OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 2

1 mark · Medium difficulty · Multiple Choice

Calculate the value of x in the formula of hydrated cobalt(II) chloride, CoCl2•xH2O, given the mass before and after heating to constant mass.

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Question

Question 2: 11.895 grams of hydrated cobalt(II) chloride, CoCl2·xH2O, is heated to remove the water of crystallisation. After heating to constant mass, the mass of the anhydrous salt was 6.495 grams. What is the value of x? Options are: A: 3, B: 5, C: 6, D: 7. Includes an answer box and is worth 1 mark.

Mark scheme

Show the mark scheme Mark scheme table showing question number 2, correct answer C, and 1 mark.

How to answer it

Water of Crystallisation in Cobalt(II) Chloride

📋 What This Question Tests

This multiple-choice question assesses your quantitative chemistry problem-solving from Module 2: Foundations in Chemistry (Amount of Substance):

  • Calculating reacting masses and lost mass from experimental data.
  • Determining relative formula masses (Mr) using the OCR Data Sheet.
  • Converting mass to amount in moles using n = m / M.
  • Finding empirical molar ratios to deduce the formula of a hydrated salt (value of x).

Question 2 — Multiple Choice (1 Mark)

Determining the value of x in CoCl₂•xH₂O

✅ Correct Answer

C — 6

The formula of the hydrated salt is CoCl₂•6H₂O (cobalt(II) chloride hexahydrate).

Mark scheme: C = [1 mark]

💡 Key Knowledge

  • Hydrated salt: Contains water molecules chemically integrated into the crystal lattice (water of crystallisation).
  • Anhydrous salt: Salt remaining after all water of crystallisation has been driven off.
  • Heating to constant mass: Ensures all water of crystallisation has evaporated and no further mass change occurs.
  • Molar masses:
    • M(H₂O) = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹
    • M(CoCl₂) = 58.9 + (2 × 35.5) = 129.9 g mol⁻¹

📐 Step-by-Step Calculation

  1. Find the mass of water lost:
    Mass of H₂O = (Mass of hydrated salt) − (Mass of anhydrous salt)
    Mass of H₂O = 11.895 g − 6.495 g = 5.400 g
  2. Calculate the amount (in moles) of anhydrous CoCl₂:
    n(CoCl₂) = m / M = 6.495 g / 129.9 g mol⁻¹ = 0.0500 mol
  3. Calculate the amount (in moles) of H₂O lost:
    n(H₂O) = m / M = 5.400 g / 18.0 g mol⁻¹ = 0.300 mol
  4. Find the simplest mole ratio (value of x):
    x = n(H₂O) / n(CoCl₂)
    x = 0.300 mol / 0.0500 mol = 6
    Therefore, the value of x = 6 (Option C).

🧠 Exam Technique & Speed Tips

  • Quick ratio shortcut: Recognise that 0.0500 mol fits cleanly into 0.300 mol exactly 6 times. Clean, integer ratios are standard in OCR questions.
  • Spot check using Molar Mass: If x = 6, total Mr of CoCl₂•6H₂O = 129.9 + (6 × 18.0) = 237.9.
    Hydrated to anhydrous mass ratio: 11.895 / 6.495 ≈ 1.831.
    Formula mass ratio: 237.9 / 129.9 ≈ 1.831. Perfectly consistent!

❌ Common Student Errors

  • Using wrong mass for CoCl₂: Dividing the total hydrated mass (11.895 g) by the Mr of CoCl₂ instead of the anhydrous mass (6.495 g).
  • Periodic Table misread: Using the wrong atomic number instead of relative atomic mass for cobalt (using 27 instead of 58.9).
  • Inverting the mole ratio: Dividing n(CoCl₂) by n(H₂O), giving 0.167 instead of 6.
  • Rounding too early: Rounding intermediate values of moles, which can cause 5.9 or 6.1 and lead candidates to second-guess their answer.

Topics

Module 2: Foundations in chemistry · Practical Activity Groups · 2.1 Atoms and reactions · PAG 1: Moles determination

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.