OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 2

14 marks · Medium difficulty · Structured Questions

Draw skeletal structures, perform mole and particle calculations, explain phenol acidity compared to carboxylic acids, and calculate the pH of a diluted barium hydroxide solution.

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Question

Question text

2 This question is about different compounds that each contain O and H atoms covalently bonded

together.

(a) The alcohol 2,3‑dimethylhexan‑1‑ol has the molecular formula C8H18O.

(i) Draw the skeletal formula of 2,3‑dimethylhexan‑1‑ol.

[1]

(ii) What is the number of H atoms in 8.45 g of 2,3‑dimethylhexan‑1‑ol?

Give your answer to 3 significant figures and in standard form.

Number of H atoms = … [3]

(b) Nuromol tablets are used to relieve pain.

Nuromol tablets contain a mixture of ibuprofen and paracetamol.

The structures of ibuprofen and paracetamol are shown.

H3C CH OH

2 O O

C C

CH3 H C

CH OH 3 N

CH3

Ibuprofen Paracetamol

A student carries out an investigation to determine the mass of ibuprofen and the mass of

paracetomol in one nuromol tablet.

• Ibuprofen reacts with aqueous sodium carbonate, Na2CO3(aq), to produce a gas.

2RCOOH + Na2CO3 2RCOONa + CO2 + H2O

• Paracetamol does not react with Na2CO3(aq).

(i) Why does paracetamol not react with Na2CO3(aq)?

… [1]

(ii) The student uses the reaction of nuromol tablets with Na2CO3(aq) to determine the mass of

ibuprofen and the mass of paracetomol in one nuromol tablet.

The student’s method is outlined below.

• Weigh four nuromol tablets.

• Add the tablets to an excess of aqueous sodium carbonate.

• Collect the gas produced and measure its volume.

Results

Mass of four nuromol tablets = 2.80 g

Volume of gas collected at RTP = 48.0 cm3

Assume that ibuprofen is the only compound in nuromol tablets that reacts with Na2CO3(aq).

Determine the mass of ibuprofen and the mass of paracetomol, in mg, in one nuromol tablet.

Mass of ibuprofen = … mg

Mass of paracetamol = … mg

[4]

A student is supplied with a solution of Ba(OH) with a concentration of 46.80 g dm–3 at 25 °C.

The student dilutes 20.0 cm3 of this solution of Ba(OH) with water and the solution is made up to

100.0 cm3 at 25 °C.

Calculate the pH of the diluted solution of Ba(OH)2 at 25 °C.

Give your answer to 2 decimal places.

pH = … [5]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

2 (a) (i) 1 Structure MUST be skeletal (NO CH3).

BUT

ALLOW O–H

DO NOT ALLOW OH–

✓ ALLOW any vertical bond to the OH group

i.e. ALLOW

Count main carbon chain first (watch for only 5 or 7 Cs)

Watch for correct reversed orientation:

IGNORE any structural formulae, etc

Treat as working

(a) (ii) FIRST, CHECK ANSWER 3

IF answer = 7.04 1023 OR 7.05 1023, award 3 marks

--------------------------------------------------------------------

8.45 ALLOW ECF throughout

n(C8H18O) = 130.0 OR 0.065(0) (mol) ✓

Alternative working

18 and 6.02 1023 in different order

n(H atoms) = 18 0.0650 OR 1.17✓

23 0.0650 6.02 1023 OR 3.913 1022 ✓

H atoms = 1.17 6.02 10

= 7.04 1023 ✓

3.913 1022 18 = 7.04 1023 ✓

3 SF and standard form required -------------------------------------------------------

Common errors

23 23 3.91 1022 → 2 marks No 18

ALLOW use of 6.022 10 → 7.05 10

2.17 1021 → 2 marks ÷18

7.0434 1023 → 2 marks Not 3 SF

5.42 1021 → 2 marks ÷130 twice

ALLOW 5.40–5.42 1021 (intermediate rounding)

(b) (i) Assume that ‘It’ means paracetamol 1 IGNORE

It is too weak an acid/not strong enough

It is a phenol It is a weak acid So is COOH

OR It does not react with Na2CO3 In Q

It is not a carboxylic acid/no carboxyl group/no carboxylic It contains amide OR benzene ring

It does not contain COOH ✓ IGNORE reference to an alcohol/hydroxyl group

A correct functional group must be in answer DO NOT ALLOW It is not acidic It is!

