Given the mass of one substance, find the mass of another. This is the core calculation of all quantitative chemistry — the 4-step method converts mass to moles, applies the mole ratio, then converts back to mass. Apply it the same way every time.
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If you burn 10 g of magnesium in oxygen, do you expect to produce exactly 10 g of magnesium oxide — or more, or less? What determines how much product forms, and why can't you simply read the answer from the masses in the balanced equation?
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📚 Core Content
Wrong: A catalyst increases the yield of products in an equilibrium reaction.
Right: A catalyst speeds up both forward and reverse reactions equally; it does not change equilibrium yield.
You cannot convert directly from the mass of one substance to the mass of another, because different substances have different molar masses. You must pass through moles. The 4-step method is the reliable algorithm for every mass–mass problem.
Consider: 12 g of carbon burns to form CO₂. You might think "12 g C gives 12 g CO₂" — but CO₂ has a molar mass of 44 g/mol while C is 12 g/mol. 1 mol C (12 g) produces 1 mol CO₂ (44 g). The actual answer is 44 g, not 12 g. The mass changes because the molar masses differ. Moles are the universal bridge between substances.
The same 4-step method works in reverse. The only difference is that Step 2 converts the given mass of product to moles, and Step 3 applies the ratio to find moles of reactant, which Step 4 then converts to mass. The direction of the problem does not change the method.
🧮 Worked Examples
🧪 Activities
1 What mass of water forms when 4.00 g of H₂ burns? 2H₂ + O₂ → 2H₂O. (H = 1.008, O = 15.999)
2 What mass of Al₂O₃ forms when 5.40 g of Al burns? 4Al + 3O₂ → 2Al₂O₃. (Al = 26.982, O = 15.999)
3 What mass of CaCO₃ is needed to produce 22.0 g of CO₂? CaCO₃ → CaO + CO₂. (Ca = 40.078, C = 12.011, O = 15.999)
4 What mass of Fe is needed to produce 46.5 g of FeCl₃? 2Fe + 3Cl₂ → 2FeCl₃. (Fe = 55.845, Cl = 35.453)
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| Reaction | Given | Find |
|---|---|---|
| N₂ + 3H₂ → 2NH₃ (Haber process) | m(H₂) = 6.05 g | m(NH₃) = ? |
| 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂ | m(FeS₂) = 24.0 g | m(SO₂) = ? |
| 2SO₂ + O₂ → 2SO₃ (Contact process) | m(SO₃) = 40.0 g | m(SO₂) needed = ? |
Atomic masses: N=14.007, H=1.008, Fe=55.845, S=32.06, O=15.999
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At the start of this lesson, you thought about whether burning 10 g of magnesium would produce exactly 10 g of magnesium oxide, and what determines the product mass.
The answer is: you produce more than 10 g — about 16.6 g of MgO. The extra mass comes from oxygen atoms being incorporated into the product. You cannot read product mass from reactant mass directly; you must pass through moles using the 4-step method, because different substances have different molar masses. The balanced equation gives you the mole ratio, not a mass ratio.
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✍️ Short Answer
6. In the thermite reaction: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. What mass of aluminium is needed to completely react with 80.0 g of Fe₂O₃? Show all four steps. (Al = 26.982, Fe = 55.845, O = 15.999) 4 MARKS
Show all 4 steps:
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7. Copper reacts with silver nitrate: Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag. (a) What mass of silver is deposited when 6.35 g of copper reacts completely? (b) What mass of AgNO₃ is consumed? (Cu = 63.546, Ag = 107.87, N = 14.007, O = 15.999) 5 MARKS
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8. A student claims: "I don't need to convert to moles — I can just use the mass ratio from the equation." Using the reaction 2Mg + O₂ → 2MgO, evaluate this claim by calculating whether 48 g of Mg produces 48 g of MgO. Explain why the student's shortcut fails. (Mg = 24.305, O = 15.999) 3 MARKS
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9. The reaction between nitrogen and hydrogen in the Haber process is: N₂ + 3H₂ → 2NH₃. A chemist starts with 28.0 g of N₂ and an unlimited supply of H₂. (a) Calculate the mass of NH₃ produced. (b) Calculate the mass of H₂ consumed. (c) Verify that mass is conserved by comparing total reactant mass with total product mass. (N = 14.007, H = 1.008) 5 MARKS
Show all working and verify conservation of mass:
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10. Two students each perform a mass–mass calculation for the reaction 2SO₂ + O₂ → 2SO₃ starting from 12.8 g of SO₂ (S = 32.06, O = 15.999). Student A uses the correct 4-step method and obtains 16.0 g of SO₃. Student B skips Step 3 (the mole ratio) and writes n(SO₃) = n(SO₂), also obtaining 16.0 g. Is Student B's answer coincidental or correct by method? Explain why the two approaches happen to give the same numerical answer in this case, and identify a reaction where B's shortcut would give the wrong answer. 4 MARKS
Explain with reasoning and a counter-example:
Answer in workbook.
