OCR A-Level Chemistry Unified chemistry (03), June 2019: Question 6

12 marks · Hard difficulty · Structured Questions

Identify compounds and write equations from an iron reaction flowchart, then determine the formulae of hydrated compound C and its thermal decomposition products D, E, F, and G using quantitative data.

Practise this question

Question

Question 6 consists of two parts. Part (a) features a reaction flowchart starting with iron(II) sulfide reacting with H+/H2O to yield pale green [Fe(H2O)6]2+(aq) and H2S(g). The hexaaquairon(II) ion reacts with Cl2(g) to form an orange-brown solution, which forms precipitate A upon adding NaOH. H2S reacts with AgNO3 to form black precipitate B (molar mass 247.9 g/mol), and with acidified permanganate to form Mn2+ and a yellow solid. Sub-questions ask for the formulae of A and B, an explanation of whether the first step is redox, and balanced equations for the reactions of [Fe(H2O)6]2+ with Cl2 and H2S with acidified MnO4-. Part (b) is a 6-mark starred extended response detailing the thermal decomposition of compound C (empirical formula FeH18N3O18) in three stages to deduce the formulae of C, D, E, F, and G with working and chemical equations.
Question text

6 This question is about reactions of iron compounds.

(a) A student carries out the reactions in the flowchart, starting with iron(II) sulfide.

H+/H O

iron(II) sulfide 2 [Fe(H O) ]2+(aq) + H S (g)

26 2

pale green

solution

Cl (g) AgNO (aq) H+(aq)/MnO −(aq)

23 4

orange-brown colourless solution

solution containing Mn2+ ions

black precipitate B +

NaOH(aq) with a molar mass yellow solid

of 247.9 g mol–1

orange-brown

precipitate A

(i) In the boxes, write the formulae of A and B. [2]

(ii) The student thinks that the reaction of iron(II) sulfide with H+/ H O is a redox reaction.

Explain, with reasons, whether the student is correct.

… [1]

(iii) Write the equation for the reaction of [Fe(H O) ]2+(aq) with Cl (g).

26 2

… [1]

(iv) Construct an equation for the reaction of H S(g) with H+(aq)/MnO −(aq).

(b)* Compound C is a hydrated ionic compound with the empirical formula: FeH18N3O18.

A student investigates the thermal decomposition of compound C as outlined below.

Stage 1

The student gently heats 0.00300 mol of compound C to remove the water of crystallisation.

0.486 g of water is collected, leaving 0.00300 mol of the anhydrous compound D.

[2]

Stage 2

The student strongly heats 0.00300 mol of compound D, which decomposes to form a solid

oxide E (molar mass of 159.6 g mol−1) and 270 cm3 of a gas mixture, measured at RTP,

containing gases F and G.

Stage 3

The student cools the 270 cm3 gas mixture of F and G.

• Gas F is a compound that condenses to form 0.414 g of a liquid.

• Gas G remains and has a volume of 54 cm3, measured at RTP.

Gas G is tested and it relights a glowing splint.

Determine the formulae of C, D, E, F and G.

Show all your working and equations for the reactions. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme Mark scheme for Question 6. (a)(i) gives A as Fe(OH)3(s) and B as Ag2S(s) (2 marks). (a)(ii) explains that the student is incorrect because no oxidation numbers change (1 mark). (a)(iii) provides the balanced equation: 2[Fe(H2O)6]2+ + Cl2 -> 2[Fe(H2O)6]3+ + 2Cl- (1 mark). (a)(iv) gives 5H2S + 2MnO4- + 6H+ -> 2Mn2+ + 5S + 8H2O (2 marks). (b) is a level of response mark scheme for 6 marks, identifying C as Fe(NO3)3·9H2O, D as Fe(NO3)3 or FeN3O9, E as Fe2O3, F as NO2, and G as O2, showing calculations of moles of water of crystallisation and gas volumes at RTP, and decomposition equations.

AO

Question Answer Marks Guidance

element

6 (a) (i) A: Fe(OH)3(s) 2 AO3.1 ALLOW Fe(OH)3(H2O)3

×2

B: Ag2S(s) IGNORE state symbols

(ii) Student is incorrect 1 AO3.2

AND

No oxidation numbers change

OR example, e,g, Fe stays as +2 ALLOW no electron transfer

(iii) 1 AO3.1 ALLOW multiples

2[Fe(H O) ]2+ + Cl → 2[Fe(H O) ]3+ + 2Cl– e.g. [Fe(H O) ]2+ + ½Cl → [Fe(H O) ]3+ + Cl–

26 2 2 6 2 6 2 2 6

ALLOW

2[Fe(H O) ]2+ + Cl → 2[Fe(H O) OH]2+ + 2HCl

26 2 2 5

OR

2[Fe(H O) ]2+ + Cl → 2[Fe(H O) Cl]2+ + 2H O

26 2 2 5 2

NOTE: equation MUST be balanced by charge

and oxidation number

IGNORE state symbols

(iv) 2 AO3.1 ALLOW multiples, e.g.

