GCSE Chemistry - How to Solve: Mole Calculations (Mass = Mr × Moles, Limiting Reactants, Yield)

How to Solve: Mole Calculations (Mass = Mr × Moles, Limiting Reactants, Yield)

Complete Guide for GCSE/A-Level Chemistry

Introduction

"Mastering mole calculations unlocks 15–20% of your GCSE Chemistry paper—and lets you predict how much medicine a factory can produce, how much CO₂ a car emits, or why your cake flopped when you mismeasured baking soda. One wrong step here costs you 4–6 marks per question. Let’s fix that."

WHAT YOU NEED TO KNOW FIRST

1. Relative atomic mass (Ar) and relative formula mass (Mr) – You must be able to calculate Mr from a formula (e.g., H₂O = 18).
2. Balanced chemical equations – You need to write and interpret them (e.g., 2H₂ + O₂ → 2H₂O).
3. Units – Mass is in grams (g), moles (mol) are unitless, and Mr is in g/mol.

KEY TERMS & FORMULAS

1. Mass, Moles, and Mr

Formula: Mass (g) = Mr (g/mol) × Moles (mol) - Mass = mass of the substance (grams) - Mr = relative formula mass (g/mol) – MEMORISE: Add up atomic masses from the periodic table - Moles = amount of substance (mol)

Example: Mr of CO₂ = 12 + (16 × 2) = 44 g/mol

2. Limiting Reactants

Definition: The reactant that runs out first, stopping the reaction. The other reactant is in excess.

How to find it:
1. Calculate moles of each reactant.
2. Compare mole ratio to the balanced equation.
3. The reactant with fewer moles than needed is limiting.

MEMORISE: "The limiting reactant determines the maximum product."

3. Percentage Yield

Formula: Percentage Yield = (Actual Yield / Theoretical Yield) × 100% - Actual Yield = mass of product actually obtained (given in question) - Theoretical Yield = mass of product expected (calculated from limiting reactant)

MEMORISE: "Yield is never 100% in real life—some product is always lost."

STEP-BY-STEP METHOD

Step 1: Write the balanced equation

• If not given, balance it first.
• Example: 2H₂ + O₂ → 2H₂O

Step 2: Calculate moles of each reactant

• Use moles = mass / Mr.
• Example: If you have 4g H₂ and 32g O₂:
• Moles of H₂ = 4 / 2 = 2 mol
• Moles of O₂ = 32 / 32 = 1 mol

Step 3: Find the limiting reactant

• Compare mole ratio to the balanced equation.
• For 2H₂ + O₂ → 2H₂O, the ratio is 2:1.
• You have 2 mol H₂ and 1 mol O₂ → exact ratio, so neither is limiting (rare in exams).
• If you had 3 mol H₂ and 1 mol O₂, O₂ would be limiting (needs 2 mol H₂ per 1 mol O₂, but you have 3 mol H₂).

Step 4: Calculate theoretical yield

• Use the limiting reactant to find moles of product.
• Example: If O₂ is limiting (1 mol), then 2 mol H₂O is produced (from the equation).
• Convert moles to mass: mass = moles × Mr.
• Mass of H₂O = 2 × 18 = 36g.

Step 5: Calculate percentage yield (if asked)

• Use % yield = (actual / theoretical) × 100%.
• Example: If you got 27g H₂O instead of 36g:
• % yield = (27 / 36) × 100 = 75%.

WORKED EXAMPLES

Example 1 – Basic: Mass from Moles

Question: Calculate the mass of 5 moles of CO₂. (Mr of CO₂ = 44 g/mol)

Solution:
1. Formula: Mass = Mr × Moles
2. Plug in numbers: Mass = 44 × 5 = 220g

What we did and why: - We used the mass-moles-Mr triangle to convert moles to mass. - Key: Always check units (g/mol × mol = g).

Example 2 – Medium: Limiting Reactant

Question: 2.4g of magnesium (Mg) reacts with 3.2g of oxygen (O₂). Equation: 2Mg + O₂ → 2MgO a) Which is the limiting reactant? b) What is the maximum mass of MgO produced?

Solution: Step 1: Calculate moles of each reactant - Moles of Mg = 2.4 / 24 = 0.1 mol - Moles of O₂ = 3.2 / 32 = 0.1 mol

Step 2: Compare to balanced equation (2Mg : 1O₂) - For 0.1 mol O₂, you need 0.2 mol Mg (from ratio). - But you only have 0.1 mol Mg → Mg is limiting.

Step 3: Calculate theoretical yield - From equation, 2 mol Mg → 2 mol MgO → 1:1 ratio. - Moles of MgO = 0.1 mol (same as limiting Mg). - Mass of MgO = 0.1 × (24 + 16) = 0.1 × 40 = 4g

What we did and why: - We compared mole ratios to find the limiting reactant. - Key: The limiting reactant always determines the product.

Example 3 – Exam-Style: Percentage Yield

Question: 10g of calcium carbonate (CaCO₃) decomposes to form calcium oxide (CaO) and CO₂. Equation: CaCO₃ → CaO + CO₂ a) Calculate the theoretical yield of CaO. b) If 4.2g of CaO is produced, what is the percentage yield?

Solution: Step 1: Calculate moles of CaCO₃ - Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100 g/mol - Moles = 10 / 100 = 0.1 mol

Step 2: Find moles of CaO (1:1 ratio) - Moles of CaO = 0.1 mol

Step 3: Calculate theoretical yield - Mr of CaO = 40 + 16 = 56 g/mol - Mass = 0.1 × 56 = 5.6g

Step 4: Calculate percentage yield - % yield = (4.2 / 5.6) × 100 = 75%

What we did and why: - We used the balanced equation to find the mole ratio. - Key: Theoretical yield is always from the limiting reactant (here, CaCO₃).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Here’s the night-before cheat sheet:
1. Mass = Mr × Moles – Memorise this. If you know two, you can find the third.
2. Limiting reactant – Always compare moles to the balanced equation’s ratio. The one that runs out first is limiting.
3. Percentage yield – Actual yield is what you get in the lab; theoretical is what the calculation says. Divide and multiply by 100.
4. Units matter – Grams for mass, g/mol for Mr, mol for moles. Don’t mix them up.
5. Practice one question now – Pick a past paper, do it step-by-step, and check your answer. You’ve got this!"