Edexcel A-Level Chemistry AS Paper 2, June 2023: Question 5

9 marks · Medium difficulty · Calculations

Calculate the enthalpy change of combustion per gram of alcohol Q using calorimetry data, explain the effect of incomplete combustion, evaluate limitations of the procedure, and identify specific mass spectrometry fragmentation peaks to distinguish between alcohol isomers.

Practise this question

Question

An exam question with multiple parts about determining the enthalpy of combustion of an unknown alcohol Q using calorimetry, followed by questions on incomplete combustion, procedural limitations, and mass spectrometry fragmentation. A diagram shows a spirit burner heating a beaker of water with a thermometer clamped above it, supported by a stand. A data table lists measurements including masses of the burner before and after combustion, mass of water, initial and final temperatures, and specific heat capacity of water.
Question text

5 A student found a bottle of a colourless liquid that was thought to be an alcohol.

However, the label was missing and the identity of the alcohol was unknown.

The student decided to attempt to identify this alcohol (Q) by measuring its

enthalpy change of combustion, and comparing the result with values in a data book.

The student used the equipment shown in the diagram to determine the

enthalpy change of combustion of alcohol Q.

thermometer

beaker

water

spirit burner

alcohol Q

Data

Mass of spirit burner + alcohol Q before combustion = 20.24g

Mass of spirit burner + alcohol Q after combustion = 19.48g

Mass of water in the beaker = 500g

Temperature of the water before the experiment = 17.8°C

Temperature of the water at the end of the experiment = 28.7°C

Specific heat capacity of water = 4.18 J g–1 °C–1

(a) (i) Calculate the enthalpy change, in kJ g–1, when 1.00 g of alcohol Q is burned.

Include a sign in your final answer.

(3)

*P71927A01828*

(ii) At the end of the experiment there was a black deposit of carbon on the

bottom of the beaker.

Explain the effect of formation of the carbon deposit on your answer to (a)(i).

(2)

(b) Give one theoretical and one practical reason why this procedure is insufficient to

identify the alcohol.

*P71927A01928* (2)

Theoretical reason

Practical reason

(c) The student concluded that alcohol Q was either propan‑1‑ol or propan‑2‑ol

because the mass spectrum had the molecular ion peak at m / z = 60.

Explain one peak that you would expect to be present in the mass spectrum of

propan‑1‑ol but not in the mass spectrum of propan‑2‑ol.

(2)

(Total for Question 5 = 9 marks)

Mark scheme

Show the mark scheme The mark scheme for the question providing acceptable answers and additional guidance for parts 5(a)(i), 5(a)(ii), 5(b), and 5(c), allocating a total of 9 marks across the sub-questions.

Question

Acceptable Answer Additional Guidance Mark

Number

5(a)(i) Example of calculation: (3)

• calculation of T T = 28.7 − 17.8 = 10.9 (K / oC)

and Accept T included in the mc T calculation

use of mc T (1) mc T = 500 × 4.18 × 10.9

= 22781 (J) / 22.781 (kJ)

Allow M1 if this number is seen

• calculation of mass of alcohol burnt 20.24 – 19.48 = 0.76 g

and 22781 ÷ 0.76 = 29975 (J g−1) / 29.975 (kJ g−1)

calculation of energy produced, g−1 (1) = 30000 (J g−1)

• final answer, in kJ g−1, including sign (1) −30 / −30.0 (kJ g−1)

Ignore SF

TE throughout

Correct final answer (inc. sign) with no working scores 3

marks

Question

Acceptable Answer Additional Guidance Mark

Number

5(a)(ii) An explanation that makes reference to the following points: (2)

• magnitude / size will be reduced / less negative Do not award change of sign of cH [alcohol]

(1)

Do not award becomes more positive

If a decrease in enthalpy is linked to less energy being

released then allow M1 but do not award just for

decrease in enthalpy

• because fewer bonds are made (in forming CO2) (1) Allow (carbon is the product of) incomplete combustion

Question

Acceptable Answer Additional Guidance Mark

Number

5(b) An answer that makes reference to the following points: (2)

Theoretical reason:

• molar mass is unknown the enthalpy change of combustion in kJ mol−1 cannot be

Or determined (for comparison with a data book listing cH

the data book values must be converted to kJ g−1 (1) in kJ mol−1)

Allow reference to unknown number of moles

Practical reason:

• experiment is (very) inaccurate (1) Accept this mark for any valid, specified source of heat loss in

the procedure: heat loss from walls of container

heat loss from surface of water

heat loss to container

heat loss to burner

heat loss from flame to air

alcohol loss by evaporation

Allow heat loss to the surroundings

Allow reference to incomplete combustion

Ignore scaffolding and award marks wherever the answer is

written

Question

Acceptable Answer Additional Guidance Mark

Number

5(c) An explanation that makes reference to the following points: (2)

