Edexcel A-Level Chemistry Paper 3, June 2018: Question 6
24 marks · Hard difficulty · Synoptic Questions
Analyze an unknown carboxylic acid X with molecular ion peak m/z = 116 using infrared spectroscopy, titration calculations, experimental error analysis, neutralization equations, chemical tests, and E/Z isomerism.
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Question text
6 This question is about the analysis of an unknown carboxylic acid X by three students.
The students analyse the mass spectrum of X and find that it has a molecular ion
peak at m/z = 116.
The three students each propose a different structural formula for compound X.
Structure 1 HOOCCH CHCOOH
Structure 2 HOCH2CH CHCH2COOH
Structure 3 CH3CH2CH2CH2CH2COOH
(a) The students are given the infrared spectrum of X.
(i) State two wavenumber ranges of the infrared absorptions providing evidence
that compound X is a carboxylic acid. Include the bonds responsible.
(2)
(ii) One of the students suggests that this infrared spectrum and the data in the
Data Booklet alone could be used to identify which of the three proposed
structures is X.
Show that this student’s suggestion is correct. Include relevant infrared data in
your answer.
(3)
(b) The students decide to carry out an acid-base titration to obtain further*P52304A01832*
information about compound X.
Each student uses solid sodium hydroxide, NaOH, to prepare a solution of
concentration 0.140 mol dm−3.
Calculate the mass, in grams, of solid sodium hydroxide that each student should
weigh out to prepare 250.0 cm3 of a 0.140 mol dm−3 solution.
(2)
(c) Each of the students makes up 250.0 cm3 of 0.140 mol dm−3 sodium hydroxide
solution in a volumetric flask and titrates this solution with the same solution of X
of known concentration.
Student A
• correctly prepares the 0.140 mol dm−3 sodium hydroxide solution and pipettes
a volume of 10.0 cm3 of the solution into a conical flask
• fills a burette with the solution of X and carries out a titration
• repeats the procedure until obtaining concordant results
• obtains a mean titre of 10.20 cm3.
Student B
• dissolves the sodium hydroxide in distilled water and transfers the solution to
a volumetric flask
• adds more distilled water to the volumetric flask and mixes the solution
• notices that the volumetric flask has been filled with distilled water several
cm3 beyond the graduation mark
• realises the mistake, removes the extra solution and discards it
• pipettes 10.0 cm3 of the sodium hydroxide solution into a conical flask and
titrates this with the solution of X.
Student C
• correctly prepares the 0.140 mol dm−3 sodium hydroxide solution
• washes a conical flask thoroughly with distilled water and pipettes 10.0 cm3 of
the sodium hydroxide solution into the wet conical flask
• titrates the contents of the conical flask with the solution of X.
(i) Explain how, if at all, Student B’s mistake affects the value of the titre.
(2)
(ii) Explain how, if at all, Student C’s use of a wet conical flask affects the value of
the titre.
(2)
… 19
… *P52304A01932*
(iii) Student A uses three pieces of apparatus to measure volumes in this experiment.
• The burette has an uncertainty of ±0.05 cm3 for each volume reading
20 • The volumetric flask has an uncertainty of ±0.30 cm3 for the volume
• The pipette has an uncertainty of ±0.04*P52304A02032*cm3for the volume
Show by calculation which volume measurement has the lowest percentage uncertainty.
(3)
(d) Student A calculates the correct value for the molar mass of compound X, using
the mean titre of 10.20 cm3. The results indicate that X has structure 1.
Structure 1 HOOCCH CHCOOH
Structure 2 HOCH2CH CHCH2COOH
Structure 3 CH3CH2CH2CH2CH2COOH
(i) Write the equation for the reaction between structure 1 and
sodium hydroxide solution. State symbols are not required.
(2)
(ii) Deduce the value that would have been obtained for the mean titre if the
structural formula of X had been structure 2.
Justify your answer.
(2)
(e) The students could have identified the three structures using chemical tests.
Complete the table to show whether or not the suggested structures react with
bromine water and when heated with acidified potassium dichromate(VI).
Use a tick ( ) if a reaction occurs.
Use a cross ( ) if no reaction occurs.
(2)
*P52304A02132*Test with acidified
Structure Test with bromine water
potassium dichromate(VI)
HOOCCH CHCOOH
HOCH2CH CHCH2COOH
CH3CH2CH2CH2CH2COOH
(f ) The structure HOOCCH*P52304A02232*CHCOOH has two stereoisomers.
