AQA GCSE Chemistry Chemistry Paper 1 (Higher), June 2025: Question 9

11 marks · High Demand difficulty · Extended Answer

Plan a method to obtain crystals of zinc iodide from zinc and iodine in ethanol, and calculate the percentage yield from reacting masses.

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Question text

09 Zinc reacts with iodine to produce the salt zinc iodide.

The reaction is done in a solution using ethanol as the solvent.

Ethanol is a flammable liquid with a boiling point of 78 °C.

Table 5 shows the solubility of zinc, iodine and zinc iodide in ethanol.

Table 5

Zinc Iodine Zinc iodide

Solubility in ethanol Insoluble Soluble Soluble

09.1 Plan a method to obtain crystals of zinc iodide from zinc and iodine.

[6 marks]

Extra space

09.2 A student used 6.35 g of iodine in the reaction.

6.29 g of zinc iodide was produced.

*27The equation for the reaction is:*

Zn + I2 ⟶ ZnI2

Calculate the percentage yield of zinc iodide.

Give your answer to 3 significant figures.

Relative formula masses (Mr): I2 = 254 ZnI2 = 319

[5 marks]

Percentage yield (3 significant figures) = %

Mark scheme

Show the mark scheme

Question 9

AO /

Question Answers Mark

Spec. Ref.

09.1 Level 3: The method would lead to the production of a valid 5–6 AO3

outcome. The key steps are identified and logically sequenced. 4.1.1.2

4.4.2.3

RPA1

Level 2: The method would not necessarily lead to a valid 3–4

outcome. Most steps are identified, but the method is not fully

logically sequenced.

Level 1: The method would not lead to a valid outcome. Some 1–2

relevant steps are identified, but links are not made clear.

No relevant content 0

Indicative content

dissolve iodine in ethanol

• in a beaker

add zinc to iodine solution

• stir

continue adding until zinc is in excess

• shown by solid remaining

filter (the reaction mixture)

• to remove the excess zinc

heat the solution

• using a water bath

using an electric heater

• to evaporate off some of the ethanol

• cool / leave remaining solution to crystallise

AO /

Question Answers Extra information Mark

Spec. Ref.

09.2 6.35 1 AO2

�moles I2 = =� 0.025 4.3.1.1

4.3.2.1

4.3.2.2

(moles ZnI2 = 0.025 allow (moles Zn = 0.025 4.3.3.1

theoretical mass ZnI2 = mass of Zn = 0.025 × 65 =

0.025 × 319 =) 1.625 g theoretical mass ZnI2 = 1

7.975 (g) 1.625 + 6.35 =) 7.975 (g)

allow correct use of an

incorrectly determined number

of moles of I2

6.29 allow correct use of an 1

(% yield =) × 100 incorrectly determined

7.975

theoretical mass of ZnI2

= 78.871 (%) 1

= 78.9 (%) allow an answer correctly 1

rounded to 3 significant figures

from an incorrect calculation

which uses both mass values

in the question

alternative approach 1:

6.35

�moles I2 = =� 0.025 (1)

�moles ZnI2 =

6.29

=� 0.019717868 (1)

(% yield =) allow correct use of an

0.019717868 incorrectly determined number

× 100 (1) of moles of I and/or ZnI

0.025 2 2

= 78.871 (%) (1)

= 78.9 (%) (1) allow an answer correctly

rounded to 3 significant figures

from an incorrect calculation

which uses both mass values

in the question

alternative approach 2: 29

�moles ZnI2 =

6.29

=� 0.019717868 (1)

(moles I2 reacted = allow correct use of an

0.019717868 incorrectly determined number

mass I2 reacted = of moles of ZnI2

0.019717868 × 254 =)

5.0083 (g) (1)

5.0083 allow correct use of an

(% yield =) × 100 (1)

6.35 incorrectly determined mass of

I2 reacted

= 78.871 (%) (1)

= 78.9 (%) (1) allow an answer correctly

rounded to 3 significant figures

from an incorrect calculation

which uses both mass values

in the question

Total Question 9 11

How to answer it

Preparing Zinc Iodide Crystals & Percentage Yield

📌 What This Question Tests

This question assesses your practical design skills and multi-step quantitative calculation skills:

Salt Preparation (Required Practical 1 adaptation): Designing a logical, step-by-step synthetic route to prepare pure, dry crystals of a soluble salt formed from an insoluble solid and a dissolved reactant.
Safety in the Laboratory: Safe heating methods for flammable solvents (ethanol) that avoid open naked flames (e.g., using a water bath or electric hotplate).
Theoretical Yield & Stoichiometry: Using reacting masses, moles ( moles = mass ÷ M_r ), and balanced molar ratios.
Percentage Yield: Calculating percentage yield ( % yield = (actual ÷ theoretical) × 100 ) and reporting to specified significant figures (3 s.f.).

