AQA A-Level Chemistry Paper 3, 2017: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the mass in mg of carbon formed when 3.0 × 10⁻³ mol of propene undergoes incomplete combustion according to the given equation.

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Question

Multiple choice question 11 asking: 'What is the mass, in mg, of carbon formed when 3.0 × 10⁻³ mol of propene undergoes incomplete combustion? 2C₃H₆ + 3O₂ → 6C + 6H₂O'. Four options are given: A (9.0 × 10⁻³), B (3.6 × 10⁻²), C (1.08 × 10²), and D (2.16 × 10²).
Question text

What is the mass, in mg, of carbon formed when 3.0 × 10−3 mol of propene undergoes

incomplete combustion?

2C3H6 + 3O2 → 6C + 6H2O

[1 mark]

A 9.0 × 10−3

B 3.6 × 10−2

C 1.08 × 102

D 2.16 × 102

Mark scheme

Show the mark scheme Mark scheme table row showing question number 11 with the correct answer as C.

11 C

How to answer it

Mass of Product in Incomplete Combustion

📋 WHAT THIS QUESTION TESTS

Core quantitative chemistry concepts tested in Paper 1 and Paper 2 multiple choice:

  • Stoichiometry & Molar Ratios: Deducing the mole ratio between a reactant and a product from a balanced chemical equation.
  • Moles to Mass Conversion: Applying the fundamental formula mass = moles × Mr .
  • Unit Conversions: Converting mass from grams ( g ) to milligrams ( mg ).
  • Standard Form: Handling numbers written in scientific notation correctly.
Question 11 (1 Mark)

Incomplete Combustion of Propene

Reaction Equation: 2C₃H₆ + 3O₂ → 6C + 6H₂O

✅ Correct Answer

Option C: 1.08 × 10²

Mark Scheme: 1 mark for identifying option C.

💡 Key Knowledge

  • Stoichiometric Ratio: The equation shows 2C₃H₆ produces 6C . Therefore, the ratio of C₃H₆ : C is 2 : 6 , which simplifies to 1 : 3 .
  • Relative Atomic Mass (Ar): For carbon, Ar(C) = 12.0 . Do not use 12.0 × 6 here, as the stoichiometric multiplier 6 is already accounted for in the mole ratio.
  • Unit Factor: 1 g = 1000 mg = 10³ mg .

📐 Step-by-Step Calculation

  1. Determine moles of carbon (C) formed:
    Moles of C = Moles of C₃H₆ × (6 / 2)
    Moles of C = (3.0 × 10⁻³ mol) × 3 = 9.0 × 10⁻³ mol
  2. Calculate mass of carbon in grams (g):
    Mass (g) = Moles × Ar
    Mass of C = (9.0 × 10⁻³ mol) × 12.0 g mol⁻¹ = 0.108 g (or 1.08 × 10⁻¹ g )
  3. Convert mass to milligrams (mg):
    Mass (mg) = Mass (g) × 1000
    Mass of C = 0.108 g × 1000 mg g⁻¹ = 108 mg = 1.08 × 10² mg

❌ Distractor Breakdown & Traps

  • Option A (9.0 × 10⁻³): This is the number of moles of carbon, not the mass! Students who rushed picked this without multiplying by Ar.
  • Option B (3.6 × 10⁻²): Omits the mole ratio completely: (3.0 × 10⁻³ mol) × 12.0 = 0.036 g = 3.6 × 10⁻² g (and left in grams).
  • Option D (2.16 × 10²): Multiplied moles of C₃H₆ by 6 instead of using the ratio 6/2 = 3: (3.0 × 10⁻³ × 6) × 12.0 × 1000 = 216 mg.

🧠 Examiner Insight & Technique

  • Underline Units First: Always highlight the target unit ( mg , not g ). Multiple choice questions intentionally place the un-converted gram answer among the options.
  • Watch Reactant Coefficients: Do not assume a 1:n ratio. Look at the balancing number in front of the reactant ( 2 C₃H₆).
  • Quick Standard Form Check: 0.108 g × 10³ = 108 mg. In standard form, move the decimal point two places left: 108 = 1.08 × 10².

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.