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1.2.6Physical Chemistry

Hydrated Salts — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 13 of 102

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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
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Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Water of Crystallisation

Hydrated
  • Hydrated A hydrated salt has water molecules as part of its crystal structure, as in CuSO₄·5H₂O.
Anhydrous
  • Anhydrous An anhydrous salt has had its water of crystallisation removed, as in CuSO₄.
Formula
  • Formula The dot formula shows the number of moles of water per mole of salt;
    • this number is called x in a general formula.
Colour change
  • Colour change Heating blue hydrated copper(II) sulfate gives white anhydrous copper(II) sulfate.

Part 2 of 6Experimental Method

Heating to constant mass drives off all the water.

1
Weigh an empty crucible, then add the hydrated salt and weigh again to find the mass of the sample.
2
Heat the crucible gently, and then more strongly, to drive off the water of crystallisation.
3
Allow the crucible to cool in a desiccator, and weigh again.
4
Reheat and reweigh until the mass no longer changes, which is heating to constant mass.
5
The mass lost is the mass of water, and the mass left is the anhydrous salt.

Part 3 of 6Calculating x

Calculating x
Moles of water
  • Moles of water Calculate moles of H₂O from the mass lost divided by 18.0.
Moles of salt
  • Moles of salt Calculate moles of anhydrous salt from the mass remaining divided by its Mr.
Ratio
  • Ratio Divide the moles of water by the moles of salt to find x.
Whole number
  • Whole number Round x to the nearest whole number, since there can only be a whole number of water molecules.

Part 4 of 6Sources of Error

Incomplete heating
  • Incomplete heating If not all the water is driven off, the mass lost is too small and x is too low.
Over-heating
  • Over-heating Strong heating can decompose the salt, which makes the mass lost too large and x too high.
Spitting
  • Spitting Material thrown out of the crucible during heating gives an inaccurate mass.
Re-absorption
  • Re-absorption A hot anhydrous salt may take in moisture from the air if it is cooled uncovered.

Part 5 of 6Worked Example

Question

  • 2.50 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated to constant mass and 1.60 g of anhydrous solid remains.
  • Find x (Mr: CuSO₄ = 159.6, H₂O = 18.0).

  1. Mass of water lost = 2.50 – 1.60 = 0.90 g, so n(H₂O) = 0.90 ÷ 18.0 = 0.0500 mol.
  2. n(CuSO₄) = 1.60 ÷ 159.6 = 0.01003 mol.
  3. x = 0.0500 ÷ 0.01003 = 4.99, which rounds to 5.

Answer x = 5, so the formula is CuSO₄·5H₂O

Part 6 of 6Exam Checklist & Tips

1.2.6 – Key Points for the Exam

  • Water of crystallisation Water molecules that are part of the crystal structure of a salt.
  • Anhydrous Containing no water of crystallisation.
  • Hydrated Containing water of crystallisation.
  • Constant mass The state reached when repeated heating and weighing give the same mass.

Common Mistakes

  • Dividing the mass of the hydrated salt by 18.0 instead of the mass of water lost.
  • Using the Mr of the hydrated salt when calculating moles of the anhydrous salt.
  • Forgetting to heat to constant mass, so the water is not all removed.
  • Giving x as a decimal, such as 4.99, instead of the nearest whole number.

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