OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 21

13 marks · Medium difficulty · Structured Questions

Analyze isotopes of sulfur and magnesium, draw and explain metallic bonding in magnesium, and determine the concentration of hydrochloric acid and collision theory factors from reaction rate gas collection data.

Practise this question

Question

Question 21 consists of five parts: (a) A table to complete number of protons, neutrons, and electrons for 34S and an unknown species with 16 neutrons and 18 electrons. (b) A mass spectrum for magnesium showing peaks at m/z 24 (75.45%), 25 (15.21%), and 26 (9.34%), asking to calculate the relative atomic mass to 2 decimal places. (c) Prompts to draw a labelled diagram of giant metallic bonding in magnesium with correct charges and explain electrical conductivity. (d) An experiment reacting excess Mg with 24.5 cm³ HCl, showing a conical flask connected to a 100 cm³ gas syringe, and a graph of gas volume vs. time plateauing at 60.0 cm³: (i) calculate the concentration of HCl to 2 significant figures in standard form, and (ii) explain the shape of the graph in terms of concentration, collisions, and time.
Question text

21 This question is about atomic structure and isotopes of two elements in Period 3.

(a) This part of the question is about sulfur.

Complete the table below.

Species Protons Neutrons Electrons

34S 16

16 18

[2]

(b) The mass spectrum of a sample of magnesium is shown below.

75.45%

Relative

abundance

15.21%

9.34%

24 25 26

m/z

Calculate the relative atomic mass of magnesium in the sample.

Give your answer to 2 decimal places.

Relative atomic mass = … [2]

(c) Magnesium has a giant metallic structure.

• Draw a labelled diagram to show the bonding and structure in magnesium.

Include the correct charges on the metal particles in your diagram.

• Explain how the structure of magnesium allows it to conduct electricity.

Diagram

Explanation …

[3]

(d) Magnesium reacts with dilute hydrochloric acid, as shown in the equation.

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

A student reacts an excess of Mg with 24.5 cm3 HCl(aq) using the apparatus below.

100 cm3 gas syringe

Dilute hydrochloric acid

Magnesium

The student measures the volume of gas collected over time. All gas volumes are measured at

RTP.

A graph of the student’s results is shown in Fig. 21.1 below.

Fig. 21.1

70.0

60.0

50.0

Volume 40.0

of gas

collected

/ cm3

30.0

20.0

10.0

0.0

05 10 15 20 25 30 35 40

Time/s

(i) Calculate the concentration of HCl that the student used.

Give your answer to 2 significant figures and in standard form.

Concentration of HCl = … mol dm–3 [3]

(ii) Explain the shape of the graph in Fig. 21.1, in terms of concentration, collisions and time.

… [3]

Mark scheme

Show the mark scheme Mark scheme for Question 21: (a) Row 1: 16 protons, 18 neutrons; Row 2: 32S2- species, 16 protons (2 marks). (b) Calculation showing (75.45*24 + 15.21*25 + 9.34*26)/100 = 24.34 (2 marks). (c) Diagram with regular arrangement of labelled 2+ ions surrounded by delocalised electrons, and explanation that delocalised electrons move/flow through the structure (3 marks). (d)(i) n(H2) = 60/24000 = 2.50 x 10^-3 mol, n(HCl) = 5.00 x 10^-3 mol, c(HCl) = 5.00 x 10^-3 / 0.0245 = 2.0 x 10^-1 mol dm^-3 (3 marks). (d)(ii) Explains acid concentration decreases, fewer collisions per second, and reaction stops when limiting reagent HCl is used up (3 marks).