It is a phenol which is not acidic is a CON

It contains ketone OR amine CON

(b) (ii) FIRST, CHECK ANSWER 4 ECF throughout

IF Mass ibuprofen = 206 mg ----------------------------------------

AND mass paracetamol = 494 mg award 4 marks Alternative order for 1st 3 marks:

-------------------------------------------------------------------- n(CO ) from 1 tablet = 5 10–4 (mol) ✓

48.0 –3

n(CO2) = = 2 10 (mol) ✓ –3

24000 n(ibuprofen) in 1 tablet = 1 10 (mol) ✓

n(ibuprofen) = 2 2 10–3 = 4 10–3 (mol) ✓ Mass of ibuprofen = 1 10–3 206 = 0.206 g ✓

In 4 tablets Common errors

Mass of ibuprofen = 4 10–3 206 = 0.824 g ✓ 103 AND 597 3 marks no 2 for n(ibuprofen)

824 AND 1976 3 marks no ÷4 for 1 tablet

In 1 tablet (÷4) 824 + 1976 = 2800

Mass ibuprofen = 206 mg -----------------------------------------------------------------

AND mass paracetamol = 494 mg ✓ ALLOW use of ideal gas equation with a sensible

temperature (290–298K)

NOTE: Mass ibuprofen + mass paracetamol in 1 tablet = 700 mg and pressure (100/101/101325 kPa)

Mass ibuprofen + mass paracetamol in 4 tablets = 2800

mg e.g. At 298 K and 100 kPa,

100 103 48.0 10–6

n(CO ) = = 1.937…. 10–3 (mol)

2 8.314 298

ALLOW use of 8.31 for R

298K → = 1.938…. 10–3 (mol)

IF pH = 13.04 award 5 marks ALLOW acceptable rounding written on scripts and take

-------------------------------------------------------------------- care as candidate may carry their calculator answer.

[Ba(OH)2]

Alternatives for 1st and 2nd marks:

46.80 mass in 20 cm3 = 100 cm3

OR 0.273….. mol dm–3 ✓

171.3 = 46.80 0.02 = 0.936 g ✓

Calculator: 0.2732049037

0.936

Concentration of diluted Ba(OH)2 [Ba(OH)2] = 10 = 0.0546…. ✓

171.3

----------------------------------------------

0.273….. –3 46.80 –3

5 = 0.0546…… (mol dm ) ✓ [Ba(OH)2] in diluted solution = = 9.36 (g dm ) ✓

Calculator: 0.05464098074

9.36

[Ba(OH)2] = = 0.0546…. ✓

– 171.3

[OH ]

–3 ----------------------------------------------

2 0.0546…… = 0.1092 … (mol dm ) ✓

For pH marks

Calculator 0.1092819615

ALLOW

–14 pOH = –log 0.1092… = 0.96…. ✓

+ 1.00 10 –14 –3

[H ] = 0.1092… OR 9.15…… 10 (mol dm ) ✓

–14 pH = 14 – 0.96 = 13.04 ✓

Calculator 9.150641026 10 ------------------------------------------------

–14 Common errors

pH = –log(9.15…… 10 ) = 13.04 to 2 DP ✓ pH = 11.04 4 marks ÷10 instead of 10 in alternative

pH = 12.44 4 marks ÷2 instead of 2 for OH–

pH = 12.74 4 marks No 2 for OH–

No pH mark without use of Kw OR pOH

pH = 13.74 4 marks No ÷5

pH = 15.27 4 marks No ÷171.3

DO NOT ALLOW a pH value ≤ 7 for last mark

pH = 13.44 3 marks No ÷5 and no 2

pH = 13.97 3 marks No ÷171.3 and no 2

pH = 14.27 3 marks No ÷171.3 and 10

pH = 14.44 2 marks No ÷5 and no 2 and 10

pH = 15.67 2 marks Kw and pH only

How to answer it

Organic Analysis, Tablet Stoichiometry & Strong Base pH

📋 What This Question Tests

This multi-topic question tests fundamental organic representation and quantitative analytical chemistry across Module 2, Module 4, and Module 5:

