Edexcel A-Level Chemistry AS Paper 2, November 2021: Question 4

10 marks · Medium difficulty · Calculations

Write a reaction equation for the thermal decomposition of calcium sulfate dihydrate, identify reaction terms, calculate water volume needed for hydration, determine enthalpy change from calorimetric data, and calculate percentage uncertainties.

Practise this question

Question

Exam question 4 about calcium sulfate dihydrate and hemihydrate. Part (a) asks for an equation with state symbols. Part (b) is a multiple-choice question with a table giving four options for enthalpy change and process type. Part (c) describes a calorimetry experiment involving calcium sulfate hemihydrate and water, containing three sub-questions: (i) calculating minimum volume of water needed, (ii) calculating enthalpy change in kJ mol-1, and (iii) deducing which measurement has the greatest uncertainty by calculating percentage uncertainties.
Question text

4 When solid calcium sulfate dihydrate, CaSO4·2H2O, is heated in a crucible, it forms

solid calcium sulfate hemihydrate, CaSO4·½H2O.

(a) Write an equation, including state symbols, for this reaction.

(1)

(b) Which two terms could be used to describe this reaction?

(1)

Enthalpy change Type of process

A endothermic hydration

B exothermic hydration

C exothermic dehydration

D endothermic dehydration

(c) When water is added to calcium sulfate hemihydrate, there is a rise

in temperature.

A student decided to investigate this reaction using the following procedure:

Step 1 10 cm3 of distilled water is measured using a measuring cylinder having

an uncertainty of ±0.5 cm3, and is placed in an insulated cup with a lid.

Step 2 A thermometer with an uncertainty of ±0.5°C is placed in the water.

Step 3 Exactly 10.00g of calcium sulfate hemihydrate is weighed out using a

balance with an uncertainty of ±0.005g.

Step 4 The weighed quantity of calcium sulfate hemihydrate is added to the

water in the insulated cup.

Step 5 The mixture in the insulated cup is stirred until no further temperature

change is observed.

Results

Temperature of the water before adding the solid = 23.5°C

Maximum temperature of the mixture after adding the solid = 26.3°C

Other data

Molar mass of calcium sulfate hemihydrate, CaSO ·½H O = 145.2 g mol–1

Density of water = 1.00 g cm–3

(i) Calculate the minimum volume of water needed to convert 10.00g of

CaSO4·½H2O into CaSO4·2H2O.

8 (2)

*P67084A0828*

(ii) Calculate the enthalpy change, in kJ mol–1, for this reaction.

Include a sign in your answer and give your answer to an appropriate number

of significant figures.

Assume that the liquid has a mass of 10.00g and a specific heat capacity

of 4.18 J g–1 °C–1.

(4)

(iii) Deduce which measurement has the greatest uncertainty in this experiment.

Justify your answer by calculating the percentage uncertainty of this piece

of apparatus.

(2)

(Total for Question 4 = 10 marks)

Mark scheme

Show the mark scheme Mark scheme for question 4 detailing the correct equation, multiple choice answer D, step-by-step calculations for moles, enthalpy change using q=mcT, and percentage uncertainty calculations.

Question Answer Additional Guidance Mark

Number

4(a) Example of equation (1)

• correct equation 2CaSO4·2H2O(s) 2CaSO4·½H2O(s) + 3H2O(l)

OR

CaSO4·2H2O(s) CaSO4·½H2O(s) + 1½H2O(l)

Allow multiples

Allow H2O(l) or H2O(g)

Question Mark

Answer

Number

4(b) The only correct answer is D (endothermic, dehydration) (1)

A is not correct because hydration involves adding water

B is not correct because a reaction that requires heat is unlikely to be exothermic and hydration involves adding water

C is not correct because a reaction that requires heat is unlikely to be exothermic

Question Answer Additional Guidance

Number Mark

4(c)(i) Example of calculation (2)

• calculation of moles of CaSO4·½H2O (1) 10.00 g CaSO4·½H2O = 10.00 ÷145.2 mol = 0.06887 mol

Allow 0.069

• calculation of volume (or mass) of water required (moles of water required = 0.06887 x 1.5 = 0.1033 mol)

(1)

volume of water required = 0.1033 x 18 ÷ 1.00 = 1.86 cm3

Allow 1.86 g

Ignore SF except 1 SF

Correct answer with no working scores (2)

Allow calculation using multiples of these moles (still gets

same final answer scores 2)

Allow alternative correct calculations: e.g. comparison of

moles of CaSO4·½H2O with moles of water in 10.00 g.