Question Answer Mark Guidance

21 (a) 2

NOTE: Look out for the 2– on 32S2–

Species Protons Neutrons Electrons ALLOW 32S–2

34S 16 18 16

32S2– 16 16 18

4 correct

3 correct

21 (b) FIRST CHECK ANSWER ON THE ANSWER LINE IF answer = 2 For 1 mark: ALLOW ECF → to 2 DP if:

24.34 (to 2 DP) award 2 marks

• %s used with wrong isotopes ONCE

(75.45 24 + 15.21 25 + 9.34 26) OR

100 • transposed decimal places for ONE %

OR 24.3389 OR 24.339

= 24.34 (to 2 DP)

21 (c) 3 Regular arrangement must have at least 2 rows of

correctly charged ions and a minimum of 2 ions per

row

ALLOW as label: positive ions, cations if correct charge

9 is seen within circle

ALLOW for labelled Mg2+ ions: circles with Mg2+ inside

DO NOT ALLOW incorrect charge for ions e.g.+ , 3+ etc

2+ DO NOT ALLOW for label of ions:

Diagram showing a regular arrangement of labelled ‘Mg ions’ or

nuclei OR positive atom OR protons

‘2+ ions’

Scattering of labelled electrons between other species ALLOW e– or ‘e’ as a label for electron

AND IGNORE “–“ for electron label (i.e. needs an additional

statement anywhere of delocalised electrons (can be in text or in label)

diagram)

ALLOW ‘electrons are mobile (charge carriers)’ OR ‘sea

Delocalised electrons move/flow (in an electric field/across a p.d.) of electrons moves’

Idea of movement required IGNORE electrons are free

DO NOT ALLOW ions can move

Question Answer Answer Mark Guidance

21 (d) (i) FIRST CHECK ANSWER ON THE ANSWER LINE IF ALLOW ECF throughout

answer = 2.0 10–1 award 3 marks 3

---------------------------------------------------------------------10 ALLOW use of ideal gas equation for all 3 marks

60 –3 provided ‘sensible’ p and T used:

n(H2) = 24000 = 2.50 10 OR 0.0025(0) (mol)

e.g.

from 101 kPa and 298 K

n(HCl) = 5.00 10–3 OR 0.005(00) (mole ratio = 2:1) → nH2 = 0.0024459 → nHCl = 0.0048919 →

cHCl 0.1996688 → 2.0 10–1 (2SF and

–3 standard form)

5.00 10 –1

c(HCl) = 0.0245 = 0.204 = 2.0 10 from 100 kPa and 273 K

2 SF and standard form required → nH2 = 0.0026435→ nHCl = 0.005287 →

0.21579656 → 2.2 10–1 (2SF and standard

form)

Examples of ‘sensible’

p = 100 kPa, 101 kPa, 101,325 Pa

T = 273 – 298 K

Common errors (final answer)

Use of 24.5cm3 for H volume

8.3 10–2 → 2 marks

Incorrect units for H2 volume (i.e. 60/24)

2.0 102 → 2 marks

Not 2SF or standard form

0.204 → 2 marks

calculator 0.2040816327

21 (d) (ii) Concentration 11 For Concentration and Collisions, ALLOW ORA

(Acid) concentration decreases ✓ 3 for initial conditions (when t = 0), but answer must

be comparative (e.g. initially the concentration is

highest/higher)

ALLOW ‘fewer particles in a given volume/space’

for concentration decreases

IGNORE amount of acid decreases

IGNORE magnesium concentration decreases

(Mg is solid)

DO NOT ALLOW (line increases so) rate

increases

IGNORE references to energy even if incorrect

Collisions IGNORE responses not linked to time, e.g.

Fewer collisions per second • fewer successful collisions

OR less frequent collisions ✓ • fewer collisions, less chance of collisions

No link to rate.