5H S + 2MnO – + 6H+ → 2Mn2+ + 5S + 8H O 2½ H S + MnO – + 3H+ → Mn2+ + 2½ S + 4H O

24 2 ×2 2 4 2

1st mark ALLOW equation with S2–-, e.g.

ALL Correct species (SIX) 5S2– + 2MnO – + 16H+ → 2Mn2+ + 5S + 8H O

OR

Equation containing Mn and S species correctly balanced

i.e. 5 H S + 2 MnO – ……. → 2 Mn2+ + 5 S … IGNORE extra electrons for 1st mark

2nd mark

Complete correct balanced equation

AO

19 element

(b)* Please refer to the marking instructions on page 4 of this mark 6 AO1.2 Indicative scientific points may include:

scheme for guidance on how to mark this question. ×2

Formula of C, D, E, F and G

Level 3 (5–6 marks) AO3.1 • C: Fe(NO ) •9H O OR FeN O •9H O

33 2 3 9 2

Reaches a comprehensive conclusion to determine the ×2

• D: FeN3O9 OR Fe(NO3)3

correct formulae of almost all of C, D, E, F, G AND 9H2O

AO3.2 • E: Fe2O3

There is a well-developed line of reasoning which is clear and ×2 • F: NO2

logically structured. • G: O

The information presented is relevant and substantiated.

• 9H2O

Level 2 (3–4 marks)

Reaches a sound conclusion to determine the correct Examples of evidence

formulae of at least half of C, D, E, F, G AND 9H2O. n(H O) = = 0.027 (mol)

There is a line of reasoning presented with some structure. 0.027 : 0.003 = 1 : 9 → 9H2O

The information presented is relevant and supported by some

evidence. n(F) = = = 0.009(00) (mol)

Level 1 (1–2 marks) M(E) = 55.8 × 2 + 16.0 × 3 = 159.6

Reaches a simple conclusion to determine the correct –1

M(F) = = 46 (g mol )

formulae of some of C, D, E, F, G AND 9H2O.

There is an attempt at a logical structure with a line of reasoning. G: oxygen linked to relighting glowing split

The information is in the most part relevant.

NOTE: Equations could include evidence

0 marks No response or no response worthy of credit. e.g

Fe(NO3)3•9H2O → Fe(NO3)3 + 9H2O

FeN3O9•9H2O → FeN3O9 + 9H2O

2Fe(NO3)3 → Fe2O3 + 6NO2 + 1½O2

Total 12

How to answer it

Reactions, Redox Chemistry & Quantitative Analysis of Iron Compounds

What this question tests

This synoptic question assesses core transition element chemistry, complex ion reactions, redox half-equations and balancing, alongside a 6-mark unstructured quantitative calculation involving thermal decomposition, molar gas volumes at RTP, water of crystallisation, and deducing chemical formulae.

Part (a) • Inorganic Flowchart & Redox Reactions • 6 Marks Total

(a)(i) Identifying Unknowns A and B

✅ Correct Answers

A: Fe(OH)₃ (or Fe(OH)₃(H₂O)₃)

B: Ag₂S

💡 Key Knowledge

  • Oxidising [Fe(H₂O)₆]²⁺ with Cl₂ forms [Fe(H₂O)₆]³⁺ (orange-brown). Adding OH⁻ causes deprotonation precipitation to form rust-orange Fe(OH)₃.
  • H₂S gas reacts with Ag⁺(aq) to form silver sulfide precipitate. Checking the molar mass: (2 × 107.9) + 32.1 = 247.9 g mol⁻¹ confirms Ag₂S.
Marks: 1 mark for formula of A, 1 mark for formula of B. State symbols are not required.

(a)(ii) Evaluating the FeS + Acid Reaction

✅ Correct Answer

Student is incorrect AND no oxidation numbers change / Fe remains +2 (and S remains -2, H remains +1) / no electrons are transferred.

❌ Common Errors

Writing "yes, because sulfur forms a gas" or stating that oxidation states change without explicitly checking: FeS + 2H⁺ → Fe²⁺ + H₂S is an acid-base / precipitation-dissolution reaction, not redox.

Mark: 1 mark (AO3.2) for both the conclusion and valid justification based on oxidation states or electron transfer.

