AQA A-Level Chemistry Paper 2, 2024: Question 7

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Question Answers Additional Comments/Guidelines Mark

2.62 ×10−3 6.56 ×10−4 Alternative method

M1 6.56 × 10−4 = 4 or 2.62 × 10−3 = 0.25

M1

M2 Hence 4CO2 + 4H2O nH in L = 5.24 × 10–3

Hence mass H = 5.24 × 10–3 g

M3 So 4C and 8H in L

M2

M4 Hence 2O so C H O n C in L = 2.62 × 10–3

48 2

Hence mass C = 2.62 × 10–3 × 12

= 3.144 × 10–2 g

C3H4O3 scores 1 if no other mark scored

M3

Mass L = 6.56 × 10–4 × 88

= 0.057728 g

mass O = 0.057728 – ( 5.24 × 10–3 + 3.144 × 10–2) 4

= 0.021048 g (4 x AO2)

M4

nO = 0.021048 / 16

= 1.3155 × 10–3

EF C H O

2.62 × 10–3 5.24 × 10–3 1.3155 × 10–3

24 1

MF = (88/44) x C2H4O

= C4H8O2

C3H4O3 scores 1 if no other mark scored

How to answer it

Finding the Formula of L from Combustion Products

What this question tests

Turning moles of CO₂ and H₂O from complete combustion into moles of C and H in the original sample, scaling to atoms per molecule using the moles of the sample, and using M_r to deduce the number of oxygen atoms. Precision with ratios and neat working are key.

Given: 6.56×10⁻⁴ mol of L produces 2.62×10⁻³ mol CO₂ and 2.62×10⁻³ mol H₂O. L contains C, H, O only and M_r = 88.

📐 Worked solution

Step 1 — Moles of atoms in the sample

n(C) = n(CO₂) = 2.62×10⁻³ mol
n(H) = 2 × n(H₂O) = 2 × 2.62×10⁻³ = 5.24×10⁻³ mol
n(sample L) = 6.56×10⁻⁴ mol

Step 2 — Atoms per molecule (divide by moles of L)

C atoms per molecule: 2.62×10⁻³ ÷ 6.56×10⁻⁴ = 4.00 → 4 C
H atoms per molecule: 5.24×10⁻³ ÷ 6.56×10⁻⁴ = 8.00 → 8 H

So far: C₄H₈O_x

Step 3 — Use M_r to get oxygen count

Mass per mole from C and H = 4×12 + 8×1 = 56 g
Given M_r = 88 → mass left for oxygen = 88 − 56 = 32 g
O atoms = 32 ÷ 16 = 2

✅ Final answer

Formula of L = C₄H₈O₂

4 marks

💡 Key knowledge

Carbon & hydrogen bookkeeping

Each CO₂ gives one C; each H₂O gives two H.
Divide by the moles of the original sample to convert “moles of atoms” → “atoms per molecule”.

Oxygen via M_r

When only C, H, O are present, get O by difference using the known M_r.

🧠 Exam technique

Fast ratio route

Compute n(C)/n(L) and n(H)/n(L) directly; this avoids unnecessary mass conversions and reduces rounding errors.

Alternative mass route (also creditworthy)

Find masses of C and H from product moles, subtract from the sample’s molar mass to get mass of O, then to atoms.

❌ Common errors

Seen in scripts

Stopping after calculating masses of CO₂/H₂O without converting to moles of C and H in the sample.
Dividing by the wrong quantity (e.g. total moles of products rather than moles of L).
Forgetting to use M_r to determine O.

Topics

Physical Chemistry · 3.1.2 Amount of Substance