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A-LevelChemistryAtoms, molecules and stoichiometry9701/21 · May/June 2025 · Q49 Marks

The structure of vitamin C is shown in Fig. 4.1. [Figure 4.1]

  1. (a)Deduce the empirical formula of vitamin C.[1]
  2. (b)(i)A vitamin C tablet is dissolved in water to produce 200.0 cm³ of vitamin C solution. 5.00 cm³ of this vitamin C solution is added to a flask with approximately 150cm³ of water and an indicator. Exactly 28.40 cm³ of 5.00 × 10–4 mol dm-3 I2(aq) reacts with the sample of vitamin C solution in the flask. [Mr: vitamin C, 176] Calculate the amount, in mol, of I2(aq) added to the flask in this titration. [Figure 4.2][1]
  3. (b)(ii)Use your answer to (b)(i) to calculate the mass, in g, of vitamin C in the tablet. Show your working. (If you were unable to calculate a value for the amount of I2(aq) in (b)(i), use the value 2.64 x 10-4 mol. This is not the correct value.)[2]
  4. (b)(iii)Deduce the role of I2(aq) in the reaction in Fig. 4.2.[1]
  5. (b)(iv)Suggest two reasons why hot concentrated acidified potassium manganate(VII) is not a suitable reagent for producing Q from vitamin C.[2]
  6. (c)Predict two absorptions that will be seen in the infrared spectra of both vitamin C and Q. Describe the relevant bond and the specific functional group that is responsible for each absorption identified. [Figure 4.3] Table 4.1 bond functional groups containing the bond characteristic infrared absorption range (in wavenumbers)/cm-1 C-O hydroxy, ester 1040-1300 C=C aromatic compound, alkene 1500-1680 C=O amide carbonyl, carboxyl ester 1640-1690 1670-1740 1710-1750 C≡N nitrile 2200-2250 C-H alkane 2850-2950 N-H amine, amide 3300-3500 O-H carboxyl hydroxy 2500-3000 3200-3650 [2]

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Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

  1. [1] C3H4O3
  2. [1] (28.40/1000 × 0.0005) = 1.42 × 10-5 mol
  3. [2] (1.42 × 10-5 × 176 × 200/5) = 0.099968 OR M1 correct method to deduce mass of vit C in 5 cm³ sample (1.42 × 10-5 × 176) = 2.49 × 10-3 M2 correct calculation for mass of vit C in original solution (2.49 × 10-3 × 200 / 5) = 0.099968 OR M1 correct method to deduce amount, mol, vit C in 5 cm³ sample (1.42 × 10-5 × 200 / 5) = 5.68 × 10-4 M2 correct calculation for mass of vit C in original solution = M1 × 176 (5.68 × 10-4) × 176 = 0.099968
  4. [1] oxidising agent
  5. [2] M1 (also) oxidises / reacts with (primary and / or secondary) alcohol groups; M2 (also) breaks the sigma and pi bond in C=C
  6. [2] Any two complete rows from: wavenumber | bond | functional group ---|---|--- 1040 – 1300 | C–O | ester / hydroxy 3200 – 3650 | O–H | hydroxy / alcohols 2850 – 2950 | C–H | alkane 1710 – 1750 | C=O | ester

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About this question

Topic
Atoms, molecules and stoichiometry, A-Level Chemistry (syllabus 9701). Structured question, 9 marks.
Source
A-Level Chemistry, May/June 2025 Paper 2 Variant 1, question 4.

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