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

18 marks · Medium difficulty · Practical Techniques & Data Analysis

Describe the alkaline hydrolysis of ethyl benzoate to produce benzoic acid, including calculations of excess reagents and percentage yield, safety considerations, solubility explanations, and the recrystallisation purification method.

Practise this question

Question

The question presents a practical method for the alkaline hydrolysis of ethyl benzoate into sodium benzoate and ethanol using sodium hydroxide, followed by acidification with hydrochloric acid to precipitate benzoic acid. Parts 03.1 to 03.9 ask: 03.1 how anti-bumping granules prevent bumping (1 mark); 03.2 a calculation showing sodium hydroxide is in excess (2 marks); 03.3 why excess NaOH is used (1 mark); 03.4 why an electric heater is used instead of a Bunsen burner (1 mark); 03.5 why heating under reflux is used (1 mark); 03.6 an equation for the reaction of sodium benzoate with HCl (1 mark); 03.7 an explanation of solubility in water of sodium benzoate versus benzoic acid (2 marks); 03.8 a description of the method to purify the crude benzoic acid solid (6 marks); 03.9 calculation of percentage yield from given quantities and a reason why the calculated yield is not 100% (3 marks).
Question text

03 Benzoic acid can be prepared from ethyl benzoate.

Ethyl benzoate is first hydrolysed in alkaline conditions as shown:

A student used the following method.

Add 5.0 cm3 of ethyl benzoate (density = 1.05 g cm–3, M = 150) to 30.0 cm3 of

r

aqueous 2 mol dm–3 sodium hydroxide in a round-bottomed flask.

Add a few anti-bumping granules and attach a condenser to the flask. Heat the mixture

under reflux for half an hour. Allow the mixture to cool to room temperature.

Pour 50.0 cm3 of 2 mol dm–3 hydrochloric acid into the cooled mixture.

Filter off the precipitate of benzoic acid under reduced pressure.

03.1 Suggest how the anti-bumping granules prevent bumping during reflux.

[1 mark]

03.2 Show, by calculation, that an excess of sodium hydroxide is used in this reaction.

[2 marks]

03.3 Suggest why an excess of sodium hydroxide is used.

*09* [1 mark]

03.4 Suggest why an electric heater is used rather than a Bunsen burner in this

hydrolysis.

[1 mark]

03.5 State why reflux is used in this hydrolysis.

[1 mark]

03.6 Write an equation for the reaction between sodium benzoate and hydrochloric acid.

[1 mark]

03.7 Suggest why sodium benzoate is soluble in cold water but benzoic acid is insoluble in

cold water.

[2 marks]

03.8 After the solid benzoic acid has been filtered off, it can be purified.

Describe the method that the student should use to purify the benzoic acid.

[6 marks]

03.9 In a similar experiment, another student used 0.040 mol of ethyl benzoate and

obtained 5.12 g of benzoic acid.

Calculate the percentage yield of benzoic acid.

Suggest why the yield is not 100%.

[3 marks]

*11* Percentage yield %

Suggestion

Mark scheme

Show the mark scheme Mark scheme for Question 03 showing: 03.1 allows smaller bubbles to form; 03.2 mass and mole calculations of ester (0.0350 mol) compared to moles of NaOH (0.06 mol); 03.3 ensuring complete hydrolysis; 03.4 organic compounds/ethanol are flammable; 03.5 allows reactant vapours to condense and return to flask preventing escape; 03.6 balanced chemical equation C6H5COONa + HCl -> C6H5COOH + NaCl; 03.7 sodium benzoate is ionic whereas benzoic acid has a non-polar benzene ring despite a polar COOH group; 03.8 6-mark recrystallisation stages: dissolve in minimum volume of hot solvent, hot filter, cool to recrystallise, filter under reduced pressure, wash with cold solvent and dry; 03.9 moles of benzoic acid = 5.12/122 = 0.042 mol, percentage yield = 105%, reason: product not fully dried or contains impurities.

allows smaller bubbles to form / prevents the formation of ALLOW provides large surface area for bubbles to form on

03.1 (very) large bubbles 1 IGNORE ‘air’

NOT no bubbles form / prevents bubbles forming

(Mass of ester = 1.05 x 5.0 = 5.25g) Mark independently

amount of ester = 5.25 / 150.0 = 0.0350 mol 1

amount of NaOH = 30 x 2 / 1000 = 0.06 mol 1

OR

(Mass of ester = 1.05 x 5.0 = 5.25g)

03.2 amount of ester = 5.25 / 150.0 = 0.0350 mol 1

Vol of 0.035 mol of NaOH = (0.035/2) x 1000 = 17.5 cm3 1

(so 30 cm3 used is an excess)

OR

amount of NaOH = 30 x 2 / 1000 = 0.06 mol 1

0.06 mol of ester = 9 g = 8.57 cm3 1

(only 5 cm3 used so NaOH in excess) (2 max)

To ensure that the ester is completely hydrolysed / to ensure

03.3 1 ALLOW to ensure the other reagent has completely reacted

all the ester reacts

03.4 Many organic compounds / the ester / ethanol are flammable 1 ALLOW prevent ignition of any flammable vapours formed

Reflux allows reactant vapours (of volatile organic compounds) – – –

03.5 to be returned to the reaction mixture / does not allow any 1 IGNORE reference to products

reactant vapour to escape

Allow ionic equation.

