Quantitative chemistry
Conservation of mass, relative formula mass, moles, and calculating masses in reactions and concentrations of solutions.
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Quantitative chemistry, explained point by point
Everything the GCSE specification expects you to be able to do, and how to actually do it - the same lesson a signed-in student studies from.
Use conservation of mass and balance equations
Atoms are never made or destroyed in a reaction, so the total mass of reactants equals the total mass of products. A balanced equation reflects this - the same number of each atom on both sides. Balance H₂ + O₂ → H₂O by adjusting numbers in front until it reads 2H₂ + O₂ → 2H₂O.
Calculate relative formula mass (Mr)
The relative formula mass is the sum of the relative atomic masses of all the atoms in a formula. For water, H₂O, that is (2 × 1) + 16 = 18. For a bracketed formula like Ca(OH)₂, multiply everything inside the bracket by the 2: 40 + 2 × (16 + 1) = 74.
Use the mole to link mass and amount
A mole is just a fixed number of particles, and one mole of a substance has a mass in grams equal to its Mr. The link is moles = mass ÷ Mr, so 36 g of water is 36 ÷ 18 = 2 moles. This equation turns masses you can weigh into amounts you can reason about.
Calculate reacting masses from balanced equations
The numbers in a balanced equation give the ratio of moles that react. Convert the known mass to moles, use the ratio to find moles of the other substance, then convert back to mass. Because mass is conserved, a quick check is that the products' masses should account for the reactants'.
Calculate concentration of solutions in g/dm³
Concentration measures how much solute is dissolved in a given volume: concentration = mass ÷ volume, in grams per cubic decimetre (1 dm³ = 1000 cm³). Dissolving 20 g of salt in 0.5 dm³ of water gives 40 g/dm³. More solute or less water both make a solution more concentrated.
Calculate percentage yield
The percentage yield compares how much product you actually got with the maximum possible: percentage yield = (actual ÷ theoretical) × 100. If a reaction could make 8 g but you collect 6 g, that is 75%. Yields fall short because reactions may not finish, some product is lost in handling, or side reactions occur.
Quantitative chemistry key terms
The words the specification and the mark schemes use, each defined the way an examiner wants it.
- Conservation of mass
- No atoms are made or destroyed in a chemical reaction, so the total mass of the products equals the total mass of the reactants.
- Relative formula mass (Mr)
- The sum of the relative atomic masses of all the atoms in a formula. For H₂O it is 1 + 1 + 16 = 18.
- The moleHigher only
- The chemist's counting unit: one mole of a substance contains 6.02 × 10²³ particles, and has a mass in grams equal to its relative formula mass.
- Avogadro constantHigher only
- The number of particles in one mole: 6.02 × 10²³ per mole.
- Limiting reactantHigher only
- The reactant that is completely used up first, stopping the reaction. The other reactants are in excess.
- Concentration
- The amount of solute dissolved in a given volume of solution, measured in g/dm³ (or mol/dm³).
Practice
Try a Quantitative chemistry question
A GCSE-style original question from this topic. Have a go before you open the working - deciding on an answer first is what makes the working stick.
Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Work out the mass of magnesium oxide made when 12 g of magnesium burns completely, in grams (g). (Ar: Mg = 24; Mr of MgO = 40)
- 10
- 40
- 20
- 8
Show the answer and the working
Answer: 20
12 g of Mg is 12 ÷ 24 = 0.5 mol; the equation shows 2 mol Mg makes 2 mol MgO (a 1 : 1 ratio), so 0.5 mol MgO forms, with mass 0.5 × 40 = 20 g.
- Moles of Mg = mass ÷ Ar = 12 ÷ 24 = 0.5 mol.
- The equation 2Mg + O₂ → 2MgO shows 2 mol of Mg make 2 mol of MgO, a 1 : 1 ratio.
- So 0.5 mol of MgO is made.
- Mass of MgO = 0.5 × 40 = 20 g.