:

    x = 0.0700/0.00997 ≈ 7.02

    So the formula is MgSO₄.7H₂O.

Check your understanding 2 Check the mass-to-mole route

Use the magnesium sulfate data above. Find the final salt amount and the water-to-salt amount ratio before rounding to an integer.

Show the steps Optional working
Unit: g
Unit: mol
Show the steps Optional working
Unit: mol
Unit: mol
Show answer

The salt amount is 1.20/120.4 = 0.00997 mol. Water amount is 1.26/18.0 = 0.0700 mol. Their ratio is 7.02, close to seven; dividing masses would give 1.05, which is not a molecule ratio.

Predict the effect of a technique error

If solid is lost by spitting, the mass loss is wrongly counted as water. The calculated water amount is too large and the calculated amount of remaining salt is too small. Both effects make the calculated x too large.

If water remains after insufficient heating, the final solid is too heavy. Water loss is underestimated and the apparent salt amount is overestimated, so x is too small. Repeated heating to constant mass addresses this particular weakness; it cannot correct solid already lost.

Check your understanding 3 Apply the method to a different hydrate

A sodium carbonate hydrate is heated without losing solid. The empty crucible and lid weigh 30.000 g; with hydrate they weigh 32.860 g; after heating to constant mass they weigh 31.060 g. Find x in Na₂CO₃ · xH₂O using molar masses 106.0 g mol⁻¹ for Na₂CO₃ and 18.0 g mol⁻¹ for water. Explain moisture uptake in the following question.

Show the steps Optional working
Unit: g
Unit: g
Unit: mol
Unit: mol
Show answer

The dry salt has mass 1.060 g and amount 0.01000 mol. Water loss is 1.800 g, or 0.1000 mol. Hence x = 10, giving Na₂CO₃.10H₂O.

Check your understanding 4 Show your method and explain moisture uptake

Show how the sodium carbonate hydrate masses above give its formula. Then explain why moisture absorbed during cooling would make the calculated number of waters too small.

Show answer

The dry salt has mass 1.060 g and amount 0.01000 mol. Water loss is 1.800 g, or 0.1000 mol. Hence x = 10, giving Na₂CO₃.10H₂O. Absorbed moisture raises the final mass: too little mass loss is assigned to water and too much final mass is assigned to salt. The resulting water-to-salt amount ratio is too small.

What to retain

Track the container separately, find a consistent final cooled mass, and convert the chemically identified masses to moles. A mass difference is meaningful only while the assumed chemical change remains valid.

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