03.6 C6H5COONa + HCl C6H5COOH + NaCl 1 ALLOW molecular formulae (C7H5O2Na and C7H6O2)

ALLOW skeletal benzene ring

15 of 19

Sodium benzoate soluble because it is ionic 1 IGNORE polar

03.7 Benzoic acid insoluble because: despite the polarity of the 1 ALLOW ‘part of molecule’ or ‘one end’ for COOH

COOH group / ability of COOH to form H-bonds, the benzene

ring is non-polar.

Dissolve crude product in hot solvent/water 1 ALLOW ethanol

If no M1 max = 4

of minimum volume 1 ALLOW reference to saturated soln as alternative to ‘min vol’

Filter (hot to remove insoluble impurities) 1 IGNORE use of Buchner funnel here

03.8

Cool to recrystallise 1 apply list principle for each additional process in an incorrect

method but IGNORE additional m.pt determination

Filter under reduced pressure / with Buchner/Hirsch apparatus 1

wash (with cold solvent) and dry 1

5.12 / 122 (= 0.042 mol) 1 method mark

(0.042/0.04) x 100 = 105 % 1 ecf for M1/0.04

03.9 or calculation that 0.04 mol of benzoic = 4.88 g (M1) so

% yield = (5.12/4.88) x100 = 105%

Product not dried / impurities present in product 1 Only allow M3 if M2>100%

Total 18

How to answer it

Alkaline Hydrolysis of Ethyl Benzoate & Purification of Benzoic Acid

📌 What this question tests

This multi-step organic synthesis question assesses core required practical skills (AQA Required Practical 10a/10b):

  • Apparatus & Techniques: Purpose of reflux, anti-bumping granules, and safe heating methods.
  • Quantitative Chemistry: Excess reactant calculations using density ( m = d × V ) and percentage yield calculations exceeding 100%.
  • Structure & Bonding: Explaining solubility differences (ionic lattice vs non-polar aromatic ring).
  • 6-Mark Practical Mastery: Step-by-step recrystallisation and Buchner filtration procedure.

Part (a) — Question 03.1

Role of Anti-Bumping Granules [1 Mark]

✅ Correct Answer

They allow smaller bubbles to form OR prevent the formation of very large bubbles.

💡 Key Knowledge

Anti-bumping granules provide nucleation sites (rough surface area) that encourage smooth, calm boiling without sudden eruptive boil-overs.

❌ Common Errors & Examiner Warning

Do not write that they "stop bubbles from forming" or "prevent boiling". Boiling must occur; the granules merely control bubble size.

Mark scheme: 1 mark for smaller bubbles / prevents large bubbles. Ignore references to "air".

Part (b) — Question 03.2

Proving Excess Reagent by Calculation [2 Marks]

📐 Calculation Breakdown

  1. Find mass of ethyl benzoate:
    Mass = Density × Volume = 1.05 g cm⁻³ × 5.0 cm³ = 5.25 g
  2. Calculate moles of ethyl benzoate:
    Moles = Mass / Mr = 5.25 g / 150.0 g mol⁻¹ = 0.0350 mol [1 mark]
  3. Calculate moles of NaOH:
    Moles = (Volume × Concentration) / 1000 = (30.0 × 2) / 1000 = 0.0600 mol [1 mark]
  4. Compare via stoichiometric ratio:
    From the equation, ethyl benzoate reacts with NaOH in a 1:1 ratio.
    Since 0.0600 mol NaOH > 0.0350 mol ester, NaOH is clearly in excess.

❌ Common Error

Forgetting to convert the 5.0 cm³ volume of ester into grams using its density (1.05 g cm⁻³) before dividing by Mr. Dividing 5.0 directly by 150 loses the first mark.

Mark scheme: 1 mark for ester moles (0.0350 mol); 1 mark for NaOH moles (0.0600 mol). Both marked independently.

Part (c) — Question 03.3

Purpose of Excess Reagent [1 Mark]

✅ Correct Answer

To ensure that the ester is completely hydrolysed (or that all the ester reacts).

🧠 Exam Technique

When asked why one reactant is in excess in organic synthesis, the goal is almost always to drive conversion to completion and ensure the limiting reactant reacts completely.

Mark scheme: 1 mark for ensuring complete reaction/hydrolysis of the ester.

Parts (d) & (e) — Questions 03.4 & 03.5

Apparatus Choices: Heating & Reflux [2 Marks Total]

✅ 03.4: Electric Heater vs Bunsen Burner [1 Mark]

Organic compounds (the ester and the ethanol produced) are flammable.

Electric heating mantles or water baths eliminate the naked flame, preventing ignition of volatile vapours.

