OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 2

1 mark · Medium difficulty · Multiple Choice

Determine the formula of an oxide of manganese given the mass reduced by hydrogen and the mass of water formed.

Practise this question

Question

Multiple choice question 2 states that 0.688 g of an oxide of manganese is reduced by hydrogen gas to form manganese metal and 0.235 g of water. Four options are provided for the formula of the oxide: A, MnO; B, MnO2; C, Mn2O3; D, Mn3O4. There is an answer box provided at the bottom with 1 mark allocated.
Question text

2 0.688 g of an oxide of manganese is reduced by hydrogen gas to form manganese metal and

0.235 g of water.

What is the formula of the oxide of manganese?

A MnO

B MnO2

C Mn2O3

D Mn3O4

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 2 showing the correct answer as C, worth 1 mark and assessing assessment objective AO2.2.

2 C 1 AO2.2

How to answer it

Determining the Formula of a Manganese Oxide

What this question tests

This question assesses your ability to apply quantitative chemistry skills—specifically calculating moles from mass, using stoichiometric mass/moles ratios, determining empirical formulas from reduction reaction data, and handling relative atomic masses (Ar) accurately.

Question 2 (Multiple Choice)

Exam Breakdown & Step-by-Step Solution

✅ Correct Answer

C: Mn₂O₃

Mark Awarded: 1 mark for selecting C (AO2.2 - Application of quantitative chemistry knowledge)

💡 Key Knowledge

  • Reduction Principle: Oxygen from the metal oxide combines with hydrogen gas to form water ( H₂O ).
  • Conservation of Mass: All manganese present in the initial oxide sample ends up as pure manganese metal.
  • Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.

🧠 Exam Technique

When given the mass of an oxide and the mass of water produced upon reduction, calculate the moles of water to find the initial moles of oxygen atoms. Then, subtract that oxygen mass from the total oxide mass to find the mass of the metal.

❌ Common Errors

  • Using the molar mass of molecular oxygen ( O₂ = 32.0 g mol⁻¹) instead of atomic oxygen ( O = 16.0 g mol⁻¹) when determining the moles of oxygen bonded to the metal.
  • Forgetting that each molecule of water ( H₂O ) contains one oxygen atom, meaning moles of oxygen atoms equal moles of water produced.

📐 Step-by-Step Calculation

  1. Find moles of H₂O produced:
    Molar mass of H₂O = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹
    Moles of H₂O = 0.235 g / 18.0 g mol⁻¹ = 0.013056 mol
  2. Determine moles of oxygen atoms (O) in the original oxide:
    Since every mole of H₂O contains 1 mole of oxygen atoms, moles of O = 0.013056 mol
  3. Calculate the mass of oxygen and mass of manganese (Mn):
    Mass of O = 0.013056 mol × 16.0 g mol⁻¹ = 0.2089 g
    Mass of Mn = Total mass of oxide - Mass of O = 0.688 g - 0.2089 g = 0.4791 g
  4. Calculate moles of manganese (Mn):
    Ar of Mn = 54.9 g mol⁻¹
    Moles of Mn = 0.4791 g / 54.9 g mol⁻¹ = 0.008727 mol
  5. Find the simplest whole-number ratio (Mn : O):
    Ratio Mn : O = 0.008727 : 0.013056
    Divide by the smaller number (0.008727):
    Mn = 1.0
    O = 0.013056 / 0.008727 = 1.496 ≈ 1.5
    Multiply by 2 to get whole numbers: Mn : O = 2 : 3 , giving Mn₂O₃.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.