OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 12
1 mark · Medium difficulty · Multiple Choice
Identify which experimental error caused the calculated enthalpy change of combustion of methanol to be more exothermic than the data book value.
Practise this questionQuestion
Question text
12 A student carried out an experiment to measure the enthalpy change of combustion of methanol.
The energy from the combustion of methanol was used to heat a beaker containing water.
The student’s calculated enthalpy change of combustion was more exothermic than the value in
data books.
Which error could have caused this difference?
A Some methanol had evaporated from the wick before the final weighing.
B In the calculation, the student used the molar mass of ethanol instead of methanol.
C There was incomplete combustion.
D The water boiled for 5 minutes before the final temperature was taken.
Your answer [1]
Mark scheme
Show the mark scheme
12 B 1
How to answer it
Evaluating Errors in Enthalpy of Combustion Experiments
What this question tests
This question assesses your understanding of calorimetry experiments used to find enthalpy changes of combustion (CH₃OH). Specifically, it tests how experimental errors, calculation mistakes, and heat losses mathematically impact the final calculated value of enthalpy change (delta H), forcing you to distinguish between experimental flaws and calculation-based errors.
Question Analysis & Solution
Question 12 (Multiple Choice)
✅ Correct Answer: Option B
Using the molar mass of ethanol ( 46.0 g mol⁻¹ ) instead of methanol ( 32.0 g mol⁻¹ ) results in a calculated enthalpy change that appears artificially more exothermic.
💡 Key Knowledge
- Enthalpy of combustion is calculated using: q = m c delta T
- Number of moles = mass / molar mass
- Enthalpy change per mole = -q / moles
- Methanol formula: CH₃OH (Mr = 32.0)
- Ethanol formula: C₂H₅OH (Mr = 46.0)
📐 Step-by-Step Mathematical Proof
- Step 1: Calculate energy released ( q ). Let's assume a fixed heat energy released of 1000 J .
- Step 2: Assume the mass of alcohol burned gives 0.100 mol of methanol.
- Step 3 (Correct calculation): -1000 J / 0.100 mol = -10,000 J mol⁻¹ ( -10.0 kJ mol⁻¹ ).
- Step 4 (Using ethanol's larger Mr): Using the larger molar mass means the calculated number of moles appears smaller for the same mass burned (e.g., 0.070 mol ).
- Step 5 (Impact): -1000 J / 0.070 mol = -14,285 J mol⁻¹ ( -14.3 kJ mol⁻¹ ) — a larger negative number, meaning more exothermic!
❌ Common Errors & Distractor Analysis
- Option A (Evaporation): If methanol evaporates from the wick, the final mass recorded is smaller than it should be. Mass burned = initial - final, so mass burned appears larger. Larger moles mean a smaller enthalpy change (less exothermic).
- Option C (Incomplete combustion): Releases less heat energy ( q is smaller), making the calculated enthalpy change less exothermic.
- Option D (Water boiling): Heat is lost to the surroundings after boiling starts, or temperature readings stop early, reducing the recorded delta T and making it less exothermic.
🧠 Exam Technique & Examiner Insight
Examiners note that students frequently confuse the direction of enthalpy changes. Remember: more exothermic means a larger negative value (further down the number line, e.g., -500 kJ mol⁻¹ is more exothermic than -200 kJ mol⁻¹). Always trace back how a larger or smaller denominator (moles) or numerator (energy, q) affects the final division.
Topics
Module 3: Periodic table and energy · Practical Activity Groups · 3.2 Physical chemistry · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.