(a)(iii) Oxidation of Hexaaquairon(II) by Chlorine

✅ Correct Equation

2[Fe(H₂O)₆]²⁺ + Cl₂ → 2[Fe(H₂O)₆]³⁺ + 2Cl⁻

Multiples allowed (e.g. dividing by 2 to use ½Cl₂).

🧠 Exam Technique

Always balance charges as well as atoms:

  • Left side: 2 × (+2) = +4
  • Right side: 2 × (+3) + 2 × (-1) = +4
Mark: 1 mark (AO3.1). Must be fully balanced for atoms, oxidation numbers, and overall charge.

(a)(iv) Redox Equation for H₂S and Acidified MnO₄⁻

✅ Correct Equation

5H₂S + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 5S + 8H₂O

💡 Derivation via Half-Equations

  1. Oxidation: H₂S → S + 2H⁺ + 2e⁻ × 5
  2. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O × 2
  3. Combine and cancel 10H⁺ on both sides to leave 6H⁺ on the left.
Marks (2 marks):
• 1st mark: All six species correct OR correct 5:2 ratio between H₂S/S and MnO₄⁻/Mn²⁺.
• 2nd mark: Fully balanced overall equation.
Part (b)* • Extended Response: Thermal Decomposition • 6 Marks

Deducing Formulae C, D, E, F, and G

Empirical formula of C given: FeH₁₈N₃O₁₈

🔨 Step-by-Step Calculation & Deduction

Step 1: Water of crystallisation (Stage 1)

  • n(H₂O) = 0.486 g / 18.0 g mol⁻¹ = 0.0270 mol
  • Ratio n(H₂O) : n(C) = 0.0270 / 0.00300 = 9
  • Compound C contains 9H₂O
  • Subtracting 9H₂O (H₁₈O₉) from FeH₁₈N₃O₁₈ leaves FeN₃O₉ → Fe(NO₃)₃
  • C = Fe(NO₃)₃•9H₂O (or FeN₃O₉•9H₂O)
  • D = Fe(NO₃)₃ (or FeN₃O₉)

Step 2: Solid Oxide E (Stage 2)

  • Molar mass of E = 159.6 g mol⁻¹
  • Iron(III) oxide: Fe₂O₃ = (2 × 55.8) + (3 × 16.0) = 111.6 + 48.0 = 159.6 g mol⁻¹
  • E = Fe₂O₃

Step 3: Identification of Gases F and G (Stage 3)

  • Gas G relights a glowing splint → G is O₂
  • V(G) = 54 cm³ → n(O₂) = 54 / 24000 = 0.00225 mol
  • V(F) = 270 - 54 = 216 cm³
  • n(F) = 216 / 24000 = 0.00900 mol
  • Molar mass of F = 0.414 g / 0.00900 mol = 46.0 g mol⁻¹
  • NO₂ = 14.0 + (2 × 16.0) = 46.0 g mol⁻¹ → F is NO₂

✅ Summary of Identified Formulae

  • C: Fe(NO₃)₃•9H₂O
  • D: Fe(NO₃)₃
  • E: Fe₂O₃
  • F: NO₂
  • G: O₂

✅ Balanced Chemical Equations

Stage 1 (Dehydration):
Fe(NO₃)₃•9H₂O → Fe(NO₃)₃ + 9H₂O

Stage 2 (Thermal Decomposition):
2Fe(NO₃)₃ → Fe₂O₃ + 6NO₂ + 1½O₂
or: 4Fe(NO₃)₃ → 2Fe₂O₃ + 12NO₂ + 3O₂

🧠 How to Hit Level 3 (5–6 Marks)

  • Comprehensive deduction: Accurately find formulae for all (or almost all) of C, D, E, F, G and show 9H₂O.
  • Logical structuring: Clearly separate each stage of the experiment in your written response.
  • Show all working: Explicitly write down n = m / M and n = V / 24000 steps with units.
  • Verification: Double-check stoichiometric ratios: 0.00300 mol D gives 0.00900 mol F (ratio 1 : 3) and 0.00225 mol G (ratio 1 : 0.75), perfectly matching the decomposition equation!

❌ Common Pitfalls

  • Forgetting to subtract 54 cm³ from 270 cm³ when finding the volume of gas F.
  • Using 24 dm³ instead of 24000 cm³ in the gas equation without converting units.
  • Writing NO instead of NO₂ by confusing molar masses (NO is 30.0 g mol⁻¹).
  • Omitting the water of crystallisation from compound C's final formula.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 2.1 Atoms and reactions · 5.2 Energy · 5.3 Transition elements

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.