Reaction stops/graph plateaus ALLOW idea that reaction stops e.g. no more

(Acid/reactant/limiting reagent) has reacted/been collisions

used up ✓ DO NOT ALLOW Magnesium/reactantS used up

(Magnesium is in excess)

How to answer it

Period 3 Elements: Structure, Bonding & Reaction Kinetics

📋 WHAT THIS QUESTION TESTS

Core AS-Level Fundamental Concepts:

  • Atomic Structure & Isotopes: Calculating subatomic particles in neutral atoms and negative ions (sulfide).
  • Mass Spectrometry: Calculating relative atomic mass (Ar) from isotopic abundances to 2 decimal places.
  • Giant Metallic Lattice: Drawing accurate metallic bonding diagrams (Mg²⁺ cations and delocalised electrons) and linking structure to conductivity.
  • Stoichiometry & Gas Volumes: Calculating molar quantities from gas volume at RTP (24.0 dm³ mol⁻¹), applying reacting ratios, and expressing final concentration in standard form to 2 significant figures.
  • Collision Theory & Rates: Explaining experimental rate curves through concentration changes, collision frequency with time, and reaction termination.

Part (a): Subatomic Particles of Sulfur Isotopes & Ions

Identifying protons, neutrons, and electrons [2 Marks]

✅ Correct Answer

Species Protons Neutrons Electrons
³⁴S 16 18 16
³²S²⁻ 16 16 18
Mark Scheme:
• 4 correct entries = [2 marks]
• 3 correct entries = [1 mark]
Note: ³²S²⁻ must include both mass number 32 and charge 2– (also allowed: S²⁻-32 or ³²S⁻²).

💡 Key Knowledge

  • Sulfur has atomic number Z = 16, so all sulfur species must possess exactly 16 protons.
  • Neutrons = Mass Number (A) − Atomic Number (Z). For ³⁴S: 34 − 16 = 18.
  • For the second row: Mass Number = Protons + Neutrons = 16 + 16 = 32.
  • Charge: Protons (16) − Electrons (18) = −2, giving the sulfide ion symbol ³²S²⁻.

❌ Common Errors

  • Writing just S²⁻ without the mass number 32.
  • Writing an incorrect charge sign such as ³²S²⁺ by subtracting protons from electrons backwards.

🧠 Exam Technique

Always cross-check the Periodic Table for the element's atomic number first. Remember: identity is defined solely by proton number.

Part (b): Relative Atomic Mass Calculation

Determining Ar of magnesium from mass spectra [2 Marks]

📐 Step-by-Step Calculation

  1. Formula:
    Ar = Σ(isotope mass × % abundance) ÷ 100
  2. Substitute values:
    Ar = [(24 × 75.45) + (25 × 15.21) + (26 × 9.34)] ÷ 100
  3. Calculate intermediate sum:
    Ar = [1810.80 + 380.25 + 242.84] ÷ 100 = 2433.89 ÷ 100 = 24.3389
  4. Round to 2 decimal places:
    24.34
Mark Breakdown:
• 1st mark: Correct mathematical expression working (or unrounded 24.3389 / 24.339).
• 2nd mark: 24.34 clearly stated to 2 decimal places.

❌ Common Traps & Examiner Insight

  • Ignoring decimal place instructions: Quoting 24.3 (1 d.p.) or 24.339 (3 d.p.) loses the second mark immediately.
  • Transcription errors: Swapping the abundances (e.g. using 15.12% instead of 15.21%).
  • Sanity check: The answer must lie strictly between 24 and 26, weighted heavily towards 24. If your answer is outside this range, check your calculation!

Part (c): Giant Metallic Structure & Electrical Conductivity

Diagram and structural explanation [3 Marks]

✅ Mark Scheme Criteria

Required Diagram Features:
  • Regular 2D lattice: At least 2 rows with a minimum of 2 ions per row neatly arranged and touching/nearly touching.
  • Cation Label & Charge: Circles clearly marked with 2+ inside (or labeled Mg²⁺ ).
  • Delocalised Electrons: Small dots or e⁻ dispersed evenly between the cations, explicitly labelled "delocalised electrons".

Explanation:

The delocalised electrons are free to move/flow through the structure (when a potential difference is applied).

• Mark 1: Regular arrangement of labelled Mg²⁺ (or 2+) ions.
• Mark 2: Labelled delocalised electrons scattered between ions.
• Mark 3: Explanation that delocalised electrons move / flow.

