AQA A-Level Chemistry Paper 1, 2017: Question 4

💡 Key knowledge (weighted mean)

• Relative atomic mass = Σ(isotopic mass × fractional abundance).
• If abundances are in %, divide by 100 at the end.
• Here, abundances add to 100%, so use Σ(m/z × %)/100.

📐 Calculation (step-by-step)

1. Multiply each m/z by its % abundance:
   (46 × 9.1) + (47 × 7.8) + (48 × 74.6) + (49 × 8.5)
2. Add:
   = 418.6 + 366.6 + 3580.8 + 416.5 = 4782.5
3. Divide by 100:
   4782.5 / 100 = 47.825
4. Round to 1 d.p.:
   47.8

✅ Correct answer

❌ Common errors

• Forgetting to divide by 100 (gives 4782.5 instead of 47.8).
• Dividing each term by 100 separately and rounding too early (can lose accuracy).
• Using 46,47,48,49 as “masses” but not weighting by abundance (just averaging → wrong).

💡 Key knowledge (electron impact)

• In electron impact, a gaseous atom loses an electron to form a +1 ion.
• General form: M(g) → M⁺(g) + e⁻
• Alternatively show the incoming electron: M(g) + e⁻ → M⁺(g) + 2e⁻

✅ Correct answers

🧠 Exam technique

• For “reaches the detector first”, think: smaller m/z → higher speed (for same kinetic energy) → shorter time of flight.
• Don’t overcomplicate: the question is about the lightest singly-charged ion.

❌ Common errors

• Writing Ti(s) instead of Ti(g) (loses the equation mark because state symbols are required).
• Producing Ti²⁺ (not asked; would change m/z and flight time).
• Choosing the most abundant peak (48) instead of the smallest m/z (46).

💡 Key knowledge

• Molar mass of ⁴⁹Ti ≈ 49 g mol⁻¹ = 49 × 10⁻³ kg mol⁻¹.
• Mass of one atom = (mass of 1 mole) ÷ (Avogadro constant).
• Use L = 6.022 × 10²³ mol⁻¹.

📐 Calculation (step-by-step)

1. Convert to kg mol⁻¹:
   49 g mol⁻¹ = 49 × 10⁻³ kg mol⁻¹
2. Divide by Avogadro constant:
   mass of one atom = (49 × 10⁻³) / (6.022 × 10²³) = 8.137… × 10⁻²⁶ kg
3. Round appropriately:
   8.14 × 10⁻²⁶ kg (≈ 8.1(37) × 10⁻²⁶ kg)

✅ Correct answer (mark scheme)

❌ Common errors

• Forgetting to convert g to kg (answer becomes 1000× too large).
• Using 47.8 instead of 49 (question is specifically ⁴⁹Ti).
• Writing units as g instead of kg (question explicitly asks kg).

📐 Calculation (top-scoring method using ratios)

1. Since d and E are constant:
   t₄₇ / t₄₉ = √(m₄₇ / m₄₉)
2. Use mass numbers (same proportional masses):
   t₄₇ = t₄₉ × √(47/49)
3. Substitute t₄₉ = 9.816 × 10⁻⁷ s:
   t₄₇ = 9.816 × 10⁻⁷ × √(47/49) = 9.816 × 10⁻⁷ × 0.97938… = 9.613… × 10⁻⁷ s
4. Significant figures: limited by 9.816 (4 s.f.), so:
   t₄₇ = 9.614 × 10⁻⁷ s (≈ 9.6(14) × 10⁻⁷ s)

✅ Correct answer (mark scheme)

Any rounding consistent with 4 s.f. is expected (e.g. 9.614 × 10⁻⁷ s).

🧠 Exam technique (how to avoid algebra overload)

• Don’t try to calculate d unless you have to. AQA rewards ratio thinking here.
• Write the proportionality clearly: t ∝ √m , then compare isotopes.
• Keep the powers of ten outside the square root where possible to reduce mistakes.

❌ Common errors (seen by examiners)

• Using a linear ratio: t₄₇ = t₄₉ × (47/49) (wrong because the equation has a square root).
• Inverting the ratio: using √(49/47) makes the lighter isotope slower (physically wrong).
• Rounding √(47/49) too early, then losing accuracy.
• Incorrect s.f. (e.g. rounding to 2 s.f. without justification).

🧠 Before you move on…

• (a) Did you divide by 100 and round to 1 d.p.?
• (b) Did your ionisation equation include (g) state symbols and Ti⁺?
• (c) Did you convert 49 g mol⁻¹ to 49 × 10⁻³ kg mol⁻¹?
• (d) Did you use t ∝ √m and give a sensible shorter time for ⁴⁷Ti⁺?

💡 Core idea linking the whole question

• Mass spectrometry connects isotopes (m/z and abundance) to atomic mass and to ion motion (TOF depends on √mass when energy is fixed).