Question 09.1

Plan a method to obtain crystals of zinc iodide from zinc and iodine. [6 marks]

💡 Key Knowledge

Carefully interpret the table and the prompt:

Iodine: Soluble in ethanol.
Zinc: Insoluble solid in ethanol.
Zinc iodide: Soluble in ethanol.
Ethanol: Flammable liquid (b.p. 78 °C). Never heat flammable liquids directly with a Bunsen burner flame!

🧠 Exam Technique (Level of Response)

Level 3 (5–6 marks): Complete, logical sequence that leads successfully to dry crystals. Must include dissolving iodine, reacting with excess zinc, filtering, safe heating, and crystallisation.
Excess Reactant Principle: Zinc must be added until in excess (unreacted solid visible) so all the iodine reacts. Unreacted zinc is insoluble, so it is easily filtered off.
Safety detail: You must state a water bath or electric heater because ethanol is flammable.

✅ Correct Step-by-Step Method

Dissolve: Add iodine to ethanol in a beaker and stir until dissolved.
React: Add zinc to the iodine solution and stir the mixture.
Ensure complete reaction: Continue adding zinc until it is in excess (visible unreacted zinc remains at the bottom).
Filter: Filter the mixture using a funnel and filter paper to remove the excess unreacted zinc solid. The filtrate is zinc iodide solution in ethanol.
Safe Evaporation: Heat the zinc iodide solution gently using a water bath or electric heating mantle (do not use a Bunsen burner because ethanol is flammable) to evaporate off some of the ethanol to reach the crystallisation point.
Crystallise: Leave the concentrated solution to cool down and crystallise. Filter and dry the crystals (e.g., pat dry between filter papers).

Marking Allocation:
• 5–6 marks: Method identifies all key stages logically in order and would successfully produce pure zinc iodide crystals.
• 3–4 marks: Most steps identified, but incomplete (e.g., forgot safe heating or excess zinc).
• 1–2 marks: Fragmented steps without clear sequencing.

❌ Common Errors & Examiner Traps

Heating with a Bunsen Burner: Ethanol is explicitly described as a flammable liquid. Heating directly over a naked flame loses marks for safe experimental planning.
Evaporating to Total Dryness: Boiling away all solvent with strong heat ruins crystals and can cause thermal decomposition or spitting. You must only evaporate some solvent, then let it cool and crystallise naturally.
Not adding zinc until in excess: If you do not use excess zinc, unreacted iodine will contaminate the solution. Because iodine is soluble in ethanol, it cannot be filtered out later!
Confusing the residue and filtrate: Forgetting that excess zinc is on the filter paper (residue) and the salt solution passes through (filtrate).

Question 09.2

Calculate the percentage yield of zinc iodide to 3 significant figures. [5 marks]

Given: 6.35 g of iodine reacted; 6.29 g of zinc iodide produced.
Equation: Zn + I₂ → ZnI₂ | M_r(I₂) = 254 | M_r(ZnI₂) = 319

📐 Step-by-Step Calculation

Step 1: Calculate moles of limiting reactant (iodine, I₂)

Moles of I₂ = mass ÷ M_r = 6.35 g ÷ 254 = 0.025 mol
[Mark 1]

Step 2: Determine maximum theoretical moles of product (ZnI₂)

From equation, the molar ratio of I₂ : ZnI₂ is 1 : 1.
Theoretical moles of ZnI₂ = 0.025 mol

Step 3: Calculate maximum theoretical mass of zinc iodide (ZnI₂)

Theoretical mass = moles × M_r = 0.025 mol × 319 = 7.975 g
[Mark 2]

Step 4: Calculate the percentage yield

% yield = (actual mass ÷ theoretical mass) × 100
% yield = (6.29 ÷ 7.975) × 100
[Mark 3]
% yield = 78.87147... %
[Mark 4]

Step 5: Round to 3 significant figures

The 4th figure is 7, which rounds up:
Percentage yield = 78.9%
[Mark 5]

Alternative Method (Mole Ratio Route):
• Moles of ZnI₂ actually formed = 6.29 ÷ 319 = 0.019718 mol
• % yield = (0.019718 ÷ 0.025) × 100 = 78.871... % = 78.9% (awards full marks).

❌ Calculation Pitfalls

Significant Figures Penalty: Writing 78.87% or 79% loses the final mark. The question explicitly specifies 3 significant figures.
Inverting the fraction: Doing 7.975 ÷ 6.29 × 100 = 126.8% . Percentage yield can never be above 100% in a correctly calculated reaction!
Using atomic iodine (I = 127) instead of molecular iodine (I₂ = 254): The question gives M_r of I₂ as 254, but students often misread the formula.

🧠 Exam Tip: Error Carried Forward (ECF)

If you make an arithmetic error in Step 1, you can still gain the subsequent 4 marks if your working is clearly laid out and your final answer is correctly rounded to 3 significant figures based on your numbers.

Always show every single step with formulas and units!

Topics

Chemistry · Required Practicals · Required Practicals · C3: Quantitative Chemistry · C4: Chemical Changes

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.