Number Mark

4(c)(ii) Example of calculation (4)

• calculation of ∆T (1) ∆T = 2.8 oC

• use of mc∆T to find Q (1) m = 10.00 g, c = 4.18 J g-1 oC-1

Q = mc∆T = 117.04 J / 0.11704 kJ

Allow M1 and M2 if figure of 117.04 J is seen

Ignore units unless converted to kJ

• calculation of ∆ H (1) 117.04 ÷ 0.06887 = −1699.4 (J mol-1)

r

−1.70 / −1.7 (kJ mol-1)

• correct final answer, with sign and 2 or 3 SF (1)

Correct answer with no working scores (4)

Allow TE throughout and from 4ci (for moles CaSO4·½H2O)

Number

4(c)(iii) Example of calculation (2)

• selection of thermometer (1)

• calculation of percentage uncertainty (1) 2 x 0.5 x 100 = 35.7 / 36 / 40 (%)

2.8

Allow selection of measuring cylinder and percentage

uncertainty is 5%, scores (1) mark

Do not award selection of balance

Ignore SF

(Total for Question 4 = 10 marks)

How to answer it

Hydrated Salts & Calorimetry Study Guide

What this question tests

This exam question assesses core AS Chemistry competencies across physical and inorganic chemistry: writing balanced chemical equations with state symbols, understanding thermal decomposition terminology, performing stoichiometric calculations involving water of crystallization, carrying out calorimetry calculations (q = mcΔT and ΔH), and evaluating experimental uncertainty.

Question Part (a)

Writing Thermal Decomposition Equations

Dehydration of Calcium Sulfate Dihydrate

✅ Correct Answer

CaSO₄·2H₂O(s) → CaSO₄·½H₂O(s) + 1½H₂O(l)

Alternatively, using integer coefficients:
2CaSO₄·2H₂O(s) → 2CaSO₄·½H₂O(s) + 3H₂O(l)

💡 Key Knowledge

  • State symbols (s) and (l) are mandatory for the mark.
  • Fractions are fully accepted in Edexcel chemical equations as long as stoichiometry is correct.
Mark allocation: 1 mark for correct formulas, balancing, and state symbols.
Question Part (b)

Reaction Terminology

Identifying Enthalpy and Process Types

✅ Correct Answer

D (endothermic, dehydration)

❌ Common Errors & Examiner Commentary

Students frequently confuse hydration with dehydration. Since heat must be supplied in a crucible to drive off water, the process is dehydration and requires heat input, making it endothermic.

Mark allocation: 1 mark for selecting option D.
Question Part (c)(i)

Stoichiometry & Water of Crystallization

Calculating Minimum Volume of Water

📐 Step-by-Step Calculation

  1. Find moles of solid:
    Moles of CaSO₄·½H₂O = 10.00 g ÷ 145.2 g mol⁻¹ = 0.06887 mol
  2. Determine moles of water needed:
    Ratio is 1 : 1.5. Moles of H₂O = 0.06887 × 1.5 = 0.1033 mol
  3. Calculate volume of water:
    Mass of H₂O = 0.1033 mol × 18.0 g mol⁻¹ = 1.86 g
    Since density = 1.00 g cm⁻³ , Volume = 1.86 cm³

🧠 Exam Technique

Always keep full calculator values through intermediate steps to avoid rounding errors. Final answers should typically be given to 2 or 3 significant figures.

Mark allocation: 1 mark for calculating moles of solid; 1 mark for final volume/mass of water.
Question Part (c)(ii)

Calorimetry and Enthalpy Change

Calculating Enthalpy of Reaction (ΔH)

📐 Step-by-Step Calculation

  1. Calculate temperature change (ΔT):
    ΔT = 26.3 °C - 23.5 °C = 2.8 °C
  2. Calculate heat energy released (q):
    q = mcΔT = 10.00 g × 4.18 J g⁻¹ °C⁻¹ × 2.8 °C = 117.04 J ( 0.117 kJ )
  3. Calculate enthalpy change per mole (ΔH):
    ΔH = -q ÷ moles = -117.04 J ÷ 0.06887 mol = -1699.4 J mol⁻¹
  4. Apply units, sign, and significant figures:
    -1.70 kJ mol⁻¹ (or -1.7 kJ mol⁻¹ to 2 sf)

❌ Common Errors & Traps

  • Sign omission: Because temperature increases, the reaction is exothermic, so a negative sign ( - ) is mandatory.
  • Unit mismatch: Energy q is calculated in Joules, but ΔH must be converted to kJ mol⁻¹ and divided by 1000.
Mark allocation: 4 marks total (1 for ΔT, 1 for q = mcΔT, 1 for numerical ΔH value, 1 for correct negative sign and appropriate 2 or 3 SF).
Question Part (c)(iii)

Experimental Uncertainty Analysis

Percentage Uncertainty Calculation

📐 Step-by-Step Calculation

  1. Identify apparatus with greatest uncertainty:
    The thermometer has the greatest percentage uncertainty due to the small temperature rise measured.
  2. Calculate percentage uncertainty for the thermometer:
    Uncertainty involves two temperature readings (initial and final), so total uncertainty = ±0.5 × 2 = ±1.0 °C .
    Percentage uncertainty = (1.0 ÷ 2.8) × 100 = 35.7% (Accept 36% or 40%).

🧠 Exam Technique

When a temperature change ( ΔT ) is calculated from two separate readings on the same instrument (like a thermometer), remember to double the instrument uncertainty because an error can occur at both the start and end points!

Mark allocation: 1 mark for identifying the thermometer; 1 mark for correct calculation including doubling the uncertainty.

Topics

Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.