AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2020: Question 3

10 marks · Standard Demand difficulty · Short Answer

Explain conservation of mass using experimental mass data, outline purification steps for an insoluble salt, and calculate the percentage atom economy to 3 significant figures.

Practise this question

Question

Question 3 presents a precipitation reaction between silver nitrate and sodium iodide forming solid silver iodide and aqueous sodium nitrate. Question 03.1 gives experimental steps and a results table showing masses of beaker A and B before and after mixing, asking to explain conservation of mass. Questions 03.2 to 03.4 ask about separating, washing, and drying the insoluble silver iodide precipitate. Question 03.5 asks to calculate percentage atom economy for silver iodide to 3 significant figures using given relative formula masses. Question 03.6 asks for one industrial reason to use reactions with high atom economy.
Question text

03 This question is about silver iodide.

Silver iodide is produced in the reaction between silver nitrate solution and

sodium iodide solution.

The equation for the reaction is:

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

03.1 A student investigated the law of conservation of mass.

This is the method used.

1. Pour silver nitrate solution into a beaker labelled A.

2. Pour sodium iodide solution into a beaker labelled B.

3. Measure the masses of both beakers and their contents.

4. Pour the solution from beaker B into beaker A.

5. Measure the masses of both beakers and their contents again.

Table 3 shows the student’s results.

Table 3

Mass before mixing in g Mass after mixing in g

Beaker A and contents 78.26 108.22

Beaker B and contents 78.50 48.54

Explain how the results demonstrate the law of conservation of mass.

You should use data from Table 3 in your answer.

[2 marks]

03.2 Suggest how the student could separate the insoluble silver iodide from the mixture at

the end of the reaction.

[1 mark]

The student purified the separated silver iodide.

This is the method used.

1. Rinse the silver iodide with distilled water.

2. Warm the silver iodide.

03.3 Suggest one impurity that was removed by rinsing with water.

[1 mark]

03.4 Suggest why the student warmed the silver iodide.

[1 mark]

03.5 Calculate the percentage atom economy for the production of silver iodide in

this reaction.

The equation for the reaction is:

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Give your answer to 3 significant figures.

Relative formula masses (Mr): AgNO3 = 170 NaI = 150 AgI = 235 NaNO3 = 85

[4 marks]

Percentage atom economy (3 significant figures) = %

03.6 Give one reason why reactions with a high atom economy are used in industry.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for Question 3 detailing points: 03.1 awards 1 mark for showing total mass before equals total mass after (156.76 g) and 1 mark for stating mass of products equals mass of reactants; 03.2 accepts filtration; 03.3 accepts sodium nitrate solution or constituent ions; 03.4 accepts removing water or drying the solid; 03.5 awards 4 marks for calculating total Mr (320), atom economy equation (235/320 x 100), intermediate value 73.4375%, and final value 73.4%; 03.6 awards 1 mark for sustainable development, economic reasons, or reducing waste.

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 (total) mass before = 156.76 (g) allow 78.26 + 78.50 = 156.76 1 AO2

and and

(total) mass after = 156.76 (g) 108.22 + 48.54 = 156.76

or

increase in mass of beaker A allow 108.22 – 78.26 = 29.96

and contents = 29.96 (g) and

and 48.54 – 78.50 = – 29.96

decrease in mass of beaker B

and contents = 29.96 (g)

(so) the mass of products allow (so) no atoms were lost or 1 AO1

equals the mass of the reactants made during the reaction

or

4.3.1.1

(so) there is no change in mass

during the reaction

03.2 filter / filtration allow a description of filtration 1 AO2

4.1.1.2

03.3 allow correct formulae AO2

4.1.1.2

sodium nitrate (solution) allow sodium / nitrate / silver / 1

or iodide ions

silver nitrate (solution)

or

sodium iodide (solution)

03.4 to remove / evaporate the water allow to dry (the solid) 1 AO3

4.1.1.2

Question 3 continued

AO /

Spec. Ref.

03.5 AO2

(total Mr = 170 + 150) = 320 allow (235 + 85) = 320 1 4.3.3.2

(% atom economy =)

235 allow correct use of incorrectly 1

12 ×100

320 calculated total Mr

= 73.4375 (%)

= 73.4 (%) allow an answer correctly 1

calculated to 3 significant figures

from an incorrect percentage

calculation which uses the

values in the question

03.6 any one from: 1 AO1

4.3.3.2

• for sustainable development allow to reduce waste

• for economic reasons

• to produce a high(er)

percentage of useful product

Total 10

How to answer it

Precipitation of Silver Iodide: Conservation of Mass & Atom Economy

What this question tests

  • Quantitative Chemistry: Explaining the Law of Conservation of Mass using experimental data.
  • Separation Techniques: Identifying filtration and purification steps (washing and drying a solid precipitate).
  • Atom Economy: Calculating percentage atom economy using relative formula masses and rounding to specified significant figures.
  • Industrial & Green Chemistry: Understanding the economic and environmental benefits of reactions with high atom economy.
Question 03.1 • 2 Marks

Law of Conservation of Mass

Explaining experimental data from a precipitation reaction

✅ Correct Answer

  • Data comparison (1 mark):
    Total mass before = 78.26 + 78.50 = 156.76 g
    Total mass after = 108.22 + 48.54 = 156.76 g
    (Alternatively: Beaker A gained 29.96 g and Beaker B lost 29.96 g)
  • Conclusion (1 mark):
    The mass of products equals the mass of reactants (or there is no overall change in mass / no atoms were lost or made).

