GCE A LEVEL MARKING SCHEME

AUTUMN 2021

A LEVEL
CHEMISTRY - COMPONENT 3
A410U30-1

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INTRODUCTION

This marking scheme was used by WJEC for the 2021 examination. It was finalised after
detailed discussion at examiners' conferences by all the examiners involved in the
assessment. The conference was held shortly after the paper was taken so that reference
could be made to the full range of candidates' responses, with photocopied scripts forming
the basis of discussion. The aim of the conference was to ensure that the marking scheme
was interpreted and applied in the same way by all examiners.

It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conference, teachers
may have different views on certain matters of detail or interpretation.

WJEC regrets that it cannot enter into any discussion or correspondence about this marking
scheme.

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 COMPONENT 3: CHEMISTRY IN PRACTICE

AUTUMN 2021 MARK SCHEME

GENERAL INSTRUCTIONS

Recording of marks

Examiners must mark in red ink.

One tick must equate to one mark, apart from extended response questions where a level of response mark scheme is applied.

Question totals should be written in the box at the end of the question.

Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate.

Extended response questions

A level of response mark scheme is applied. The complete response should be read in order to establish the most appropriate band. Award the
higher mark if there is a good match with content and communication criteria. Award the lower mark if either content or communication barely
meets the criteria.

Marking rules

All work should be seen to have been marked.

Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer.

Crossed out responses not replaced should be marked.

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Marking abbreviations

The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded.

cao = correct answer only

ecf = error carried forward
bod = benefit of doubt

Credit should be awarded for correct and relevant alternative responses which are not recorded in the mark scheme.

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
1 (a) (i) C3H8O(l) + 4½O2(g) → 3CO2(g) + 4H2O(l)

balanced equation (1)

state symbols (1) 2 2

(ii) spirit burner containing propan-1-ol and metal can containing water
accept beaker in place of metal can (1)
2
known volume / mass of water added to metal can (1)

award (1) for both measurements

• initial and final temperature of the water (allow temperature change 2 4 4

of the water)
• mass of the burner before and after heating (allow change in mass
of the burner)
! × $ × ∆&
use of ΔH = – (1)
'

(b) (i) suitable scale selected on x-axis and y-axis (1)

all points plotted correctly (tolerance ±½ square) and line of best fit 2 2 2
drawn (1)

(ii) as the number of moles of ethanol (in the mixture) increases the heat
evolved (in kJ) decreases 1 1 1

linear relationship – neutral answer

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
(iii) –2005 1 1

accept 2005

(iv) n(CH3CH2OH) from graph = 0.0072 (1) 1 1

2
mass of CH3CH2OH = 0.0072 × 46.06 = 0.332 (1) 1

(v) line starting from same point on right-hand side as plotted line (1)
2 2 2
gradient greater than plotted line (1)

Question 1 total 4 6 4 14 6 4

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
2 (a) award (1) for either of following

• reacts with phenol to produce the (nucleophilic) phenoxide ion 1 1 1

• removes the proton from phenol

(b) (i) benzoic anhydride (accept benzoic acid) 1 1 1

(ii) award (1) for either of following

benzoyl chloride more reactive 1 1 1

not an equilibrium process

(c) dissolve solid in minimum volume of hot ethanol (1)

award (1) each for any two of following

• heat ethanol using a hot plate / hot water bath / electric heating
3 3 3
mantle
• filter off any insoluble impurities whilst hot
• allow to cool (to crystallise solid)
• filter and dry

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
(d) (.*
n(C6H5OH) = +,.-. = 0.016 (1)

mass of C7H5OCl = 3.0 × 1.2 = 3.6 g

/..
n(C7H5OCl) = (,-.** = 0.026 ð in excess (1) 1 4 2
3
theoretical mass of (C13H10O2) = 0.016 × 198.1 = 3.2 g (1)
0.+
percentage yield = /.0 × 100 = 91 (1)

(e) award (1) each for any two of following

NMR
HPLC 2 2 2
GC mass spectra
melting temperature

(f) (i) CHI3 (1)

2 2 2
(1)

(ii) acidified potassium dichromate solution (1)

2 2 2
colour change from orange to green (1)

