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.
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.
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
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
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
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.
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?
Worked examples · reveal as you go
What mass of CO₂ is produced when 12.0 g of carbon burns completely? C + O₂ → CO₂. (C = 12.011, O = 15.999)
m(CO₂) = 0.9991 × 44.009 = 44.0 g ✓
What mass of iron is produced from 80.0 g of Fe₂O₃ in the reaction: Fe₂O₃ + 3CO → 2Fe + 3CO₂? (Fe = 55.845, O = 15.999)
n(Fe₂O₃) = 80.0 ÷ 159.69 = 0.5010 mol
Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. What mass of Mg must be burned to produce 20.0 g of MgO? (Mg = 24.305, O = 15.999)
n(MgO) = 20.0 ÷ 40.304 = 0.4962 mol
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.
n(N₂) = 56.0 ÷ 28.014 = 1.999 mol
m(NH₃) = 3.998 × 17.031 = 68.1 g ✓
Odd one out three of these are steps in the mass→mass stoichiometry chain. Which one doesn't belong?
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.
- MM(CaCO₃) = 40.078 + 12.011 + 3(15.999) = 100.086 g mol⁻¹
- n(CaCO₃) = 10.00 ÷ 100.086 = 0.09991 mol
- n(CO₂) = n(CaCO₃) × 2 = 0.1998 mol (ratio 1 : 2)
- MM(CO₂) = 12.011 + 2(15.999) = 44.009 g mol⁻¹
- m(CO₂) = 0.1998 × 44.009 = 8.79 g
Common errors · the 3 traps that cost marks
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.
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.
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.
Quick-fire practice · 5 reps +2 XP per reveal
What mass of water forms when 4.00 g of H₂ burns? 2H₂ + O₂ → 2H₂O. (H = 1.008, O = 15.999)
What mass of Al₂O₃ forms when 5.40 g of Al burns? 4Al + 3O₂ → 2Al₂O₃. (Al = 26.982, O = 15.999)
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)
What mass of Fe is needed to produce 46.5 g of FeCl₃? 2Fe + 3Cl₂ → 2FeCl₃. (Fe = 55.845, Cl = 35.453)
What mass of NH₃ is produced from 28.0 g of N₂ in the Haber process? N₂ + 3H₂ → 2NH₃. (N = 14.007, H = 1.008)
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.