Skeletal Drawing: Accurate skeletal representations of branched aliphatic alcohols without showing CH₃ or CH₂ vertices.
Avogadro & Moles: Calculating the absolute number of atoms within a covalent compound using molar mass, Avogadro's constant (N_A), and stoichiometric ratios.
Acid-Base Organic Properties: Differentiating weak acids (phenols vs carboxylic acids) via their reaction with aqueous metal carbonates.
Stoichiometric Mixture Analysis: Using gas molar volume at RTP (24 000 cm³ mol⁻¹), reacting mole ratios (2:1), mass calculations, and per-tablet unit conversions (g to mg).
Strong Base Dilution & pH: Calculating molarity from g dm⁻³, applying dilution factors, accounting for diprotic hydroxide dissociation ([OH⁻] = 2 × [Ba(OH)₂]), and using K_w to find pH to 2 decimal places.

Part (a)(i) — Drawing Skeletal Formula of 2,3-dimethylhexan-1-ol

1 Mark

✅ Correct Answer

A 6-carbon zigzag main chain with:

An -OH group attached to carbon 1 (end of the chain).
A single line (methyl group) branching off carbon 2.
A single line (methyl group) branching off carbon 3.

Examiner Drawing Advice: Draw a 6-carbon zigzag (5 vertex/end points after C1). Extend carbon 1 with a bond to OH (or O-H ). At carbon 2 (the vertex adjacent to C1) and carbon 3, draw single downward/upward sticks representing the two methyl branches.

❌ Common Errors & Examiner Traps

Writing carbon labels: Writing CH₃ at terminal ends loses the mark immediately—skeletal structures must not show terminal carbon atoms.
Miscounting chain length: Drawing 5 or 7 carbons in the parent chain instead of 6. Always number your backbone (C1 to C6).
Incorrect connectivity: Writing HO- bonded through the hydrogen atom when drawn on the right-hand side. The bond from carbon must attach directly to oxygen: C-OH .

Mark Scheme: 1 mark for correct skeletal formula. Main chain must contain 6 carbons. Allow vertical or angled bond to OH. Allow O-H bond shown. Do NOT allow OH-.

Part (a)(ii) — Number of H Atoms in 8.45 g of 2,3-dimethylhexan-1-ol

3 Marks

📐 Step-by-Step Calculation

Step 1: Calculate molar mass of C₈H₁₈O

M(C₈H₁₈O) = (8 × 12.0) + (18 × 1.0) + 16.0 = 130.0 g mol⁻¹

Step 2: Calculate moles of C₈H₁₈O

n(C₈H₁₈O) = 8.45 / 130.0 = 0.0650 mol

[Mark 1]

Step 3: Account for number of H atoms per molecule

Each molecule of C₈H₁₈O contains 18 hydrogen atoms:

n(H atoms) = 0.0650 × 18 = 1.17 mol of H atoms

[Mark 2]

Step 4: Multiply by Avogadro's constant (N_A = 6.02 × 10²³ mol⁻¹)

Number of H atoms = 1.17 × (6.02 × 10²³) = 7.0434 × 10²³

Rounding to 3 significant figures and standard form: 7.04 × 10²³ (or 7.05 × 10²³ if using 6.022 × 10²³).

[Mark 3]

❌ Common Errors

Forgetting to multiply by 18: Calculating 0.0650 × 6.02 × 10²³ = 3.91 × 10²² gives the number of molecules, not the number of H atoms (loses 1 mark).
Incorrect Significant Figures: Writing 7.043 × 10²³ loses the final mark; the prompt strictly demanded 3 sig figs.
Dividing instead of multiplying: Writing 0.0650 / 18 .