🧠 Exam Technique

The prompt explicitly says: "You should use data from Table 3 in your answer."

Whenever you see this instruction, you must calculate values. Stating only "mass is neither created nor destroyed" earns zero marks without calculating and comparing total masses before and after.

❌ Common Errors

  • Only calculating the mass before (156.76 g) but failing to calculate and write down the total mass after.
  • Looking only at Beaker A and concluding "mass increased from 78.26 g to 108.22 g, so mass was gained". You must consider both beakers together!

💡 Key Knowledge

In a closed or precipitation system where no gas escapes, total mass of reactants = total mass of products because atoms are simply rearranged, not created or destroyed.

Mark breakdown: 1 mark for showing both totals equal 156.76 g (or matching transfer masses of 29.96 g); 1 mark for stating that total mass does not change / mass of reactants equals mass of products.
Question 03.2 • 1 Mark

Separating Insoluble Products

Technique for collecting a solid precipitate

✅ Correct Answer

Filtration (or filter / filtering through filter paper and a funnel).

💡 Key Knowledge

Look at the state symbols in the equation:
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq)

Because silver iodide is a solid (s) and sodium nitrate is dissolved in solution (aq), filtration is the standard method to separate an insoluble solid from a liquid mixture.

❌ Common Errors

Confusing crystallisation or evaporation with filtration. Evaporating the mixture would leave behind sodium nitrate mixed with the silver iodide instead of separating them.

Mark breakdown: 1 mark for naming "filter" or "filtration" (or an accurate description of pouring through filter paper).
Questions 03.3 & 03.4 • 2 Marks Total

Purifying the Precipitate

Rinsing and warming the separated solid

✅ 03.3: Impurity Removed by Rinsing

Any one of:

  • Sodium nitrate (or NaNO₃ solution)
  • Silver nitrate (excess AgNO₃ reactant)
  • Sodium iodide (excess NaI reactant)
  • Allow individual ions: sodium / nitrate / silver / iodide ions

✅ 03.4: Reason for Warming

  • To remove / evaporate the water
  • To dry the solid precipitate

🧠 Exam Technique: Practical Insight

When an insoluble precipitate is filtered, the filter paper retains wet solid (residue). The liquid clinging to it contains dissolved soluble salts. Rinsing with distilled water washes these dissolved salts away without dissolving the precipitate. Gentle warming then removes the water to obtain a dry sample.

❌ Common Errors

  • 03.3: Writing "silver iodide" (that is the desired product, not an impurity!).
  • 03.4: Stating "to speed up the reaction" — the reaction is already finished!
Mark breakdown: 1 mark for identifying a soluble substance from the reaction (03.3); 1 mark for stating evaporation of water or drying (03.4).
Question 03.5 • 4 Marks

Calculating Percentage Atom Economy

Step-by-step quantitative calculation

💡 Formula Reference

Percentage Atom Economy = (Mr of desired product / Total Mr of all reactants) × 100

Desired product: AgI (Mr = 235)

📐 Step-by-Step Calculation

Step 1: Calculate total relative formula mass (Mr) of reactants
Reactants = AgNO₃ + NaI
Total Mr = 170 + 150 = 320
(Notice: Products total Mr = 235 + 85 = 320 as well, confirming conservation of mass!)
✓ Award Mark 1
Step 2: Set up the percentage atom economy expression
% Atom Economy = (235 / 320) × 100
✓ Award Mark 2
Step 3: Calculate the unrounded percentage
% Atom Economy = 73.4375%
✓ Award Mark 3
Step 4: Round to 3 significant figures
73.4375 → Fourth figure is 3, so round down.
Final Answer = 73.4%
✓ Award Mark 4

❌ Common Calculation Traps

  • Significant figures penalty: Leaving the answer as 73% (2 s.f.) or 73.44% (4 s.f.) loses the 4th mark. Always check the required precision!
  • Wrong denominator: Dividing 235 by only one reactant (e.g. 235 / 170) instead of the sum of all reactants.

🧠 Error-Carried-Forward (ECF)

If you made an addition error in Step 1 (e.g. finding a total Mr of 310), you could still earn Marks 2, 3, and 4 if you correctly applied your number and rounded your final answer to 3 significant figures.

Mark breakdown: 1 mark for total Mr = 320; 1 mark for correct fraction × 100; 1 mark for 73.4375; 1 mark for correct rounding to 3 sig figs (73.4).
Question 03.6 • 1 Mark

Importance of High Atom Economy

Industrial and environmental advantages

✅ Acceptable Answers (Give ONE)

  • For sustainable development (conserves raw materials / resources).
  • For economic reasons (reduces cost of purchasing raw materials / higher profit).
  • To reduce waste (less unwanted by-product to treat or dispose of).
  • To produce a higher percentage of useful product.

❌ Common Errors & Misconceptions

  • Confusing atom economy with yield: Saying "it gives a higher yield" is incorrect. Yield refers to practical recovery; atom economy is theoretical waste from the chemical equation.
  • Vague answers: Simply writing "it is cheaper" without clarifying why (e.g. less wasted materials, less waste disposal costs) might not gain full credit.
Mark breakdown: 1 mark for any valid reason linking high atom economy to sustainability, economics, or reduced waste.

Topics

Chemistry · C1: Atomic Structure and the Periodic Table · C3: Quantitative Chemistry

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.