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1.5.6Principles of Chemistry

Experiment- Finding Formulae of Compounds — Edexcel International GCSE Science Double Award 4SD0 - Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 28 of 50

📚 Edexcel International GCSE Science Double Award 4SD0 - Chemistry Hub / Topical Revision
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Part 1 of 5Practical Method

1.5.6 Experiment- Finding Formulae of Compounds

Heating a hydrated salt drives off its water of crystallisation, and the mass lost gives the amount of water.

1
Weigh a clean, dry evaporating basin, add a known mass of hydrated copper(II) sulfate crystals, and weigh again.
2
Heat gently while stirring until the blue crystals turn completely white, which shows that the water of crystallisation has left.
3
Avoid overheating, because strong heating can decompose the anhydrous salt and cause a mass loss that is not due to water.
4
Allow the basin to cool, reweigh it, then heat again and repeat until the mass stays constant.

Part 2 of 5Key Measurements

Mass of hydrated salt
  • The starting mass of the blue crystals, measured accurately before any heating takes place.
Mass of anhydrous salt
  • The mass of white powder left at constant mass, which is the salt without its water.
Mass of water
  • Found by subtraction, hydrated salt mass minus anhydrous salt mass, which gives the mass of water of crystallisation.
Water of crystallisation
  • Water molecules held within the crystal structure, written after a dot followed by the number of molecules.

Part 3 of 5Calculating x

Calculating x
Moles of salt
  • Divide the mass of anhydrous salt by its molar mass, such as 159.5 g/mol for copper(II) sulfate.
Moles of water
  • Divide the mass of water by 18 g/mol, since each water molecule has a Mr of 18.
Simplest ratio
  • Divide both mole values by the smaller value, which is usually the salt, to find the number of water molecules.
Rounding
  • The ratio is rarely a whole number exactly, so round to the nearest whole number when the value is close to one.

Part 4 of 5Worked Example

Question

  • A sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, has a mass of 5.00 g.
  • After heating to constant mass, the anhydrous salt has a mass of 3.20 g.
  • Find x.
  • Mr: CuSO₄ = 159.5, H₂O = 18.

  1. Mass of water = 5.00 − 3.20 = 1.80 g.
  2. Moles of CuSO₄ = 3.20 ÷ 159.5 = 0.0201 mol; moles of H₂O = 1.80 ÷ 18 = 0.100 mol.
  3. Ratio0.100 ÷ 0.0201 ≈ 4.98, which rounds to 5.

Answer The formula is CuSO₄·5H₂O, so x = 5.

Part 5 of 5Exam Checklist & Tips

1.5.6 – Key Points for the Exam

  • Hydrated salt A crystalline salt that contains water molecules as part of its structure.
  • Anhydrous salt A salt with no water of crystallisation.
  • Water of crystallisation Water molecules that are fixed within the crystal structure of a hydrated salt.
  • Constant mass The state reached when repeated heating and weighing gives the same mass.
  • Empirical formula The simplest whole number ratio of the atoms in a compound.

Common Mistakes

  • Using the mass of the hydrated salt instead of the anhydrous salt when calculating the moles of salt.
  • Stopping before constant mass is reached, which leaves water in the sample and gives a value of x that is too small.
  • Writing the water part without the dot, which shows that the water is water of crystallisation.
  • Leaving a non-integer ratio in the answer rather than rounding to the nearest whole number.

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