🧠 Exam Technique

Always double-check what the question specifies: "number of H atoms" vs "number of molecules". Also, circle formatting constraints like 3 significant figures and standard form immediately so you don't drop the final mark at the very end.

Mark Scheme:
• Mark 1: Moles of C₈H₁₈O = 0.065(0) mol.
• Mark 2: Moles of H atoms = 1.17 mol (or multiplying molecules by 18).
• Mark 3: 7.04 × 10²³ (or 7.05 × 10²³). Requires 3 SF and standard form.

Part (b)(i) — Why Paracetamol Does Not React with Na₂CO₃(aq)

1 Mark

✅ Correct Answer

Any one of the following:

Paracetamol contains a phenol group (which is not acidic enough to react with carbonates).
Paracetamol is not a carboxylic acid / contains no carboxyl group (-COOH).

💡 Key Knowledge: Acidity of Organic Compounds

Relative acidities follow this hierarchy:

Carboxylic acids > Phenols > Alcohols / Water

Carboxylic acids: React with carbonates (Na₂CO₃), hydrogencarbonates (NaHCO₃), and strong bases (NaOH).
Phenols: React with strong bases (NaOH), but are too weakly acidic to react with weak bases like carbonates (Na₂CO₃).
Alcohols: Do not react with either NaOH or Na₂CO₃.

❌ Examiner Trap & Rejected Answers

DO NOT say "paracetamol is not acidic": Phenol is acidic (it has a pK_a around 10), so claiming it is "not an acid" is scientifically incorrect and penalized.
DO NOT just say "it is too weak an acid": The mark scheme requires naming the functional group responsible (phenol or lacks carboxyl group).
DO NOT confuse with alcohol: Referring to the phenolic -OH as an "alcohol group" is ignored or rejected.

Mark Scheme: 1 mark for stating paracetamol is a phenol OR does not contain a carboxylic acid / carboxyl group / -COOH. A correct functional group must be in the answer.

Part (b)(ii) — Determining Mass of Ibuprofen & Paracetamol in One Tablet

4 Marks

📐 Step-by-Step Calculation

Step 1: Moles of CO₂ gas collected (from 4 tablets)

n(CO₂) = Volume / Molar gas volume = 48.0 / 24 000 = 2.00 × 10⁻³ mol

[Mark 1]

Step 2: Moles of ibuprofen in 4 tablets using stoichiometry

From the equation: 2 RCOOH + Na₂CO₃ → 2 RCOONa + CO₂ + H₂O

The reacting ratio is 2 ibuprofen : 1 CO₂:

n(ibuprofen in 4 tablets) = 2 × (2.00 × 10⁻³) = 4.00 × 10⁻³ mol

[Mark 2]

Step 3: Mass of ibuprofen

Find molar mass of ibuprofen (C₁₃H₁₈O₂):

M = (13 × 12.0) + (18 × 1.0) + (2 × 16.0) = 206.0 g mol⁻¹

Mass in 4 tablets = 4.00 × 10⁻³ × 206.0 = 0.824 g

Mass in ONE tablet = 0.824 / 4 = 0.206 g = 206 mg

[Mark 3]

Step 4: Mass of paracetamol in ONE tablet

Total mass of 1 tablet = 2.80 g / 4 = 0.700 g = 700 mg

Mass of paracetamol = 700 mg - 206 mg = 494 mg

(Alternatively: total paracetamol in 4 tablets = 2.80 - 0.824 = 1.976 g = 1976 mg; per tablet = 1976 / 4 = 494 mg)

[Mark 4]

❌ Common Errors

Forgetting the 2:1 ratio: Using a 1:1 ratio yields 103 mg ibuprofen and 597 mg paracetamol (loses Mark 2).
Forgetting to divide by 4: Leaving answers as 824 mg and 1976 mg gives the mass in four tablets instead of one tablet (loses Mark 4).
Forgetting unit conversion: Leaving values in grams (0.206 g and 0.494 g) when the question specifies mg.

