The mole (Higher)
- Chemical amounts are measured in moles (mol).
- One mole of a substance contains the same number of particles as one mole of any other substance: the Avogadro constant, 6.02 × 10²³ per mole.
- The mass of one mole of a substance in grams equals its relative formula mass (Mr) or relative atomic mass (Ar). For example, one mole of CO₂ has a mass of 44 g.
- Number of moles = mass (g) ÷ Mr.
Reacting masses (Higher)
- A balanced equation shows how many moles react. 2Mg + O₂ → 2MgO means 2 mol of Mg react with 1 mol of O₂ to make 2 mol of MgO.
- To find a mass: work out moles of the known substance, use the equation's ratio to find moles of the other, then multiply by its Mr.
- Example: 12 g of Mg is 12 ÷ 24 = 0.5 mol, which makes 0.5 mol of MgO = 0.5 × 40 = 20 g.
Balancing equations from masses (Higher)
- Convert the masses of reactants and products to moles, then divide by the smallest number of moles to get the simplest whole-number ratio.
- Example: 4.8 g Mg, 3.2 g O₂ and 8.0 g MgO are 0.2, 0.1 and 0.2 mol, a ratio of 2 : 1 : 2, so 2Mg + O₂ → 2MgO.
Limiting reactants (Higher)
- In a reaction, one reactant may be completely used up. It is the limiting reactant, because it limits the amount of product that can form.
- The other reactant is in excess: some of it is left over.
- The mass of product depends on the amount of the limiting reactant.
Volumes of gases (Higher, separate sciences)
- One mole of any gas occupies 24 dm³ at room temperature and pressure (20 °C and 1 atmosphere).
- Volume of gas (dm³) = moles × 24. For example, 0.25 mol of any gas occupies 6 dm³ at RTP.
Empirical formulae (Edexcel and OCR)
- The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. For example, the empirical formula of glucose, C₆H₁₂O₆, is CH₂O.
- To find it from masses: divide the mass of each element by its Ar to get moles, then divide by the smallest number of moles to get the simplest whole-number ratio.
- Example: 2.4 g of magnesium reacts with 1.6 g of oxygen. Mg: 2.4 ÷ 24 = 0.1 mol; O: 1.6 ÷ 16 = 0.1 mol. The ratio is 1 : 1, so the empirical formula is MgO.
- Core practical (Edexcel): heat a weighed piece of magnesium ribbon in a crucible, lifting the lid now and then, until it stops glowing. Reweigh to find the mass of oxygen that combined, then work out the empirical formula.
- The molecular formula is a whole-number multiple of the empirical formula: divide the Mr by the mass of the empirical formula to find the multiple.
Key terms
- Mole
- The unit for amount of substance; one mole contains 6.02 × 10²³ particles.
- Avogadro constant
- The number of particles in one mole: 6.02 × 10²³.
- Molar mass
- The mass of one mole of a substance in grams, equal to its Mr.
- Limiting reactant
- The reactant that is completely used up, limiting the amount of product.
- In excess
- More than enough of a reactant; some is left over.
- RTP
- Room temperature and pressure: 20 °C and 1 atmosphere.
- Molar gas volume
- The volume of one mole of any gas at RTP: 24 dm³.
- Empirical formula
- The simplest whole-number ratio of atoms of each element in a compound.