A balanced equation gives the ratio in which substances react and form, in moles. To relate two substances, find the amount of the one you know, multiply by the ratio, and you have the amount of the other. This is the middle step of most calculations in relative masses and amounts.
How does the ratio work?
Take N₂ + 3H₂ → 2NH₃. It says 1 mol of N₂ reacts with 3 mol of H₂ to form 2 mol of NH₃. Any multiple of those amounts reacts in the same proportion.
To go from substance A (known) to substance B (wanted):
amount of B = amount of A × (coefficient of B ÷ coefficient of A)
The coefficient of the wanted substance goes on top.
Steps
- Check the equation is balanced.
- Convert the given mass to moles (see the previous lesson).
- Apply the ratio using the coefficients.
- Convert to a mass or other quantity if the question asks.
- Write units at every stage.
Worked example
What mass of magnesium oxide forms when 6.0 g of magnesium burns completely in oxygen? This is a model calculation for learning. Use Mg = 24, O = 16.
The equation is 2Mg + O₂ → 2MgO.
Step 1, moles of Mg: 6.0 ÷ 24 = 0.25 mol.
Step 2, ratio: 2 mol of Mg gives 2 mol of MgO, so the ratio is 2 : 2, or 1 : 1. Then MgO = 0.25 × (2 ÷ 2) = 0.25 mol.
Step 3, mass: Mr of MgO = 24 + 16 = 40. Mass = 0.25 × 40 = 10 g.
Check: the product mass (10 g) is greater than the magnesium mass (6.0 g), as expected, because oxygen has joined the magnesium. The extra 4.0 g is the oxygen: 0.125 mol of O₂ × 32 = 4.0 g, so 6.0 + 4.0 = 10.0 g.
A plausible mistake
A student writes: the ratio is 2 : 2, so 6.0 g of Mg gives 6.0 g of MgO.
Mistaken answer: 6.0 g
The ratio 2 : 2 applies to moles, not to grams. Mg and MgO have different molar masses, so equal moles are not equal masses.
The correction is always to go through moles: mass of A → moles of A → moles of B → mass of B.
A second example of a ratio that is not 1 : 1: with 0.60 mol of H₂ in N₂ + 3H₂ → 2NH₃, NH₃ = 0.60 × (2 ÷ 3) = 0.40 mol. With Mr of NH₃ = 14 + 3 = 17, mass = 0.40 × 17 = 6.8 g.
Check yourself
1. In 2H₂ + O₂ → 2H₂O, how many moles of water form from 0.50 mol of oxygen? What mass is that? (H = 1, O = 16)
Show answer
Ratio O₂ : H₂O = 1 : 2, so H₂O = 0.50 × 2 = 1.0 mol. Mr = 18, so mass = 1.0 × 18 = 1.0 mol, 18 g.
2. In CaCO₃ → CaO + CO₂, what mass of CaO forms from 50 g of CaCO₃? (Ca = 40, C = 12, O = 16)
Show answer
Mr of CaCO₃ = 40 + 12 + 48 = 100, so n = 50 ÷ 100 = 0.50 mol. Ratio 1 : 1, so CaO = 0.50 mol. Mr of CaO = 56, so mass = 0.50 × 56 = 28 g.
3. In 4Al + 3O₂ → 2Al₂O₃, how many moles of O₂ react with 0.80 mol of Al, and how many moles of Al₂O₃ form?
Show answer
O₂ = 0.80 × (3 ÷ 4) = 0.60 mol. Al₂O₃ = 0.80 × (2 ÷ 4) = 0.40 mol.
Where this leads next
The next lesson asks what happens when both reactants are given: identifying a limiting quantity. Use the mole and equation-ratio tutor to see each ratio step written out, and the equation balance reasoning trainer to confirm the equation first.
If you understand the idea but the setup still slows you down, working through your own questions with a teacher can help. See our online one-to-one Chemistry tuition.