Module 2 · L12 of 20~40 min⚡ +50 XP in Learn · +25 to complete
Mass–Mass Stoichiometry
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.
Today's hook, If you burn 10 g of magnesium in oxygen, do you expect exactly 10 g of magnesium oxide, or more, or less? The balanced equation has the answer, but masses don't read directly from it.
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Warm up and think first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
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?
⚠️ Every mass–mass problem uses these four steps in this order. Never skip Step 1 (balancing) or Step 3 (the mole ratio). The most common error is jumping from mass directly to mass.
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What you'll master
Know
Key facts
The 4-step stoichiometry method, steps 1 to 4 in order
Mole ratio comes from coefficients of the balanced equation
All mass calculations pass through moles, no shortcuts
Understand
Concepts
Why mass cannot be directly converted to mass without moles
How the mole ratio acts as a conversion factor between species
Why the method is identical regardless of what substances react
Can do
Skills
Perform mass–mass calculations for any reaction
Correctly apply non-1:1 mole ratios
Work backwards: given mass of product, find mass of reactant
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Key terms
mass–mass stoichiometry
A calculation that converts the mass of one substance to the mass of another using moles as an intermediary.
given substance
The substance whose mass (or moles) is provided in the problem and used to start the calculation.
wanted substance
The substance whose mass (or moles) you are asked to find at the end of the calculation.
molar mass
The mass of one mole of a substance in g mol⁻¹, calculated by summing the atomic masses of all atoms in the formula.
mole ratio
The ratio of coefficients in the balanced equation used to convert moles of the given substance to moles of the wanted substance.
theoretical mass
The maximum mass of product predicted by stoichiometry, assuming complete reaction with no losses.
balanced equation
The equation that must be written and verified before any mass–mass calculation can begin.
reverse problem
A stoichiometry problem where the mass of a product is given and the mass of a reactant must be found, using the same 4-step method.
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The 4-step stoichiometry method
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The 4-Step Stoichiometry Method
core concept · +3 XP at end
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.
Why You Can't Skip Moles
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 ratio step is the heart of stoichiometry.
In C + O₂ → CO₂, the ratio is 1:1:1, easy. But in 2Al + 3Cl₂ → 2AlCl₃, the ratio of Al to Cl₂ is 2:3, and the ratio of Al to AlCl₃ is 2:2 = 1:1. Always read the ratio from the balanced equation for the specific pair of substances in the question.
Reverse Problems, Given Product, Find Reactant
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.
4-step stoichiometry method: (1) balance the equation; (2) n(given) = m ÷ MM; (3) n(wanted) = n(given) × coeff(wanted) ÷ coeff(given); (4) m(wanted) = n × MM. Mass cannot convert directly to mass, moles are always the bridge. The same steps apply in reverse (product → reactant).
Pause, copy the highlighted method into your book before moving on.
Did you get this? True or false: you can convert directly from mass of A to mass of B using the coefficients in the balanced equation, without going through moles.
Quick check: Which statement best describes the 4-step method?
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Four worked examples
Worked examples · reveal as you go
Worked example 1 · Carbon combustion (1:1 ratio)+5 XP on full reveal
What mass of CO₂ is produced when 12.0 g of carbon burns completely? C + O₂ → CO₂. (C = 12.011, O = 15.999)
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$\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}$ · Ratio $\mathrm{C}:\mathrm{CO_2} = 1:1$
Worked example 4 · Haber process (industrial)+5 XP on full reveal
In the Haber process: N₂ + 3H₂ → 2NH₃. What mass of NH₃ is produced from 56.0 g of N₂ (assuming complete reaction)? (N = 14.007, H = 1.008) The Haber process is used here only as a stoichiometry context; its real industrial conditions and equilibrium yield are studied in Year 12.
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Balanced. Ratio $\mathrm{N_2} : \mathrm{NH_3} = 1 : 2$
Odd one out three of these are steps in the mass→mass stoichiometry chain. Which one doesn't belong?
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Spot the error
Spot the slip-up a student calculates the mass of CO₂ produced when 10.00 g of CaCO₃ decomposes. One line has a chemistry error, click the line that's wrong.
Problem: CaCO₃ → CaO + CO₂. Calculate m(CO₂) from 10.00 g CaCO₃. (Ca=40.078, C=12.011, O=15.999)
Skipping the mole ratio (assuming everything is 1:1)
In Fe₂O₃ + 3CO → 2Fe + 3CO₂, a student who assumes a 1:1 ratio for Fe₂O₃:Fe gets half the correct answer for iron. The 1:2 ratio doubles the moles of iron. Students who rush to Step 2 before Step 1 almost always make this mistake.
Fix: Always write the balanced equation first. Circle the two species in the question. Write the ratio explicitly (e.g. "Fe₂O₃:Fe = 1:2") before touching numbers.
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Using the wrong molar mass, elements vs compounds
For diatomic elements (O₂, H₂, N₂, Cl₂, Br₂, I₂, F₂) the molar mass is double the atomic mass. O₂ has MM = 32.00 g/mol, not 16.00. Students regularly use the atomic mass when the equation calls for the diatomic molecule, halving or doubling their answer.
Fix: Always use the formula in the equation to determine MM. If the equation says O₂, use MM = 2 × 15.999 = 31.998 g/mol.
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Applying the mole ratio in the wrong direction
In N₂ + 3H₂ → 2NH₃, the ratio of N₂ to H₂ is 1:3. Given moles of N₂ wanting moles of H₂ → multiply by 3. Given moles of H₂ wanting moles of N₂ → divide by 3. Students sometimes apply the ratio backwards.
Fix: Use n(wanted) = n(given) × coeff(wanted) ÷ coeff(given) every time. Plug in the coefficients, don't guess the direction.
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Quick-fire drill, then revisit your thinking
Quick-fire practice · 5 reps +2 XP per reveal
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What mass of water forms when 4.00 g of H₂ burns? 2H₂ + O₂ → 2H₂O. (H = 1.008, O = 15.999)
Earlier you were asked: If you burn 10 g of magnesium in oxygen, do you expect exactly 10 g of magnesium oxide?
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.
Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. If 0.40 mol of Mg reacts completely, how many moles of MgO are produced? (Use the mole ratio from the balanced equation.)
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Practice questions
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Multiple choice
+2 XP per correct · +5 bonus if perfect
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Short answer
ApplyLower-order4 marks
Q1. 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) (Calculation only: thermite exceeds 2500 °C and is never a student practical, do not prepare or ignite it.)
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ApplyLower-order5 marks
Q2. 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)
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EvaluateHigher-order3 marks
Q3. 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)
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AnalyseMid-order5 marks
Q4. 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. (N = 14.007, H = 1.008)
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EvaluateHigher-order4 marks
Q5. 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.
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📖 Comprehensive answers (click to reveal)
Multiple choice, drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
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 g
The 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 extra mass comes from oxygen incorporated into the product (16.0 g/mol of O₂ accounts for the extra mass). Mass conservation applies to atoms, not to individual substances.
Q4 (5 marks):
(a) MM(N₂) = 28.014; n(N₂) = 28.0÷28.014 = 0.9995 mol; ratio N₂:NH₃ = 1:2; n(NH₃) = 1.999 mol MM(NH₃) = 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 ✓
Q5 (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 Al:Al₂O₃ = 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.
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Retrieve, then finish
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