When the arithmetic is right and the answer is wrong, the mistake is almost always the ratio step: which substance you compared with which, and which way up you used the numbers. The routine below makes that step visible so you can check it before you calculate.
Mole calculations have three stages. First turn the given quantity into moles, then use the equation to move from one substance to another, then turn the new amount into the quantity the question asks for. Students usually practise the first and third stages and rush the middle one.
Where does the ratio go wrong?
There are three common ways to choose the wrong ratio.
- Reading the coefficient of the wrong substance. The student takes the 2 in front of Mg and uses it for oxygen, because it is the first number in the equation.
- Turning the ratio upside down. The student multiplies by 2/3 when the equation needs 3/2.
- Using masses instead of moles. The student compares 2 and 1 as though they were grams.
All three produce a calculation that looks tidy. The mole and equation-ratio tutor lets you rehearse the middle stage with the amounts and units shown, and the equation balance reasoning trainer checks that the equation is balanced before you read any ratio.
Worked example 1: the wrong substance
Question. Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. What mass of oxygen reacts with 4.8 g of magnesium? (Ar: Mg = 24, O = 16.)
The mistaken solution.
- Moles of Mg = 4.8 ÷ 24 = 0.20 mol.
- The student reads “2” from the equation and writes: moles of O₂ = 0.20 mol.
- Mass of O₂ = 0.20 × 32 = 6.4 g.
Every line of arithmetic is correct. The ratio is the problem: 2 is the coefficient of Mg, not of O₂.
The repaired solution. Write the ratio as a labelled line first.
- From the equation: n(Mg) : n(O₂) = 2 : 1.
- So n(O₂) = 0.20 × 1/2 = 0.10 mol.
- Mass of O₂ = 0.10 × 32 = 3.2 g.
Check. 3.2 g of oxygen plus 4.8 g of magnesium gives 8.0 g of magnesium oxide. As a separate check, n(MgO) = 0.20 mol and its Mr is 40, so 0.20 × 40 = 8.0 g. The masses agree, which confirms the ratio.
Worked example 2: the upside-down ratio
Question. 2Al + 6HCl → 2AlCl₃ + 3H₂. What volume of hydrogen forms from 5.4 g of aluminium with excess acid? Take the molar volume as 24 dm³/mol at room temperature and pressure. (Ar: Al = 27.)
The mistaken solution.
- n(Al) = 5.4 ÷ 27 = 0.20 mol.
- The student writes n(H₂) = 0.20 × 2/3 = 0.133 mol.
- Volume = 0.133 × 24 = 3.2 dm³.
The repaired solution.
- n(Al) : n(H₂) = 2 : 3. Hydrogen has the larger number in the equation, so it must have the larger amount.
- n(H₂) = 0.20 × 3/2 = 0.30 mol.
- Volume = 0.30 × 24 = 7.2 dm³.
Check. The expectation from the equation is that H₂ has more moles than Al. The mistaken answer gave fewer moles, which contradicts it. That single sanity question would have caught the error.
The routine that fixes the ratio step
Use these five lines on every calculation, in this order.
- Balance first. Count atoms on both sides. Do not read a ratio from an unbalanced equation.
- Name the two substances. Write “I know Mg, I want O₂”.
- Write the ratio line. n(Mg) : n(O₂) = 2 : 1, with the substance names attached to the numbers.
- Predict the direction. Which substance should have more moles? Say it before calculating.
- Attach units to each number. Write “0.10 mol O₂”, never just “0.10”.
Keep the ratio line even when you are short of time. It takes ten seconds and is the line most likely to earn or protect a mark.
Check yourself
1. 2H₂ + O₂ → 2H₂O. If 0.40 mol of hydrogen reacts completely, how many moles of oxygen react, and how many moles of water form?
Show answer
Ratio H₂ : O₂ : H₂O = 2 : 1 : 2.
n(O₂) = 0.40 × 1/2 = 0.20 mol. n(H₂O) = 0.40 × 2/2 = 0.40 mol.
0.20 mol O₂ and 0.40 mol H₂O. Check: water has the same number of moles as hydrogen because both have the coefficient 2.
2. N₂ + 3H₂ → 2NH₃. How many moles of ammonia form from 0.60 mol of hydrogen, and how many moles of nitrogen are needed?
Show answer
Ratio N₂ : H₂ : NH₃ = 1 : 3 : 2.
n(NH₃) = 0.60 × 2/3 = 0.40 mol. n(N₂) = 0.60 × 1/3 = 0.20 mol.
0.40 mol NH₃ and 0.20 mol N₂. Check: hydrogen has the largest coefficient, so it has the largest amount.
3. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. What amount of hydrochloric acid reacts with 0.050 mol of calcium carbonate, and what volume of carbon dioxide forms at 24 dm³/mol?
Show answer
Ratio CaCO₃ : HCl : CO₂ = 1 : 2 : 1.
n(HCl) = 0.050 × 2 = 0.10 mol. n(CO₂) = 0.050 mol, so volume = 0.050 × 24 = 1.2 dm³.
0.10 mol HCl and 1.2 dm³ CO₂.
Where to go next
The lesson on using an equation ratio to relate reacting amounts builds this step from the start, and balancing atoms without changing subscripts secures the equation you read the ratio from. To practise a mixed set, use the Chemistry original practice page. If your balanced equations are also unreliable, see changing subscripts while balancing.
Some students repair this once and then slip again when a question adds concentration or gas volume. That pattern is where online one-to-one Chemistry tuition can help, because a teacher hears which substance you name at the ratio step and stops you there.