1. Percentage composition
Percentage composition: the percentage by mass of each element in a compound.
Method: find the molar mass → mass of the element in 1 mol → divide → × 100. All the percentages add up to 100.
A pure compound always has the same percentage composition, whatever its source (law of constant proportions).
Example: ammonium nitrate, NH₄NO₃ (M = 80) contains 28 g of nitrogen per mole, so 35% nitrogen; urea, CO(NH₂)₂ (M = 60), contains 46.67% nitrogen.
| Compound | Molar mass / g | % of the named element |
|---|---|---|
| NH₄NO₃ | 80 | N: 35% |
| CaCO₃ | 100 | O: 48% |
| CH₄ | 16 | C: 75% |
Formulas
- Percentage of an element
Key terms
- Percentage composition
The percentage by mass of each element present in a compound.
- Law of constant proportions
A pure compound always contains the same elements combined in the same proportion by mass.
2. Water of crystallisation
Water of crystallisation: the fixed number of water molecules that are part of the crystal structure of a hydrated salt, written after a dot (CuSO₄·5H₂O).
- Include the water in the molar mass: CuSO₄·5H₂O = 64 + 32 + 64 + 5 × 18 = 250 (Cu = 64).
- % water = (n × 18 ÷ molar mass of the hydrate) × 100; for CuSO₄·5H₂O, 90 ÷ 250 × 100 = 36%.
- Heating drives the water off: mass lost = water; mass left = anhydrous salt. The number of water molecules, x, = moles of water ÷ moles of anhydrous salt.
| Hydrated salt | Molar mass / g | % water |
|---|---|---|
| CuSO₄·5H₂O | 250 | 36% |
| MgSO₄·7H₂O | 246 | 51.2% |
| CaSO₄·2H₂O | 172 | 20.9% |
Key terms
- Water of crystallisation
The fixed number of molecules of water present in one formula unit of a hydrated crystalline salt.
- Anhydrous
Without water of crystallisation.
3. Empirical and molecular formulae
- Empirical formula: the simplest whole-number ratio of atoms of the elements in a compound.
- Molecular formula: the actual number of atoms of each element in one molecule.
- Molecular formula = n × empirical formula, where n = molar mass ÷ empirical formula mass (a whole number, often 1).
- For a gas or vapour: molar mass = 2 × vapour density.
Examples: glucose C₆H₁₂O₆ → CH₂O (n = 6); ethyne C₂H₂ and benzene C₆H₆ both → CH; water H₂O → H₂O (n = 1).
| Molecular formula | Empirical formula | n |
|---|---|---|
| C₆H₁₂O₆ | CH₂O | 6 |
| H₂O₂ | HO | 2 |
| N₂O₄ | NO₂ | 2 |
Formulas
- Multiplying factor
Key terms
- Empirical formula
The formula showing the simplest whole-number ratio of the atoms of each element in a compound.
- Molecular formula
The formula showing the actual number of atoms of each element in one molecule of a compound.
4. Empirical formula from percentage composition
| Step | What to do |
|---|---|
| 1 | Write the percentage of each element |
| 2 | Divide each by its atomic mass (relative number of moles of atoms) |
| 3 | Divide every result by the smallest |
| 4 | Make whole numbers: 1.5 → × 2; 1.33 or 1.67 → × 3; 1.25 → × 4 |
| 5 | Write the empirical formula |
Then, if the vapour density or molar mass is given, find n and the molecular formula. Example: 85.7% C and 14.3% H gives CH₂ (formula mass 14); VD 28 → M = 56 → n = 4 → C₄H₈.
| Element | % | ÷ atomic mass | ÷ smallest |
|---|---|---|---|
| N | 82.35 | 5.88 | 1 |
| H | 17.65 | 17.65 | 3 |
Key terms
- Empirical formula mass
The sum of the atomic masses of the atoms in the empirical formula.
5. Empirical formula from combining masses
When an experiment gives masses (for example, a metal burnt to its oxide, or an oxide reduced to the metal):
- mass of the second element = mass of the compound − mass of the first element;
- divide each mass by its atomic mass, divide by the smallest and make whole numbers, exactly as with percentages.
Example: 8.0 g of an iron oxide contains 5.6 g of iron, so 2.4 g of oxygen; Fe 5.6 ÷ 56 = 0.1, O 2.4 ÷ 16 = 0.15; ratio 1 : 1.5 = 2 : 3 → Fe₂O₃.
| Data | Element masses | Empirical formula |
|---|---|---|
| 0.24 g Mg gives 0.40 g oxide | Mg 0.24 g, O 0.16 g | MgO |
| 8.0 g oxide contains 5.6 g Fe | Fe 5.6 g, O 2.4 g | Fe₂O₃ |
Key terms
- Combining masses
The masses of the elements that combine with each other to form a compound.
6. Calculations from balanced equations
The coefficients of a balanced equation give the mole ratio (and the volume ratio of gases at the same temperature and pressure). They are not a mass ratio.
Three steps: (1) given quantity → moles (mass ÷ M, or volume at STP ÷ 22.4); (2) × mole ratio from the equation; (3) moles → wanted quantity (× M for mass, × 22.4 for volume at STP, × Avogadro's number for particles).
| Type | Example |
|---|---|
| Mass–mass | 10 g NaOH → 0.25 mol → 0.25 mol NaCl → 14.6 g |
| Mass–volume | 2.7 g Al → 0.1 mol → 0.15 mol H₂ → 3.36 dm³ |
| Volume–volume | 6.72 dm³ Cl₂ → 13.44 dm³ HCl (ratio 1 : 2) |
Total mass of reactants used = total mass of products (conservation of mass).
| Given | Wanted | Last step |
|---|---|---|
| Mass of A | Mass of B | × molar mass of B |
| Mass of A | Volume of gas B at STP | × 22.4 dm³ |
| Volume of gas A | Volume of gas B (same conditions) | Volume ratio only |
Formulas
- Moles of B from moles of A
- Decomposition of potassium chlorate
Key terms
- Stoichiometry
The calculation of the quantities of reactants and products in a chemical reaction from its balanced equation.
- Mole ratio
The ratio of the coefficients of two substances in a balanced equation.
Exam tips
Common mistakes
- Rounding a ratio such as 1.5 or 1.33 instead of multiplying through.
- Leaving out the water of crystallisation from the molar mass of a hydrate.
- Treating the coefficients of an equation as a mass ratio.
- Forgetting the mole ratio (for example 2 : 3 for Al : H₂).
- Using 22.4 dm³ for a gas that is not at STP, or for a solid or liquid.
How marks are usually awarded
- Empirical formula (2–3 marks): the division table, the whole-number ratio, the formula.
- Molecular formula (1 mark): molar mass from vapour density and the value of n.
- Equation calculations (3 marks): moles of the given substance, the mole ratio, the final answer with a unit.