(i) Draw the structures of these stereoisomers.
(2)
E-isomer
Z-isomer
(ii) State why HOOCCH CHCOOH has E/Z isomers.
(2)
(Total for Question 6 = 24 marks)
Mark scheme
Show the mark scheme
64 marks, and 3 or 4 indicative points would score
5–4 3 1 reasoning mark. A total of 2, 1 or 0 indicative
3–2 2 points would score 0 marks for reasoning.
00 If there is any incorrect chemistry, deduct
mark(s) from the reasoning. If no reasoning
mark(s) awarded do not deduct mark(s).
The following table shows how the marks should be
awarded for structure and lines of reasoning.
Indicative content (IPs) Allow omission of square brackets throughout
IP1: Allow for IP1
[Cu(H O) ]2+(aq) + 2OH−(aq) → [Cu(OH) (H O) ](s) + 2H O(l) Cu2+(aq) + 2OH−(aq) → Cu(OH) (s)
26 2 2 4 2 2
Only penalise incorrect or missing state symbols
IP2: in this equation (IP1)
blue ppt / blue solid (when [Cu(OH)2(H2O)4](s) is formed)
IP3:
[Cu(H O) ]2+(aq) + 4NH (aq) → [Cu(NH ) (H O) ]2+(aq) + 4H O(l) Allow for IP3
26 3 3 4 2 2 2
Cu2+(aq) + 4NH (aq) → [Cu(NH ) ]2+(aq)
33 4
[Cu(OH)2(H2O)4](s) + 4NH3 (aq) →
[Cu(NH ) (H O) ]2+(aq) + 2H O(l) + 2OH−(aq)
34 2 2 2
[Cu(OH)2(H2O)4](s) + 6NH3 (aq) →
[Cu(NH ) (H O) ]2+(aq) + 2NH + (aq) 2H O(l) +
34 2 2 4 2
2OH−(aq)
Ignore formation of initial precipitate Cu(OH)2(s)
Do not award [Cu(NH ) ]2+(aq)
IP4:
Deep blue solution / dark blue solution (when [Cu(NH ) (H O) ]2+(aq) is
34 2 2
formed)
IP5:
[Cu(H O) ]2+(aq) + 4Cl−(aq) → [CuCl ]2−(aq) + 6H O(l)
26 4 2
IP6:
Yellow / green (solution when [CuCl ]2−(aq) is formed) Do not award ‘yellow precipitate’
Allow equilibrium sign ⇌ in any reaction
Ignore any initial colours, even if incorrect
(Total for Question 5 = 16 marks)
Question
Acceptable Answers Additional Guidance Mark
Number
6(a)(i) An answer that makes reference to the following (2)
points:
Allow any value(s) within the range 3300 — 2500
3300 — 2500 (cm−1) and O-H (bond) (1) (cm−1)
Allow -OH
Allow any value(s) within the range 1725 — 1700
1725 — 1700 (cm−1) and C=O (bond) (1) (cm−1)
Allow 1320 – 1210 (cm-1) and C-O
Question
Acceptable Answers Additional Guidance Mark
Number
6(a)(ii) An answer that makes reference to the following points: (3)
structures 1 and 2 will have an absorption at
Either
C=C at 1669 — 1645 (cm−1)
or
C—H in an alkene at 3095 — 3010 (cm−1) (1) Reject C=C at 3010 (cm−1)
only structure 2 will have an absorption due to the
presence of an alcohol / O—H at 3750 —3200 (cm−1) (1)
structure 3 will have none of these absorptions / will not
show C=C absorption / C-H absorption for an alkene (1)
Question
Acceptable Answers Additional Guidance Mark
Number
6(b) Example of calculation: (2)
calculation of moles of NaOH (1) (moles NaOH = 0.140 x 250)
1000
= 0.035(0) (mol)
calculation of mass of NaOH (1) = 40(.0) x 0.035(0) = 1.4(0) (g)
Correct answer with or without working
scores 2 marks
Allow TE for M2 on moles of NaOH
Alternative route, allow M1 for conversion
of concentration to 5.6 g dm-3
Ignore SF
Question
Acceptable Answers Additional Guidance Mark
Number
6(c)(i) An explanation that makes reference to the following points: (2)
(because the) sodium hydroxide has been diluted (1) Allow
Fewer moles of sodium hydroxide present /
some sodium hydroxide will have been
removed
(the titre will be) smaller (1) M2 dependent on M1
Question
Acceptable Answers Additional Guidance Mark
Number
6(c)(ii) An explanation that makes reference to the following points: (2)
M1 no effect (on the titre) (1) M2 depends on M1
M2 because the (number of) moles of sodium hydroxide is