✅ 03.5: Why Reflux is Used [1 Mark]

Reflux allows reactant vapours to condense and return to the reaction flask, preventing volatile reactants from escaping before they have reacted.

❌ Common Misconceptions

  • 03.4: Stating "it provides more even heating" is not accepted—the primary safety reason is flammability.
  • 03.5: Saying "prevents products from escaping" is ignored by examiners. Reflux is specifically used to keep reactants in the vessel over prolonged heating.
Mark scheme: 03.4 = 1 mark for flammable; 03.5 = 1 mark for returning reactant vapours / preventing reactant escape.

Part (f) — Question 03.6

Acidification Equation [1 Mark]

✅ Balanced Equation

C₆H₅COONa + HCl → C₆H₅COOH + NaCl

Ionic equation is also accepted: C₆H₅COO⁻ + H⁺ → C₆H₅COOH

Mark scheme: 1 mark for correct balanced equation. Molecular formula (C₇H₅O₂Na + HCl → C₇H₆O₂ + NaCl) is allowed.

Part (g) — Question 03.7

Solubility Comparison: Salt vs Carboxylic Acid [2 Marks]

💡 Why Sodium Benzoate is Soluble [Mark 1]

Sodium benzoate is an ionic compound. Its ions (C₆H₅COO⁻ and Na⁺) readily hydrate and interact strongly with polar water molecules via ion-dipole interactions.

💡 Why Benzoic Acid is Insoluble [Mark 2]

Although the –COOH group is polar and can form hydrogen bonds, the large benzene ring is non-polar and hydrophobic, disrupting water's hydrogen-bonding network.

🧠 Examiner Insight

To secure the second mark, you must mention the benzene ring (or "hydrophobic hydrocarbon part") being non-polar. Simply stating "benzoic acid forms weak bonds with water" will not score.

Mark scheme: M1 for "sodium benzoate is ionic" (ignore polar); M2 for "benzene ring is non-polar" despite polar/H-bonding –COOH group.

Part (h) — Question 03.8

Method to Purify Benzoic Acid: Recrystallisation [6 Marks]

This is a classic required practical extended response. Learn these 6 logical stages in order:

✅ Standard 6-Step Recrystallisation Protocol

  1. Dissolve: Dissolve the crude solid in the minimum volume of hot solvent (water).
  2. Hot filtration: Filter the hot solution through fluted filter paper to remove insoluble impurities.
  3. Crystallise: Allow the hot filtrate to cool slowly to room temperature, then place in an ice bath to recrystallise.
  4. Filter under reduced pressure: Filter off the purified crystals using a Büchner funnel and flask connected to a vacuum pump.
  5. Wash: Wash the collected crystals with a small amount of ice-cold solvent (removes soluble surface impurities).
  6. Dry: Dry the crystals between absorbent filter papers or in a desiccator / low-temperature oven.

🧠 Top-Grade Keywords

Examiners award marks directly for these specific qualifiers: hot, minimum volume, cool, reduced pressure / Büchner, cold solvent wash, and dry.

❌ Critical Pitfalls

Never wash crystals with warm solvent—they will redissolve and drastically reduce your yield! Also, remember not to use Büchner apparatus for hot filtration as rapid cooling causes premature crystallisation in the funnel.

Mark scheme: 1 mark per bullet point (up to 6 marks). If 'hot' is omitted in M1, maximum mark achievable is 4.

Part (i) — Question 03.9

Percentage Yield & Analysis of Anomalous Yield [3 Marks]

📐 Step-by-Step Calculation

  1. Find theoretical moles of benzoic acid:
    Mole ratio is 1:1, so 0.040 mol ethyl benzoate gives a theoretical yield of 0.040 mol benzoic acid.
    Alternatively in grams: Theoretical mass = 0.040 mol × 122.0 g mol⁻¹ = 4.88 g
  2. Calculate actual moles obtained:
    Moles = 5.12 g / 122.0 g mol⁻¹ = 0.0420 mol [Mark 1]
  3. Calculate percentage yield:
    % Yield = (Actual / Theoretical) × 100
    % Yield = (5.12 / 4.88) × 100 = 105% (or from moles: (0.0420 / 0.040) × 100 = 105%) [Mark 2]

✅ Explanation for Yield > 100% [Mark 3]

The product was not completely dry (contains water/solvent) OR contains impurities.

❌ Common Error

Students often default to generic loss reasons like "reaction was incomplete" or "product lost in transfer". But here the yield is over 100%! The only valid reasons are excess mass from water/solvent or unseparated impurities.

Mark scheme: M1 for calculating 0.042 mol or 4.88 g theoretical mass; M2 for 105%; M3 for product wet / impurities (only awarded if % yield > 100%).

Topics

Organic Chemistry · Physical Chemistry · Required Practicals · 3.3.9 Carboxylic Acids and Derivatives · 3.1.2 Amount of Substance · 3.1.3 Bonding · Required Practical 12: Separation and purification techniques

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.