🧠 Exam Technique

Always highlight the number of units used in the experiment ("four tablets") versus the quantity required in the prompt ("in one tablet"). You can either divide the CO₂ moles by 4 at the start, or scale down by 4 at the end.

Mark Scheme:
• Mark 1: n(CO₂) = 2 × 10⁻³ mol.
• Mark 2: n(ibuprofen) = 4 × 10⁻³ mol (or 1 × 10⁻³ mol in 1 tablet).
• Mark 3: Mass of ibuprofen in 4 tablets = 0.824 g (or 0.206 g in 1 tablet).
• Mark 4: Both final answers in mg: Ibuprofen = 206 mg AND Paracetamol = 494 mg.

Part (c) — pH Calculation for Diluted Barium Hydroxide Solution

5 Marks

📐 Step-by-Step Calculation

Step 1: Calculate concentration of stock Ba(OH)₂ in mol dm⁻³

M(Ba(OH)₂) = 137.3 + (2 × 16.0) + (2 × 1.0) = 171.3 g mol⁻¹

[Ba(OH)₂]_stock = 46.80 / 171.3 = 0.2732... mol dm⁻³

[Mark 1]

Step 2: Calculate concentration after dilution

Dilution factor = 20.0 cm³ / 100.0 cm³ = 1/5 = 0.2

[Ba(OH)₂]_diluted = 0.2732... × (20.0 / 100.0) = 0.05464... mol dm⁻³

[Mark 2]

Step 3: Calculate [OH⁻] taking stoichiometry into account

Ba(OH)₂ is a strong diprotic base: Ba(OH)₂ → Ba²⁺ + 2 OH⁻

[OH⁻] = 2 × 0.05464... = 0.10928... mol dm⁻³

[Mark 3]

Step 4: Calculate [H⁺] using K_w (at 25 °C, K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶)

[H⁺] = K_w / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.10928... = 9.1506... × 10⁻¹⁴ mol dm⁻³

(Alternatively, use pOH: pOH = -log(0.10928...) = 0.9614...)

[Mark 4]

Step 5: Calculate pH to 2 decimal places

pH = -log[H⁺] = -log(9.1506... × 10⁻¹⁴) = 13.0385... ≈ 13.04

(Or pH = 14 - pOH = 14 - 0.9614... = 13.04)

[Mark 5]

❌ Common Errors & Lost Marks

Forgetting ×2 for OH⁻ (pH = 12.74): Forgetting that barium hydroxide releases 2 moles of OH⁻ per mole of Ba(OH)₂ is the most common error. Scores max 4 marks.
Forgetting dilution (pH = 13.74): Calculating pH of the undiluted solution scores max 4 marks.
Inverting dilution factor (pH = 11.04): Multiplying by 5 instead of dividing by 5.
Rounding too early: Rounding intermediate values to 2 SF leads to rounding drift (e.g. 13.03 or 13.05). Keep all values in your calculator memory!
Incorrect Decimal Places: Writing 13 or 13.0 loses the final mark. All pH values must be given to 2 decimal places.

🧠 Top Exam Tip

Whenever you see a Group 2 hydroxide like Ba(OH)₂, Ca(OH)₂, or Sr(OH)₂, immediately circle the subscript '2'. Train yourself to write [OH⁻] = 2 × [base] before doing anything else.

Mark Scheme:
• Mark 1: [Ba(OH)₂] stock = 0.273... mol dm⁻³.
• Mark 2: [Ba(OH)₂] diluted = 0.0546... mol dm⁻³.
• Mark 3: [OH⁻] = 0.1092... mol dm⁻³ (accounts for 2 × OH⁻).
• Mark 4: [H⁺] = 9.15... × 10⁻¹⁴ mol dm⁻³ OR pOH = 0.96...
• Mark 5: pH = 13.04 (strictly 2 decimal places; DO NOT allow pH ≤ 7).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis · 5.1 Rates, equilibrium and pH · 6.1 Aromatic compounds, carbonyls and acids

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.