3.1.7Chemical Reactions
Using Moles to Balance Equations — OCR Gateway GCSE Combined Science A J250 - Chemistry Revision Notes & Topical Questions
5 min read 7 parts Lesson 8 of 20
Part 1 of 5Why masses can balance an equation (HT only)
Coefficients show mole ratios
- Coefficients show mole ratios The large numbers in front of formulae in a balanced equation tell you the ratio in which amounts of substances react and form.
- Because these numbers count moles, experimental masses can be converted into moles to recover them.
What the data must give
- What the data must give You need the mass of each reactant and product, plus relative atomic masses from the Periodic Table.
- Every substance in the reaction must be included, otherwise the ratio is incomplete.
Why the answer must be whole numbers
- Why the answer must be whole numbers You cannot have half a molecule taking part,
- so the final coefficients are the smallest set of whole numbers in the correct ratio.
Part 2 of 5Method: from masses to an equation (HT only)
1
Work out the relative formula mass (Mr) of each substance by adding the relative atomic masses of its atoms.
2
Convert each mass to moles using: amount (mol) = mass (g) ÷ Mr (g/mol).
3
Divide every answer by the smallest number of moles to get a simple ratio.
4
If any value is not a whole number (for example 2.5), multiply every value by the same small number (here 2) until all are whole.
5
Write these whole numbers in front of the correct formulae, reactants on the left and products on the right.
6
Check that each element has the same number of atoms on both sides.
Part 3 of 5Worked example
| Substance | Mass (g) | Mr | Moles | Ratio |
|---|---|---|---|---|
| CH3OH | 64 | 32 | 64 ÷ 32 = 2 | 2 |
| O2 | 96 | 32 | 96 ÷ 32 = 3 | 3 |
| CO2 | 88 | 44 | 88 ÷ 44 = 2 | 2 |
| H2O | 72 | 18 | 72 ÷ 18 = 4 | 4 |
Part 4 of 5Reading the result
2CH3OH + 3O2 → 2CO2 + 4H2O
Ratio matches coefficients
- Ratio matches coefficients The mole ratio 2 : 3 : 2 : 4 gives the coefficients directly.
Atom check
- Atom check Carbon: 2 left, 2 right. Hydrogen: 8 left, 8 right. Oxygen: 2 + 6 = 8 left, 4 + 4 = 8 right.
Conservation of mass
- Conservation of mass Reactant masses (64 + 96 = 160 g) equal product masses (88 + 72 = 160 g), which confirms the data are consistent.
Part 5 of 5When a product formula is unknown (HT only)
1Find element masses
- Find element masses Write down the mass of each element that combined, from the experiment.
2Convert to moles
- Convert to moles Divide each element's mass by its relative atomic mass.
3Simplest ratio
- Simplest ratio Divide all mole values by the smallest one, then scale to whole numbers.
4Empirical formula
- Empirical formula Use the ratio as subscripts, then insert the formula into the balanced symbol equation.
Key terms
| Term | Meaning |
|---|---|
| Coefficient | The number written in front of a formula in a balanced equation, showing the relative number of moles. |
| Mole | The amount of substance containing 6.02 × 1023 particles; mass in grams equal to the relative formula mass is one mole. |
| Relative formula mass (Mr) | The sum of the relative atomic masses of all atoms in a formula. |
| Empirical formula | The simplest whole-number ratio of atoms of each element in a compound. |
| Stoichiometry | The quantitative relationship between the amounts of reactants and products in a reaction. |
Avoid these mistakes
1
Dividing masses by atomic number instead of relative atomic mass or Mr — always use the mass values.
2
Changing the small subscripts in formulae to balance an equation; only the large coefficients may be changed.
3
Stopping at a ratio such as 1 : 1.5 : 1 — multiply through by 2 to reach whole numbers.
4
Forgetting to use Mr for molecules: O2 has Mr 32, not 16.
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