OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 6

Calculating Electron Removal in Ionisation

What this question tests

This question assesses your ability to interconvert between mass, moles, and the number of particles using the Avogadro constant, alongside understanding the stoichiometry of single-electron ionisation (forming Ne⁺ ions).

Question Multiple Choice Analysis

Question 6

How many electrons are removed from 2.02 × 10⁻² g of Ne(g) atoms to form Ne⁺(g) ions?

✅ Correct Answer: C

6.02 × 10²⁰

Option C is the only correct choice based on proper stoichiometric and mole calculations.

💡 Key Knowledge

Molar Mass: Neon has a relative atomic mass (Ar) of 20.2 g mol⁻¹ .
Avogadro Constant (N_A): 6.022 × 10²³ mol⁻¹ .
Ionisation Ratio: Forming one Ne⁺ ion requires the removal of exactly 1 electron per atom: Ne → Ne⁺ + e⁻ .

🧠 Exam Technique

Break multi-step calculation questions down systematically. Find moles first, scale to the number of atoms, and then factor in the ionisation ratio before checking standard form indices.

❌ Common Errors

Using the molar mass of neon incorrectly (e.g., confusing it with fluorine or sodium).
Forgetting that Ne is monatomic, and mistakenly treating it as a diatomic gas like Ne₂ .
Failing to adjust powers of 10 correctly when dividing mass by molar mass.

📐 Step-by-Step Calculation Guide

Calculate the moles of Neon atoms:
Moles = Mass / Molar Mass = (2.02 × 10⁻² g) / (20.2 g mol⁻¹) = 1.00 × 10⁻³ mol
Calculate the number of Ne atoms:
Number of atoms = Moles × Avogadro constant = 1.00 × 10⁻³ mol × 6.02 × 10²³ mol⁻¹ = 6.02 × 10²⁰ atoms
Determine the number of electrons removed:
Since the equation is Ne(g) → Ne⁺(g) + e⁻ , exactly 1 electron is lost per atom.
Total electrons removed = 6.02 × 10²⁰ atoms × 1 = 6.02 × 10²⁰ electrons (Matching option C).

Mark Scheme Allocation: 1 mark awarded for selecting C.