OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 1

Calculating the Greatest Number of Molecules

OCR A-Level Chemistry • Amount of Substance

What this question tests

This question assesses your understanding of the mole concept, molar mass calculations, and the direct proportionality between the number of moles and the total number of particles (Avogadro's constant, L). To master this, you must be comfortable converting between mass and moles using moles = mass / molar mass without getting distracted by larger initial mass values.

Question 1 Breakdown

Determining the sample with the greatest number of molecules

✅ Correct Answer

B (45.0 g of C₂H₆)

Mark Awarded: 1 / 1 (AO1.2)

💡 Key Knowledge

Number of molecules = Moles × Avogadro constant (L).
Since Avogadro's constant is a fixed number for all substances, the sample with the greatest number of moles will automatically contain the greatest number of molecules.
Molar mass (g mol⁻¹) is calculated by summing the relative atomic masses (Aᵣ) from the Periodic Table.

🧠 Exam Technique

In multiple-choice questions involving calculations, avoid guessing based on the largest mass value (e.g., dismissing option A because 35.0 g seems small). Systematically calculate the number of moles for every option before selecting your final answer.

❌ Common Errors

Inverting the formula: Calculating mass × molar mass instead of mass / molar mass.
Arithmetical slips: Miscalculating the molar mass of hydrocarbons (e.g., getting C₂H₆ wrong by miscounting hydrogen atoms).
Stopping halfway: Calculating moles and forgetting that the question asks for the greatest number of molecules (though proportional, ensure you understand why moles equal molecules in relative terms here).

📐 Step-by-Step Calculation Guide

To find the correct option, calculate the number of moles ( n = m / M ) for each choice:

A: Molar mass of C₂H₂ = (2 × 12.0) + (2 × 1.0) = 26.0 g mol⁻¹. Moles = 35.0 / 26.0 = 1.35 mol
B: Molar mass of C₂H₆ = (2 × 12.0) + (6 × 1.0) = 30.0 g mol⁻¹. Moles = 45.0 / 30.0 = 1.50 mol (Highest value!)
C: Molar mass of C₄H₁₀ = (4 × 12.0) + (10 × 1.0) = 58.0 g mol⁻¹. Moles = 60.0 / 58.0 = 1.03 mol
D: Molar mass of C₆H₆ = (6 × 12.0) + (6 × 1.0) = 78.0 g mol⁻¹. Moles = 100.0 / 78.0 = 1.28 mol

Conclusion: Option B yields 1.50 moles, which is the highest value among all choices, meaning it contains the greatest number of molecules.