Either

• presence of a peak at 29 (1) Do not award peak at 15 due to CH +

Do not award peak at 17 / 43 due to OH+ or C H +

+ + (1) Do not award peak at 59 due to C H O+

• which is due to C2H5 / CH3CH2 3 7

Or

• presence of a peak at 31 (1)

• which is due to CH OH+ (1)

(Total for Question 5 = 9 marks)

How to answer it

Determining an Unknown Alcohol via Combustion & Mass Spectrometry

What this question tests

This question assesses core AS practical and analytical chemistry skills: calculating enthalpy changes of combustion using experimental calorimetry data, understanding sources of experimental error and incomplete combustion, recognising limitations of standard enthalpy comparisons, and interpreting mass spectrometry fragmentation patterns to distinguish between structural isomers.

Question 5 (a)(i) — Enthalpy Calculation

Calculating Enthalpy Change per Gram

📐 Step-by-Step Calculation

  1. Find temperature change ( ΔT ):
    28.7 - 17.8 = 10.9 °C (or K)
  2. Calculate heat energy released ( q = mcΔT ):
     m = 500 g (mass of water)
    c = 4.18 J g⁻¹ °C⁻¹
    q = 500 × 4.18 × 10.9 = 22781 J = 22.781 kJ
  3. Find mass of alcohol burned:
    20.24 - 19.48 = 0.76 g
  4. Calculate enthalpy change per gram:
    22.781 kJ ÷ 0.76 g = 29.975 kJ g⁻¹
    Round/format with negative sign: -30 or -30.0 kJ g⁻¹

✅ Correct Answer & Mark Scheme

Final Answer: -30 or -30.0 kJ g⁻¹

Mark Breakdown (3 marks):
1. Correct calculation of ΔT and substitution into mcΔT .
2. Calculation of mass of alcohol burned and energy produced per gram.
3. Final numerical value with the correct negative sign ( - ).

❌ Common Errors & Traps

  • Missing the negative sign: Enthalpy of combustion is always exothermic, so a negative sign is mandatory.
  • Using burner mass instead of water mass: Using 0.76 g for m in mcΔT instead of 500 g of water.
  • Inverted mass subtraction: Calculating 19.48 - 20.24 leading to a negative mass.
Question 5 (a)(ii) — Practical Evaluation

Effect of Carbon Deposit (Incomplete Combustion)

✅ Correct Answer

Magnitude / Size: The calculated enthalpy change will be reduced (less negative).

Reason: Fewer bonds are made (in forming CO₂ / complete combustion products) because some carbon remains unoxidised as soot.

Mark Breakdown (2 marks):
1. Statement that the magnitude is reduced / less negative.
2. Explanation linking this to incomplete combustion / fewer bonds formed making CO₂.

💡 Key Knowledge

Enthalpy of combustion assumes complete combustion to carbon dioxide and water. When black soot (carbon) forms, less energy is released overall because carbon is not fully oxidised to CO₂ .

Question 5 (b) — Experimental Limitations

Theoretical and Practical Flaws

✅ Correct Answer

Theoretical Reason: The molar mass of alcohol Q is unknown, so the experimental value ( kJ g⁻¹ ) cannot be converted into kJ mol⁻¹ to compare directly with data books.

Practical Reason: The experiment suffers from severe heat loss to the surroundings / incomplete combustion / evaporation of alcohol.

Mark Breakdown (2 marks):
1. Theoretical mark: Unknown molar mass / unknown number of moles / data books use kJ mol⁻¹ .
2. Practical mark: Any valid source of heat loss (e.g. heat loss from container, water surface, flame to air) or incomplete combustion.

🧠 Exam Technique

Read the question carefully to ensure you address both aspects: one theoretical reason and one practical reason. Do not give two practical points like "heat loss" and "drafts".

Question 5 (c) — Mass Spectrometry

Distinguishing Propan-1-ol and Propan-2-ol

✅ Correct Answer

Either of the following valid fragment peaks:

  • Peak at m / z = 29, due to fragment C₂H₅⁺ (or CH₃CH₂⁺ ) present in propan-1-ol.
  • Peak at m / z = 31, due to fragment CH₂OH⁺ .
Mark Breakdown (2 marks):
1. Identifying the correct peak ( m / z = 29 or 31 ).
2. Identifying the correct corresponding ionic formula.

❌ Common Errors to Avoid

  • Do not credit common unhelpful fragments like m / z = 15 ( CH₃⁺ ) which can appear in both isomers due to terminal methyl groups.
  • Ensure charges are clearly indicated on formulae (e.g., C₂H₅⁺ not just C₂H₅ ).

Topics

Physical Chemistry · Organic Chemistry · Core Practicals · Topic 8: Energetics I · Topic 6: Organic Chemistry I · Topic 7: Modern Analytical Techniques I · Core Practical 8: Determine the enthalpy change of a reaction using Hess’s law

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.