AQA A-Level Chemistry Paper 2, 2021: Question 1

15 marks · Easy difficulty

This question explores the chemistry of triesters found in coconut oil, focusing on their hydrolysis with KOH. Students must complete a reaction to show the formation of glycerol, identify and give a use for carboxylate salts (soaps), and deduce the structure of the alkyl chains from molecular mass data. There is a calculation to find the percentage by mass of triester in a sample using titration data, and a final question on solvent choice and safety precautions during heating.

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Question

AQA A-Level Chemistry Paper 2, 2021: Question 1
Question text

01 Coconut oil contains a triester with three identical R groups.

This triester reacts with potassium hydroxide.

01.1 Complete the equation by drawing the structure of the other product of this reaction in

the box.

Name the type of compound shown by the formula RCOOK

Give one use for this type of compound.

[3 marks]

Type of compound

Use

01.2 The triester in coconut oil has a relative molecular mass, Mr = 638.0

In the equation shown at the start of Question 01, R represents an alkyl group that

can be written as CH3(CH2)n

Deduce the value of n in CH3(CH2)n

Show your working.

[3 marks]

n

01.3 A 1.450 g sample of coconut oil is heated with 0.421 g of KOH in aqueous ethanol

until all of the triester is hydrolysed.

The mixture is cooled.

The remaining KOH is neutralised by exactly 15.65 cm3 of 0.100 mol dm–3 HCl

Calculate the percentage by mass of the triester (Mr= 638.0) in the coconut oil.

[6 marks]

Percentage by mass

01.4 Suggest why aqueous ethanol is a suitable solvent when heating the coconut oil

with KOH.

Give a safety precaution used when heating the mixture.

Justify your choice.

[3 marks]

*03* Reason

Safety precaution

Justification

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2021: Question 1

Question Answers Additional Comments/Guidelines Mark

CH2OHCH(OH)CH2OH 1

(Potassium) Carboxylate salt Allow fatty acid salt / salt 1

01.1 Salt of a carboxylic acid

Soap Allow detergent / surfactant 1

638 = 173 + 3(15 + 14n) M1

Mr ester fragment = 173

M2

01.2 Show substract 638 – (M1 + 45)

Division of M2 by 42 n must be an integer–A-LEVEL CHEMISTRY – – M3

n = 10

Amount HCl = 0.100 × 0.01565 = 1.565 ×10‒3 mol

M1

0.421 ‒3

Initial amount KOH = = 7.50 ×10 mol

56.1

M2

Amount KOH used = M2 – M1 = 5.939 ×10‒3 mol

5.935× 10‒3 M3

01.3 Amount ester = = 1.980 ×10‒3 mol (M3 / 3)

Mass ester = (1.980 ×10‒3 ) × 638 = 1.263 g (M4 x 638) M411

1.263 M5

%age by mass = × 100 = 87.1 % ( (M5 / 1.45) x 100) Allow 87.0 to 87.1

1.45 M6

Allow 2 sf

Don’t allow M6 for an answer >100%

Allow to dissolve both oil and KOH To act as a mutual solvent OR To ensure reactants M1

are miscible

Precaution must be linked to heating Allow electrical heater / mantle M2

01.4

e.g. Use a water bath for heating mixture Allow sand bath

Allow KOH is corrosive/caustic/damages eyes if M3

Prevents risk of fire / Ethanol is flammable matches alternative precaution given

How to answer it

Study Guide: Hydrolysis of Coconut Oil Triester

Q1. Completing the Equation and Identifying the Compound

Step-by-step:

  • The triester reacts with 3 KOH to produce 3 RCOOK and a glycerol backbone.
  • Glycerol structure: CH2OH–CHOH–CH2OH
  • Type of compound: Carboxylate salt
  • Use: Soap / detergent / surfactant
Common Error: Forgetting to name the glycerol product or only saying “alcohol” without structure.

Q2. Determining the Value of n in CH3(CH2)n

Step-by-step:

  • The full triester’s relative molecular mass is 638.0
  • Each ester “arm” = 173 + (15 + 14n)
  • Subtract glycerol (Mr = 92) → Remaining ester mass = 638 – 92 = 546
  • 546 ÷ 3 = 182 per ester chain → 182 = 173 + 14n ⇒ n = 10
Know your ester group fragment (173) — it saves time in questions like this!
Common Error: Trying to divide 638 by 3 directly without accounting for glycerol.

Q3. Calculating Percentage by Mass of Triester

Step-by-step:

  1. Find moles of HCl: 0.100 × 0.01565 = 1.565 × 10⁻³ mol
  2. Find initial moles of KOH: 0.421 g ÷ 56.1 = 7.50 × 10⁻³ mol
  3. KOH used in hydrolysis: 7.50 – 1.565 = 5.935 × 10⁻³ mol
  4. Moles of triester: divide by 3 → 1.980 × 10⁻³ mol
  5. Mass of triester: 1.980 × 10⁻³ × 638 = 1.263 g
  6. Percentage: (1.263 ÷ 1.450) × 100 = 87.1%
Watch out: Don’t assume the coconut oil is pure triester—only part of it is!

Q4. Solvent & Safety in Hydrolysis Reaction

Reason: Aqueous ethanol acts as a mutual solvent — it helps dissolve both the oil (non-polar) and KOH (polar).

Precaution: Use a water bath or electric heater when heating ethanol mixtures.

Justification: Ethanol is flammable — open flames are dangerous. Also, KOH is corrosive, so eye protection is key!

Link safety to the actual risk (e.g. flammability, caustic nature of KOH) to secure the marks.
Common Error: Students often miss the point about ethanol helping mix the reactants — say it’s a “mutual solvent”.

Topics

Physical Chemistry · Organic Chemistry · Required Practicals · 3.1.2 Amount of Substance · 3.3.9 Carboxylic Acids and Derivatives · Required Practical 5: Distillation of a product from a reaction

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.