A mole calculation has two parts: the arithmetic and the ratio from the equation. Students practise the arithmetic a lot, so it is usually right. The ratio is the part you have to choose, and it is where correct-looking working goes wrong.
This skill appears wherever an equation is used to link amounts, in formulas and equations, relative masses and amounts and the later calculation topics. It is the first lesson of the extended explanation and error repair module.
Why does a wrong ratio survive correct arithmetic?
Arithmetic is faithful. If you tell it “one mole of A gives one mole of B”, it will carry that claim through perfectly. The error is in the claim, not in the carrying.
That is why these mistakes feel unfair. Every line of working checks out, and only the chemistry is wrong.
How to choose the ratio, step by step
- Write the balanced equation and check it. A wrong equation gives a wrong ratio.
- Circle the two substances you are linking: the one you know and the one you want.
- Read their coefficients and write the ratio as moles of known : moles of wanted.
- Convert the known mass to moles first, using mass ÷ molar mass.
- Multiply by the ratio to get moles of the wanted substance, then convert to mass.
- Run a mass check: total mass of reactants used should equal total mass of products.
Worked example
All data here are invented for practice. Aluminium reacts with oxygen: 4Al + 3O₂ → 2Al₂O₃. What mass of aluminium oxide forms from 5.4 g of aluminium? Ar: Al = 27, O = 16.
Step 1, equation check: Al 4 and 4 (2 × 2). O 6 and 6 (2 × 3). Balanced.
Step 2, moles of Al: 5.4 ÷ 27 = 0.20 mol.
Step 3, ratio Al : Al₂O₃ is 4 : 2, which is 2 : 1. So moles of Al₂O₃ = 0.20 ÷ 2 = 0.10 mol.
Step 4, mass: Mr of Al₂O₃ = 2 × 27 + 3 × 16 = 102. Mass = 0.10 × 102 = 10.2 g.
Step 5, mass check: oxygen needed = 0.20 × 3/4 = 0.15 mol, which is 0.15 × 32 = 4.8 g. Reactants: 5.4 + 4.8 = 10.2 g. Product: 10.2 g. They agree.
The mistake to watch for
Mistaken answer: moles of Al = 0.20, so moles of Al₂O₃ = 0.20, so mass = 0.20 × 102 = 20.4 g.
The arithmetic is perfect. The student treated the ratio as 1 : 1 because the formulas “both contain aluminium”.
The mass check exposes it at once. The product cannot weigh 20.4 g when only 5.4 g of aluminium and 4.8 g of oxygen were used.
The correction is to read the coefficients, 4 and 2, and not the subscripts or the feeling of “one to one”. Write the ratio before the multiplication, every time.
Check yourself
Do these on paper, then open each answer. All data are invented.
1. 2H₂ + O₂ → 2H₂O. What mass of water forms from 0.50 mol of oxygen? (Mr of H₂O = 18)
Show answer
Ratio O₂ : H₂O is 1 : 2, so moles of water = 0.50 × 2 = 1.00 mol. Mass = 1.00 × 18 = 18 g.
Check: hydrogen needed is 1.00 mol = 2.0 g, oxygen is 0.50 × 32 = 16 g, total 18 g. The masses agree.
2. N₂ + 3H₂ → 2NH₃. What mass of ammonia forms from 0.60 mol of hydrogen? (Mr of NH₃ = 17)
Show answer
Ratio H₂ : NH₃ is 3 : 2, so moles of NH₃ = 0.60 × 2/3 = 0.40 mol. Mass = 0.40 × 17 = 6.8 g.
3. CH₄ + 2O₂ → CO₂ + 2H₂O. A student says 1.6 g of methane needs 3.2 g of oxygen to burn completely. Find the error and correct it. (Mr: CH₄ = 16, O₂ = 32)
Show answer
Moles of CH₄ = 1.6 ÷ 16 = 0.10 mol. The student used 0.10 mol of O₂, a 1 : 1 ratio. The equation gives CH₄ : O₂ = 1 : 2, so O₂ = 0.20 mol = 0.20 × 32 = 6.4 g.
Check: CO₂ is 0.10 × 44 = 4.4 g and H₂O is 0.20 × 18 = 3.6 g. Products total 8.0 g, and reactants total 1.6 + 6.4 = 8.0 g.
Where this leads next
Next, make your explanations as careful as your ratios with repairing a particle-level explanation that names only a trend. The mole and equation-ratio tutor lets you test a ratio step by step, and the mixed practice set mixes all five skills.
Some students understand every rule here but reach for 1 : 1 under time pressure. That is a habit pattern a teacher can see quickly in online one-to-one Chemistry tuition.