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
(iii) aldehyde 1 1 1

(iv) award (2) for all four ester structures

award (1) for any two ester structures

4 4 2

award (1) for each of A and B correctly identified

Question 2 total 2 13 6 21 2 15

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
3 Indicative content

1. Fe(NO3)2 identified from pale green solution

2. Fe(NO3)2 added to other five solutions
3. Pale green precipitate formed identifies NaOH
4. NaOH added to remaining four solutions
• two solutions give white precipitate which are soluble in
excess NaOH indicating Pb(NO3)2 and Al2(SO4)3
• two solutions give no observable reaction indicating
2 4 6 4
Ba(NO3)2 and KI
5. Take one of the solutions which gave a white precipitate with
NaOH and add in turn to both solutions that gave no observable
reaction with NaOH
• If white precipitate formed then Al2(SO4)3 and Ba (NO3)2 both
identified
• If (bright) yellow precipitate formed then Pb(NO3)2 and KI both
identified

5-6 marks
Method outlined leads to the identification of all six solutions
The candidate constructs a relevant, coherent and logically structured account including all key elements of the indicative
content. A sustained and substantiated line of reasoning is evident and scientific conventions and vocabulary is used
accurately throughout.

3-4 marks
Method outlined leads to the identification of Fe(NO3)2 and NaOH and solutions containing amphoteric and non-amphoteric
metals
The candidate constructs a coherent account including many of the key elements of the indicative content. Some reasoning
is evident in the linking of key points and use of scientific conventions and vocabulary is generally sound.

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
1-2 marks
Method leads to the identification of Fe(NO3)2 and NaOH
The candidate attempts to link at least two relevant points from the indicative material. Coherence is limited by omission
and/or inclusion of irrelevant materials. There is some evidence of appropriate use of scientific conventions and vocabulary.

0 marks
The candidate does not make any attempt or give an answer worthy of credit.

Question 3 total 2 0 4 6 0 4

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
4 (a) (i) award (3) for all five points
award (2) for point 4 and any three others
award (1) for any three points

1. Record the mass of an empty crucible + lid

2. Add known mass of CaSO4.2H2O and replace the crucible lid 3 3 3
3. Heat the crucible (with lid) containing the solid above a Bunsen
burner flame / carefully lift lid but do not remove completely
4. Heat to constant mass
5. Weigh crucible + lid + solid (after cooling)

(ii) *.0-
n(CaSO4.2H2O) used = = 0.0302 (1)
(10.0,

mass of anhydrous CaSO4 formed must be measured

mass = 136.2 × 0.0302 = 4.11 g (1) 2 2 2

accept mass of water lost as an alternative

mass = 0.0302 × 2 × 18.02 = 1.09 g

(b) 0.,+
n(CaSO4.2H2O) = (10.0, = 0.0145 mol (1)

(.*+ 2
Mr(M2SO4) = -.-(,* = 109.90 (1)
3 2
((-+.+- 3 +..() 1
Mr(M) = 0
= 6.91 ð M is lithium (1)

Question 4 total 0 7 1 8 4 3

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
– –
5 (a) 2I – 2e → I2 1 1

(b) PbO2 + 4H+ + 2e– → Pb2+ + 2H2O 1 1

(c) starch indicator

1 1 1
blue/black to colourless

(d) (i) 0(.+*

n(S2O32–) = 0.0510 × (---
= 0.00112

n(I2) in 25.0 cm3 = 0.000560 1 1 1

n(I2) formed in step 1 = 8 × 0.000560 = 0.00448

(ii) n(PbO2) = 0.00448 in 200 cm3 of solution (1) 1

mass of PbO2 = 0.00448 × 239 = 1.071 g

2 2
(.-1( 1
percentage PbO2 = (.(5
× 100 = 90.7 (1)

(e) 0 × -.--* 1 1 1
× 100 = 0.85
(.(5

(f) (i) 2PbO2(s) → 2PbO(s) + O2(g) 1 1

ignore state symbols

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 Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
(ii) award (1) for conversion / correct units for p, V and T

67 (.-( × (-9 × (.0/ × (-:;

n(O2) = 8& = 5./( × 51/
= 0.00171 (1)

3 3 3
n(PbO2) = 0.00342

mass of PbO2 = 0.00342 × 239 = 0.819 (1)

Question 5 total 3 6 2 11 7 1

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 COMPONENT 3: CHEMISTRY IN PRACTICE

SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES

Question AO1 AO2 AO3 Total Maths Prac

1 4 6 4 14 6 4
2 2 13 6 21 2 15
3 2 0 4 6 0 4
4 0 7 1 8 4 3
5 3 6 2 11 7 1
Totals 11 32 17 60 19 27

A410U30-1 EDUQAS GCE A Level Chemistry – Component 3 MS A21/CB

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