Contents: 8 sections
Syllabus points
- Define and use relative atomic, isotopic, molecular and formula mass.
- Define and use the mole and the Avogadro constant.
- Calculate empirical and molecular formulae from composition data.
- Write and balance equations, including ionic equations with state symbols.
- Perform calculations involving reacting masses, volumes of gases and volumes and concentrations of solutions, including limiting reagent and percentage yield.
Relative masses
Relative atomic mass (Aᵣ) is the weighted mean mass of an atom of an element, relative to 1/12 of the mass of an atom of carbon-12.
The phrase weighted mean is what makes the calculation work. Chlorine is not 35 or 37; it is the average of its isotopes in the proportions in which they occur.
Worked example
Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. Its relative atomic mass is
Relative molecular mass (Mᵣ) is the sum of the Aᵣ values in a molecule; relative formula mass is the same idea for an ionic compound, which has no molecules. Neither has units, because both are ratios.
The mole
One mole is the amount of substance containing the same number of particles as there are atoms in 12 g of carbon-12, which is 6.02 × 10²³, the Avogadro constant.
Three relationships do almost all the work in this topic:
moles = mass ÷ Mᵣ
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moles = concentration × volume (with volume in dm³)
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moles = volume ÷ 24.0 (for a gas at room temperature and pressure, volume in dm³)
The molar gas volume of 24.0 dm³ mol⁻¹ applies at room conditions only. Always check whether the question gives conditions before using it.
Two conversions cause more lost marks than any concept here:
- cm³ to dm³: divide by 1000
- g to kg or tonnes: be sure Mᵣ is in the same mass unit as the mass
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms. The molecular formula is the actual number in a molecule, and it is always a whole-number multiple of the empirical formula.
The method never changes:
- Divide each mass or percentage by the element's Aᵣ.
- Divide every answer by the smallest of them.
- Multiply up if needed to reach whole numbers.
Worked example
A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and its Mᵣ is 180.
| Element | % | ÷ Aᵣ | ÷ smallest |
|---|---|---|---|
| C | 40.0 | 40.0 ÷ 12.0 = 3.33 | 3.33 ÷ 3.33 = 1 |
| H | 6.7 | 6.7 ÷ 1.0 = 6.7 | 6.7 ÷ 3.33 = 2 |
| O | 53.3 | 53.3 ÷ 16.0 = 3.33 | 3.33 ÷ 3.33 = 1 |
The empirical formula is CH₂O, which has a mass of 12.0 + 2.0 + 16.0 = 30.0.
So the molecular formula is C₆H₁₂O₆.
Equations and ionic equations
A balanced equation must have the same number of each type of atom on both sides, and the total charge must balance too.
An ionic equation shows only the species that change. Spectator ions, which appear unchanged on both sides, are left out. For the reaction of silver nitrate with sodium chloride:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
The sodium and nitrate ions are spectators. State symbols are expected and are frequently worth a mark on their own.
Reacting quantities
Every calculation of this kind has the same three steps, and writing them out in order is the reliable way through:
- Convert what you are given into moles.
- Use the balancing numbers in the equation as a ratio to get moles of what you want.
- Convert those moles into the quantity the question asks for.
Worked example
What volume of 0.100 mol dm⁻³ hydrochloric acid reacts exactly with 2.50 g of calcium carbonate?
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Step 1. Mᵣ of CaCO₃ = 40.1 + 12.0 + 48.0 = 100.1
Step 2. The ratio is 1 : 2, so n(HCl) = 2 × 0.0250 = 0.0500 mol.
Step 3.
Limiting reagent and percentage yield
When quantities of two reactants are given, one runs out first and limits the product. Find the moles of each, divide each by its balancing number, and the smaller answer is the limiting reagent.
percentage yield = (actual yield ÷ theoretical yield) × 100
The theoretical yield comes from the limiting reagent, never from the reactant in excess. A yield below 100% is normal: reactions may be reversible, side reactions occur, and product is lost in transfer and purification.
Common mistakes
- Using 24.0 dm³ mol⁻¹ for a gas that is not at room conditions, or for a liquid or solid.
- Forgetting to divide cm³ by 1000 before using moles = c × V.
- Calculating theoretical yield from the reactant in excess rather than the limiting one.
- Rounding intermediate steps. Carry extra figures through and round only at the end.
- Leaving state symbols out of an ionic equation.
- Giving relative atomic mass a unit. It is a ratio and has none.