Row 1 (Haber):
n(H₂) = 6.05÷2.016 = 3.001 mol; ratio H₂:NH₃ = 3:2 n(NH₃) = 3.001×(2÷3) = 2.001 mol; m(NH₃) = 2.001×17.031 = 34.1 gRow 2 (FeS₂):
MM(FeS₂) = 55.845+2(32.06) = 119.97; n(FeS₂) = 24.0÷119.97 = 0.2001 mol ratio FeS₂:SO₂ = 4:8 = 1:2; n(SO₂) = 0.2001×2 = 0.4002 mol MM(SO₂) = 32.06+2(15.999) = 64.058; m(SO₂) = 0.4002×64.058 = 25.6 gRow 3 (Contact):
MM(SO₃) = 32.06+3(15.999) = 80.057; n(SO₃) = 40.0÷80.057 = 0.4997 mol ratio SO₂:SO₃ = 2:2 = 1:1; n(SO₂) = 0.4997 mol MM(SO₂) = 64.058; m(SO₂) = 0.4997×64.058 = 32.0 g1. C — Step 3 applies the mole ratio: n(wanted) = n(given) × coeff(wanted) ÷ coeff(given).
2. B — 51.0 g. n(Al) = 27.0÷26.982 = 1.001 mol; ratio Al:Al₂O₃ = 4:2 = 2:1; n(Al₂O₃) = 1.001×(2÷4) = 0.5003 mol; MM(Al₂O₃) = 101.96; m = 0.5003×101.96 = 51.0 g.
3. A — The working is correct. This is a valid reverse problem: given mass of CO₂ (product), find mass of CaCO₃ (reactant). The 1:1 ratio is correct for CaCO₃ → CaO + CO₂.
4. D — 15.9 g. MM(H₂) = 2.016; n(H₂) = 2.00÷2.016 = 0.992 mol; ratio H₂:O₂ = 2:1; n(O₂) = 0.992×(1÷2) = 0.496 mol; MM(O₂) = 31.998; m(O₂) = 0.496×31.998 = 15.9 g.
5. C — The mole ratio must always come from the balanced equation. A and B are false (mass can change; you cannot skip moles even for equal molar masses if ratio ≠ 1:1). D is false — the method works for any ratio.
6. C — Total mass increases because the other reactants (CO, O₂, etc.) also contribute mass to the products. Mass is always conserved across all species. Option D is false — the 4-step method predicts product mass based on moles and molar masses, which need not equal reactant mass.
7. B — 14.3 g. n(Fe) = 10.0 ÷ 55.845 = 0.1791 mol; ratio Fe:Fe₂O₃ = 2:1 (reverse ratio from equation); n(Fe₂O₃) = 0.1791 ÷ 2 = 0.08954 mol; MM(Fe₂O₃) = 2(55.845)+3(15.999) = 159.69; m = 0.08954×159.69 = 14.3 g. Option A uses the wrong molar mass for Fe alone instead of working backwards. Option D skips moles entirely.
Q6 (4 marks):
Ratio Al:Fe₂O₃ = 2:1 MM(Fe₂O₃) = 2(55.845)+3(15.999) = 159.69; n(Fe₂O₃) = 80.0÷159.69 = 0.5010 mol n(Al) = 0.5010×(2÷1) = 1.002 mol m(Al) = 1.002×26.982 = 27.0 gQ7 (5 marks):
(a) n(Cu) = 6.35÷63.546 = 0.09992 mol; ratio Cu:Ag = 1:2 n(Ag) = 0.09992×2 = 0.1998 mol; m(Ag) = 0.1998×107.87 = 21.6 g (b) ratio Cu:AgNO₃ = 1:2; n(AgNO₃) = 0.09992×2 = 0.1998 mol MM(AgNO₃) = 107.87+14.007+3(15.999) = 169.88; m(AgNO₃) = 0.1998×169.88 = 33.9 gQ8 (3 marks):
n(Mg) = 48÷24.305 = 1.976 mol; ratio 1:1; n(MgO) = 1.976 mol MM(MgO) = 40.304; m(MgO) = 1.976×40.304 = 79.6 g — not 48 gThe shortcut fails because Mg (MM = 24.3) and MgO (MM = 40.3) have different molar masses. Even though 1 mol of Mg produces 1 mol of MgO, 1 mol of each weighs a different mass. The mass changes because each oxygen atom (16.0 g/mol) is incorporated into the product. Mass conservation applies to atoms, not to individual substances — the 16.0 g/mol of O₂ consumed accounts for the extra mass in MgO.
Q9 (5 marks):
(a) MM(N₂) = 28.014; n(N₂) = 28.0÷28.014 = 0.9995 mol; ratio N₂:NH₃ = 1:2; n(NH₃) = 0.9995×2 = 1.999 mol MM(NH₃) = 14.007+3(1.008) = 17.031; m(NH₃) = 1.999×17.031 = 34.0 g (b) ratio N₂:H₂ = 1:3; n(H₂) = 0.9995×3 = 2.999 mol; MM(H₂) = 2.016; m(H₂) = 2.999×2.016 = 6.05 g (c) Total reactants = 28.0+6.05 = 34.0 g; Total products = 34.0 g — mass is conserved ✓Q10 (4 marks): Student B's answer is coincidentally correct in this case because the ratio SO₂:SO₃ = 2:2 = 1:1. When the mole ratio is 1:1, skipping Step 3 produces the same mole count and therefore the same product mass. This is not a valid general method — it only works when the ratio is exactly 1:1. For a reaction with a different ratio (e.g., 4Al + 3O₂ → 2Al₂O₃, ratio 4:2 = 2:1), skipping the ratio step would give n(Al₂O₃) = n(Al) instead of the correct n(Al₂O₃) = n(Al)÷2, leading to a product mass that is double the true answer. Student B's shortcut is conceptually wrong even when it gives the right number.
Answer questions on mass-to-mass stoichiometry using mole ratios and molar mass before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–12.
Tick when you've finished all activities and checked your answers.