Edexcel A-Level Chemistry AS Paper 2, June 2017: Question 3
8 marks · Medium difficulty · Calculations
Calculate the empirical and molecular formula of a hydrocarbon from combustion data, determine the enthalpy change of combustion using calorimetry data, and explain the choice of a copper beaker.
Practise this questionQuestion
Question text
3 (a) In an experiment, 1.000 g of a hydrocarbon, A, was burned completely in oxygen
to produce 3.143 g of carbon dioxide and 1.284 g of water.
In a different experiment, the molar mass of the hydrocarbon, A, was found to be
84.0 g mol–1.
Calculate the empirical formula and the molecular formula of the hydrocarbon, A.
(4)
(b) A spirit burner was filled with the liquid hydrocarbon, A. The burner was weighed,
lit and then used to raise the temperature of a quantity of water in a beaker, as
shown in the diagram. The burner was then reweighed.
thermometer
beaker
water
spirit burner
hydrocarbon A
Results
Mass of spirit burner + hydrocarbon A before use 112.990 g
Mass of spirit burner + hydrocarbon A after use 112.732 g
Volume of water in the beaker 250 cm3
Temperature of water before heating 21.3°C
Temperature of water after heating 29.5°C
Other data
Density of water 1.00 g cm–3
Specific heat capacity of water 4.18 J g–1 °C–1
Molar mass of hydrocarbon A 84.0 g mol–1
(i) Use these results to calculate the enthalpy change of combustion of
hydrocarbon A in kJ mol–1.
Give your answer to an appropriate number of significant figures and include a sign.
(3)
*P49857A0324*
*P49857A0424*
(ii) The beaker used in this experiment was made of copper rather than glass.
Give a reason for this.
(1)
(Total for Question 3 = 8 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional Guidance Mark
Number
3(a) example of calculation (4)
moles of CO2 /moles of C (1) moles of CO2 = 3.143/44 (= 0.07143/0.071)
= moles of C
moles of H (1) moles of H2O = 1.284/18 (= 0.07133)
moles of H = 2 x moles of H2O = 0.1427
empirical formula (1) C:H = 0.07143:0.1427 = 1:2
hence C1H2 or CH2
allow TE from first and/or second mark
point(s)
Allow any workable calculation
Ignore SF in intermediate stages of calculation
Award 3 marks for correct C:H ratio, with or
without working.
calculates molecular formula C6H12 (1) 84/14 = 6
6 x CH2 = C6H12
Mark independently of M1, M2, M3
Question Acceptable Answer Additional Guidance Mark
Number
3(b)(i) example of calculation (3)
use of Q= m c T
Q = 250 x 4.18 x 8.2
calculation of Q (1) = 8569 (J) / 8.569 kJ
ignore any sign at this stage
mass of hydrocarbon burnt and value of cH (1) = 112.990 – 112.732
= 0.258 g
cH = (-) 8569 x 84/0.258
= (-) 2789907 (J mol-1) /(-)
2789.907 (kJ mol-1)
TE on incorrect value from M1
sign and significant figures (1)
= -2790/-2800 (kJ mol-1)
allow -2790000/-2800000 J mol-1
final answer to 2 or 3 sig figs only
Do not award M3 for incorrect
method used in M2
correct final answer without working
scores 3
Question Acceptable Answer Additional Guidance Mark
Number
3(b)(ii) an answer that makes reference to the following point: Allow copper is a good conductor (of (1)
improved/better (thermal/heat) conduction heat)
Allow reverse argument in terms of
(thermal) insulators
Ignore references to heat capacity/
heat lost to surroundings/ heat
absorbed by container.
Ignore any mention of glass
breakage
(Total for Question 3 = 8 marks)
How to answer it
Formulae, Equations and Amounts / Energetics Study Guide
What this question tests
This question assesses core quantitative chemistry and energetics skills: determining empirical and molecular formulae from combustion mass data, calculating enthalpy changes of combustion using experimental calorimetry data (Q = mcΔT), applying correct sign conventions and significant figures, and evaluating basic calorimeter design choices.
