Quantitative ChemistryWorked Example
Worked Example — Full Method
Part of Moles & Calculations — GCSE Chemistry
This worked example covers Worked Example — Full Method within Moles & Calculations for GCSE Chemistry. Revise Moles & Calculations in Quantitative Chemistry for GCSE Chemistry with 22 exam-style questions and 20 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 9 of 15 in this topic. Treat this as a marking guide for what examiners are looking for, not just a fact list.
Topic position
Section 9 of 15
Practice
22 questions
Recall
20 flashcards
🧮 Worked Example — Full Method
Question: What mass of magnesium oxide forms when 4.8g of magnesium burns completely?
Equation: 2Mg + O₂ → 2MgO (Ar: Mg = 24, O = 16)
Step 1: Balanced equation
2Mg + O₂ → 2MgO ✓
2Mg + O₂ → 2MgO ✓
Step 2: Identify known and unknown
Known: 4.8g of Mg
Unknown: mass of MgO
Known: 4.8g of Mg
Unknown: mass of MgO
Step 3: Calculate moles of known (Mg)
n = m ÷ Mr
n = 4.8 ÷ 24
n = 0.2 mol of Mg
n = m ÷ Mr
n = 4.8 ÷ 24
n = 0.2 mol of Mg
Step 4: Use the ratio
From equation: 2Mg : 2MgO = 1:1 ratio
So 0.2 mol Mg produces 0.2 mol MgO
From equation: 2Mg : 2MgO = 1:1 ratio
So 0.2 mol Mg produces 0.2 mol MgO
Step 5: Convert to mass
Mr of MgO = 24 + 16 = 40
m = n × Mr
m = 0.2 × 40
m = 8.0g
Mr of MgO = 24 + 16 = 40
m = n × Mr
m = 0.2 × 40
m = 8.0g
Step 6: Check
Started with 4.8g Mg, added oxygen → got 8.0g product
Mass increased (oxygen added) ✓ Makes sense!
Started with 4.8g Mg, added oxygen → got 8.0g product
Mass increased (oxygen added) ✓ Makes sense!
Answer: 8.0g of magnesium oxide