🧠 Examiner Commentary: Diagram Dos & Don'ts

  • DO NOT write simply + for magnesium. Magnesium is in Group 2, so ions are 2+.
  • DO NOT label the circles as "protons", "nuclei", or "atoms". They must be labelled cations, positive ions, or Mg²⁺ ions.
  • DO NOT just write "electrons are free" for the explanation mark. You must mention their movement/flow or state they act as mobile charge carriers.
  • Never state that the "ions move" when explaining metallic conductivity — the cations are fixed in their lattice positions!

Part (d)(i): Gas Syringe Stoichiometry Calculation

Finding the concentration of hydrochloric acid [3 Marks]

📐 Step-by-Step Calculation

Reaction: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

  1. Read final volume of H₂ gas from Fig. 21.1:
    The plateau occurs at exactly 60.0 cm³.
  2. Calculate moles of H₂ gas produced (at RTP):
    n(H₂) = Volume ÷ 24000 cm³ mol⁻¹
    n(H₂) = 60.0 ÷ 24000 = 2.50 × 10⁻³ mol (0.00250 mol)
  3. Use stoichiometric ratio to find moles of HCl reacted:
    From the equation, 1 mol H₂ : 2 mol HCl
    n(HCl) = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol (0.00500 mol)
  4. Calculate concentration of HCl:
    Volume of HCl = 24.5 cm³ = 0.0245 dm³
    c = n ÷ V
    c(HCl) = (5.00 × 10⁻³) ÷ 0.0245 = 0.20408... mol dm⁻³
  5. Format as specified (2 SF and standard form):
    2.0 × 10⁻¹ mol dm⁻³

❌ Common Errors & Lost Marks

  • Forgetting the 1:2 stoichiometry: Many students miss multiplying by 2, finding c = 1.0 × 10⁻¹ mol dm⁻³.
  • Volume confusion: Using 24.5 cm³ for the gas formula instead of 60.0 cm³.
  • Failing format requirements: Writing 0.20 (not standard form) or 2.04 × 10⁻¹ (3 SF instead of 2 SF). Both lose the final mark!
  • Unit conversion errors: Dividing 60 by 24 without converting cm³ to dm³.
Marks Awarded:
• Mark 1: n(H₂) = 2.50 × 10⁻³ mol
• Mark 2: n(HCl) = 5.00 × 10⁻³ mol
• Mark 3: 2.0 × 10⁻¹ mol dm⁻³ (strictly 2 SF in standard form)

Part (d)(ii): Rates & Collision Theory Explanation

Explaining the curve in terms of concentration, collisions, and time [3 Marks]

✅ Model Answer (3 Essential Points)

  1. 1. Concentration:
    As the reaction proceeds, the concentration of acid decreases (or is highest at the start).
  2. 2. Collisions:
    This causes fewer collisions per second (or less frequent collisions / reduced collision rate between reactant particles).
  3. 3. Plateau / Time:
    The graph levels off / reaction stops because the acid is completely used up (acid is the limiting reagent).
• 1 mark for decrease in acid concentration.
• 1 mark for fewer collisions per second / less frequent collisions.
• 1 mark for reaction stops because acid / limiting reagent is used up.

🧠 Exam Technique: Examiner Warnings

  • The "Per Second" Rule: Never just say "there are fewer collisions" or "less chance of collisions". You must link collision frequency to time: write "fewer collisions per unit time" or "less frequent collisions".
  • Identify the Limiting Reagent: The question states magnesium is in excess. Saying "the reactants are used up" or "magnesium is used up" is factually incorrect and scores 0 for the plateau mark. Specify that the acid is used up.
  • Avoid talking about energy: This question is about changing concentration at constant temperature. Talking about particles having less kinetic energy is incorrect.

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 1: Development of practical skills in chemistry · Practical Activity Groups · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 3.2 Physical chemistry · 1.1 Practical skills assessed in a written examination · PAG 9: Rates of reaction – continuous monitoring method

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.