unaffected (1) Allow base / alkali / hydroxide (ions)
Allow amount / mass of sodium hydroxide
is unaffected
Question
Acceptable Answers Additional Guidance Mark
Number
6(c)(iii) Example of calculation: (3)
calculation of percentage uncertainty in burette volume (1) 2 x (±)0.05 x 100% = (±)0.980392156%
10.20
calculation of percentage uncertainty in volumetric flask (±)0.30 x 100% = (±)0.12%
volume 250.0
and and
in pipette volume (1) (±)0.040 x 100% = (±)0.4%
10.0
identification of volume with the lowest percentage Volumetric flask has the lowest
uncertainty (1) uncertainty
Allow TE for identification in M3
Allow ANY number of SF in answer, from
1 SF up to calculator value
Question
Acceptable Answers Additional Guidance Mark
Number
6(d)(i) Example of equation (2)
left-hand side of equation correct (1)
HOOCCH=CHCOOH + 2NaOH → NaOOCCH=CHCOONa + 2H2O
right-hand side of equation correct (1)
ALLOW use of molecular formulae or ionic equation:
C4H4O4 + 2NaOH → Na2C4H2O4 + 2H2O
HOOCCH=CHCOOH + 2OH- (+ 2Na+) →
-OOCCH=CHCOO- + 2H O (+ 2Na+)
ALLOW
Multiples
Correct charges
Do not award if O−Na covalent bond drawn
IGNORE
State symbols, even if incorrect
Question
Acceptable Answers Additional Guidance Mark
Number
6(d)(ii) An answer that makes reference to the following points: Mark M1 and M2 independently (2)
(New mean titre) = 20.4(0) (cm3) / double (the original value) (1)
For structure 2, mole ratio / reacting ratio is 1:1 (with NaOH) (1) Allow structure 2 has 1 COOH / 1
acid group
Question
Acceptable Answers Additional Guidance Mark
Number
6(e) (2)
Structure Test with Test with 3 correct ticks with no crosses scores 1
Br2 water acidified K2Cr2O7
Ignore descriptions of result in terms of
HOOCCH=CHCOOH x colour (changes) / reactions occurring
HOCH2CH=CHCH2COOH
CH3CH2CH2CH2CH2COOH x x
Left hand column correct (1)
Right hand column correct (1)
Question
Acceptable Answers Additional Guidance Mark
Number
6(f)(i) E-isomer: ALLOW skeletal or displayed structures (2)
HOOC H ALLOW −CO2H
C C
IGNORE
H COOH Connectivity to the –COOH group
(1)
IGNORE
Z-isomer: bond angles
HOOC COOH Award one mark if correct structures are drawn, but
E- and Z-isomers labelled the wrong way round
C C
H H Award 1 mark if incorrect molecule used but E - and
(1) Z- isomers are correct
Question
Acceptable Answers Additional Guidance Mark
Number
6(f)(ii) An answer that makes reference to the following points: (2)
restricted / limited rotation (about the C=C double bond)(1) Allow “no rotation”
each carbon atom in the double bond is attached to (two)
different atoms / different groups (of atoms) / to a H (atom)
and a COOH group (1) Do not award the carbons are attached to 2
“different molecules”
Mark points M1 and M2 independently
(Total for Question 6 = 24 marks)
How to answer it
Analysis of Carboxylic Acid Isomers
What this question tests
This comprehensive multi-part question tests your ability to interpret infrared spectroscopy data, perform solution calculations (moles, mass, and percentage uncertainty), evaluate laboratory errors in titrations, write neutralisation equations, predict chemical test outcomes (bromine water and acidified potassium dichromate), and identify E/Z stereoisomerism in alkenes.
Infrared Spectroscopy & Structure Identification
✅ Correct Answers (Part a)(i)
- 3300 – 2500 cm⁻¹ for the O-H bond in carboxylic acids.
- 1725 – 1700 cm⁻¹ for the C=O bond.
✅ Correct Answers (Part a)(ii)
- Structures 1 and 2 show a C=C absorption (1669 – 1645 cm⁻¹) or alkene C-H (3095 – 3010 cm⁻¹).
