AQA A-Level Chemistry Paper 2, 2022: Question 9

12 marks · Hard difficulty · Practical Techniques & Data Analysis

Analysis of Olive Oil Composition This question involves determining the amount of the unsaturated fat Y in olive oil using bromine titration. It asks for the justification of titration volumes, calculation of the amount of bromine in the target titre, and mass of olive oil needed. Additional steps include ensuring accurate mass measurement of olive oil and deducing the molecular formula of a compound identified via mass spectrometry, using the empirical formula C₅H₁₀O.

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Question

AQA A-Level Chemistry Paper 2, 2022: Question 9
Question text

09 This question is about olive oil.

A sample of olive oil is mainly the unsaturated fat Y mixed with a small amount of inert

impurity.

The structure of Y in the olive oil is shown.

Y has the molecular formula C57H100O6 (Mr = 880).

The amount of Y is found by measuring how much bromine water is decolourised by a

sample of oil, using this method.

• Transfer a weighed sample of oil to a 250 cm3 volumetric flask and make up to the

mark with an inert organic solvent.

• Titrate 25.0 cm3 samples of the olive oil solution with 0.025 mol dm−3 Br (aq).

09.1 A suitable target titre for the titration is 30.0 cm3 of 0.025 mol dm−3 Br (aq).

Justify why a much smaller target titre would not be appropriate.

Calculate the amount, in moles, of bromine in the target titre.

[2 marks]

Justification

Amount of bromine mol

09.2 Calculate a suitable mass of olive oil to transfer to the volumetric flask using your

answer to Question 09.1 and the structure of Y.

*24* Assume that the olive oil contains 85% of Y by mass.

(If you were unable to calculate the amount of bromine in the target titre, you should

assume it is 6.25 ×10−4 mol. This is not the correct amount.)

[5 marks]

Mass of olive oil g

The olive oil solution can be prepared using this method.

• Place a weighing bottle on a balance and record the mass, in g, to 2 decimal places.

• Add olive oil to the weighing bottle until a suitable mass has been added.

• Record the mass of the weighing bottle and olive oil.

*25* • Pour the olive oil into a 250 cm3 volumetric flask.

• Add organic solvent to the volumetric flask until it is made up to the mark.

• Place a stopper in the flask and invert the flask several times.

09.3 Suggest an extra step to ensure that the mass of olive oil in the solution is recorded

accurately.

Justify your suggestion.

[2 marks]

Extra step

Justification

09.4 State the reason for inverting the flask several times.

[1 mark]

09.5 A sample of the olive oil was dissolved in methanol and placed in a

mass spectrometer. The sample was ionised using electrospray ionisation.

Each molecule gained a hydrogen ion (H+) during ionisation.

m

The spectrum showed a peak for an ion with = 345 formed from an impurity in the

z

olive oil.

m

The ion with = 345 was formed from a compound with the empirical formula C5H10O

z

Deduce the molecular formula of this compound.

[2 marks]

Show your working.

Molecular formula

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2022: Question 9

Question Answers Additional Comments/Guidelines Mark

Smaller titre will increase (%) uncertainty / error 1

09.1

amount Br = 0.025 × 30/ = 7.5 × 10–4 mol Or 0.00075 1

2 1000

(2 x AO3)

Ratio Y :bromine Alternative calc using supplied answer M1

M1 1 : 5

M2 n Y in 25 cm3 oil = 7.5 x 10–4 = 1.5 × 10–4 n Y in 25 cm3 oil = 6.25 x 10–4 = 1.25 × 10–4 M2

If no ratio must state n Y for M2

09.2 3 – 3 3 –4 – 3

M3 n Y in 250 cm = M2 × 10 = (1.5 × 10 ) n Y in 250 cm = 1.25 × 10 × 10 = (1.25 × 10 ) M3

M4 Mass = M3 × 880 = (1.32 g) Mass = 1.25 × 10– 3 × 880 = (1.1 g) M4

M5 Total mass oil needed = M4 × 100/ = 1.55 g Total mass oil needed = 1.1 × 100/ = 1.29g M5

85 85

(3 x AO2,

If wrong ratio used treat as AE and mark ECF 2 x AO3)

Extra step: Weigh the bottle after oil transfer (and record the mass) OR Rinse the bottle with solvent after transfer and M1

add the washings (to the volumetric flask)

09.3

30 Justification: Not all of the oil is transferred To ensure all the oil is transferred M2

Or so that the mass of oil left in the bottle is accounted for Or find the (2 x AO3)

exact mass of oil used M2 is dependent on M1

To ensure the solution is homogeneous Allow evenly mixed/ distributed OWTTE 1

09.4

Uniform solution (AO3)

Mr = 345 – 1 Must show workings in both M1 and M2 M1

09.5

Mr (C5H10O) = 86 M2

M1/ = 4 Hence C H O (2 x AO2)

86 20 40 4

How to answer it

Study Guide: Olive Oil Titration and Analysis

Part (a)

What to do: Justify why a much smaller titre would not be appropriate and calculate the amount of bromine in the target titre.

Calculation:

Amount of Br2 = 0.025 × (30 ÷ 1000) = 7.5 × 10−4 mol

Common errors:

  • Failing to mention that a smaller titre increases percentage uncertainty.
Common Misconception: Many students only discuss accuracy and not uncertainty when addressing smaller titres.

Part (b)

What to do: Calculate the mass of olive oil required based on the reaction ratio and the composition of the oil.

Key Steps:

Ratio of Y : Br2 = 1 : 5
Moles of Y in 25 cm3 = 7.5 × 10−4 ÷ 5 = 1.5 × 10−4
Mass of Y = Moles × Mr = 1.5 × 10−4 × 880 = 1.32 g
Total mass of oil = (Mass of Y × 100) ÷ 85 = 1.55 g

Common errors:

  • Failing to apply the 1:5 ratio for the reaction between Y and bromine.
  • Calculating 85% of the mass instead of scaling up to 100%.
Teacher Tip: Emphasise reaction ratios and scaling calculations for mixed compositions.

Part (c)

What to do: Suggest an additional step to ensure the mass of oil is recorded accurately.

Suggested Step: Weigh the bottle after oil transfer to account for any residue.

Justification: Ensures that the exact mass of transferred oil is known, accounting for any remaining oil in the bottle.

Common errors:

  • Not considering residue in the bottle after transfer.

Part (d)

What to do: Explain why the volumetric flask must be inverted several times.

Correct Answer: To ensure the solution is homogeneous (evenly mixed).

Common errors:

  • Failing to mention uniform distribution of solutes.
Teacher Tip: Discuss practical techniques for ensuring solution uniformity in titration experiments.

Part (e)

What to do: Deduce the molecular formula of the compound with m/z = 345.

Calculation:

Mr = 345 − 1 = 344
Empirical formula mass = (C5H10O) = 86
Number of units = 344 ÷ 86 = 4
Molecular formula = C20H40O4

Common errors:

  • Not subtracting 1 for the hydrogen ion added during electrospray ionisation.
  • Failing to show working despite the question requiring it.
Common Misconception: Some students incorrectly treat the m/z value as the molecular ion without considering ionisation details.

Final Notes

This question requires attention to detail in calculations, practical procedures, and understanding spectroscopic data. Practice applying reaction ratios and interpreting experimental setups to avoid common pitfalls.

Topics

Physical Chemistry · Required Practicals · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance · Required Practical 6: Tests for alcohol, aldehyde, alkene and carboxylic acid

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.