Empirical and Molecular Formula Calculation
Calculate the empirical formula and the molecular formula of hydrocarbon A. (4 marks)
📐 Step-by-Step Calculation
- Find moles of CO₂ and carbon:
Moles of CO₂ = 3.143 g / 44.0 g mol⁻¹ = 0.07143 mol.
Since each CO₂ contains 1 carbon, moles of C = 0.07143 mol. - Find moles of H₂O and hydrogen:
Moles of H₂O = 1.284 g / 18.0 g mol⁻¹ = 0.07133 mol.
Since each H₂O contains 2 hydrogens, moles of H = 0.07133 × 2 = 0.1427 mol. - Find empirical formula ratio:
C : H = 0.07143 : 0.1427 = 1 : 2.00
Empirical formula = CH₂ . - Find molecular formula:
Mass of empirical unit (CH₂) = 12.0 + (2 × 1.0) = 14.0 g mol⁻¹.
Ratio = Molar mass / Empirical mass = 84.0 / 14.0 = 6.
Molecular formula = 6 × (CH₂) = C₆H₁₂ .
✅ Final Answers & Mark Scheme Breakdown
- Mark 1: Correct calculation of moles of CO₂ (or moles of C).
- Mark 2: Correct calculation of moles of H (accounting for the 2:1 ratio in water).
- Mark 3: Correct empirical formula ( CH₂ or C₁H₂ ).
- Mark 4: Correct molecular formula ( C₆H₁₂ ).
❌ Common Errors
- Forgetting to multiply the moles of water by 2 to get the moles of hydrogen atoms.
- Using the mass of oxygen consumed instead of determining elements from products.
🧠 Exam Technique
- Keep unrounded numbers in your calculator during intermediate steps to prevent rounding errors.
- Always clearly state your mole ratios before converting them to whole numbers.
Enthalpy Change of Combustion
Use these results to calculate the enthalpy change of combustion of hydrocarbon A in kJ mol⁻¹. Give your answer to an appropriate number of significant figures and include a sign. (3 marks)
📐 Step-by-Step Calculation
- Calculate heat energy transferred to water (Q):
Mass of water (m) = 250 cm³ × 1.00 g cm⁻³ = 250 g.
Temperature change (ΔT) = 29.5 °C - 21.3 °C = 8.2 °C.
Q = m × c × ΔT = 250 × 4.18 × 8.2 = 8569 J (= 8.569 kJ). - Calculate mass of hydrocarbon burned:
Mass = 112.990 g - 112.732 g = 0.258 g . - Calculate moles burned and ΔH_c:
Moles = 0.258 g / 84.0 g mol⁻¹ = 0.003071 mol.
Enthalpy change per mole = - (Energy / Moles) = - (8.569 kJ / 0.003071 mol) = -2789.9 kJ mol⁻¹ . - Apply sign and significant figures:
Since combustion is exothermic, sign must be negative ( - ).
The temperature and masses are given to 2 or 3 sig figs, so round final answer to 2 or 3 sig figs: -2790 kJ mol⁻¹ or -2800 kJ mol⁻¹ .
✅ Mark Scheme Breakdown
- Mark 1: Correct calculation of Q (8569 J or 8.569 kJ).
- Mark 2: Correct calculation of moles burned and scaling up to find enthalpy change per mole.
- Mark 3: Negative sign included AND appropriate number of significant figures (2 or 3 sf).
❌ Common Errors
- Failing to include the minus sign for the exothermic enthalpy change.
- Giving too many significant figures (e.g., writing 4 or 5 sig figs when data is given to 2 or 3).
- Using the molar mass incorrectly when finding moles of fuel burned.
💡 Key Knowledge
- Definition: Enthalpy change of combustion is the enthalpy change when 1 mole of a substance is burned completely in oxygen under standard conditions.
- Always check whether your final unit is in J mol⁻¹ or kJ mol⁻¹.
Calorimeter Material Evaluation
The beaker used in this experiment was made of copper rather than glass. Give a reason for this. (1 mark)
✅ Acceptable Answer
Copper is an improved/better thermal conductor (or a good conductor of heat).
🧠 Exam Technique & Context
In calorimetry questions, equipment choices are designed to either minimize heat loss to surroundings or maximize heat transfer into the system being measured. Copper transfers thermal energy much more efficiently from the flame to the water than glass would.
Topics
Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics 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 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.