- Only structure 2 shows an alcohol O-H absorption (3750 – 3200 cm⁻¹).
- Structure 3 lacks all of these specific functional group absorptions, proving the identity is Structure 1.
🧠 Exam Technique
Always quote wavenumber ranges directly from your Data Booklet rather than memorising arbitrary numbers. Clearly link each absorption range to the specific bond responsible.
❌ Common Errors
Students frequently confuse broad alcohol O-H stretches with carboxylic acid O-H stretches, or incorrectly quote C=C absorptions at 3010 cm⁻¹ instead of the C-H alkene stretch range.
Solution Preparation & Mass Calculation
📐 Step-by-Step Calculation
- Find moles of NaOH required:
Moles = (concentration × volume) / 1000
Moles = (0.140 × 250.0) / 1000 = 0.0350 mol - Calculate molar mass of NaOH:
Mr(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹ - Calculate mass of solid NaOH:
Mass = moles × Mr = 0.0350 × 40.0 = 1.40 g
❌ Common Calculation Traps
- Forgetting to divide the volume by 1000 when converting from cm³ to dm³.
- Using incorrect relative atomic masses (e.g., omitting decimal values for sodium or oxygen).
Titration Errors & Percentage Uncertainty
✅ Correct Answers (Parts c)(i) & (ii)
- Student B's mistake: Filling past the graduation mark dilutes the solution, meaning fewer moles of NaOH are present in the aliquot. Consequently, the titre will be smaller.
- Student C's use of a wet flask: Has no effect on the titre. Although water is added, the total *number of moles* of NaOH pipetted into the flask remains completely unchanged.
📐 Part (c)(iii): Percentage Uncertainty Calculation
- Burette: (2 × 0.05 / 10.20) × 100 = ±0.980% (Note: 2 readings taken)
- Volumetric Flask: (0.30 / 250.0) × 100 = ±0.120%
- Pipette: (0.04 / 10.0) × 100 = ±0.400%
- Lowest uncertainty: Volumetric flask.
🧠 Exam Technique for Titration Theory
When evaluating water in conical flasks, examiners look for the distinction between concentration changes (which water causes) and mole constancy (why the titration endpoint is unaffected because moles of alkali pipetted stay identical).
Neutralisation Equations & Stoichiometry
✅ Correct Answers
- (d)(i) Equation:
HOOCCH=CHCOOH + 2NaOH → NaOOCCH=CHCOONa + 2H₂O
(Allow molecular or ionic formulas; state symbols optional). - (d)(ii) Mean titre with Structure 2:
New titre = 20.40 cm³ (it doubles).
Justification: Structure 2 has only one carboxylic acid group (COOH), meaning a 1:1 reacting ratio with NaOH instead of the 1:2 ratio found in Structure 1.
💡 Key Knowledge
Dicarboxylic acids like but-2-enedioic acid react with two moles of NaOH per mole of acid due to the presence of two acidic proton-donating groups.
Chemical Tests Table
✅ Correct Test Outcomes
Complete grid mapping functional groups to chemical tests:
| Structure | Test with Bromine Water | Test with Acidified K₂Cr₂O₇ |
|---|---|---|
| HOOCCH=CHCOOH | ✔ (decolourises) | ✖ (no reaction) |
| HOCH₂CH=CHCH₂COOH | ✔ (decolourises) | ✔ (turns green) |
| CH₃CH₂CH₂CH₂COOH | ✖ (no reaction) | ✖ (no reaction) |
E/Z Stereoisomerism
✅ Correct Structures & Explanation
- E-isomer: Trans-configuration where high-priority groups (e.g., COOH groups) are on opposite sides across the C=C double bond.
- Z-isomer: Cis-configuration where similar/high-priority groups are on the same side across the C=C double bond.
- Why E/Z exists:
1. Restricted or limited rotation about the C=C double bond.
2. Each carbon atom in the double bond is attached to two different groups or atoms.
❌ Common Pitfalls
Avoid saying "carbons are attached to different molecules." Examiners strictly require that each carbon of the C=C bond is bonded to two different groups or atoms.
Topics
Organic Chemistry · Physical Chemistry · Core Practicals · Core Practical 2: Preparation of a standard solution from a solid acid · Core Practical 3: Find the concentration of a solution of hydrochloric acid · Core Practical 15: Analyse organic and inorganic unknowns · Topic 7: Modern Analytical Techniques I · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.