Quantitative chemistry
Conservation of mass, relative formula mass, moles, and calculating masses in reactions and concentrations of solutions.
Watch the lesson
Every point below, worked through in one video, and each point links straight to its own part.
Learn
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 amountHigher only
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 equationsHigher only
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 yieldChemistry only
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).
Common mistakes in Quantitative chemistry
The slips that cost marks year after year, and what to do instead.
Doubling only the hydrogen in Ca(OH)₂, so its relative formula mass comes out as 58 instead of 74.
The 2 after a bracket multiplies everything inside it. Ca(OH)₂ has one Ca, two O and two H: 40 + (2 × 16) + (2 × 1) = 74.
Balancing an equation by changing a formula, such as turning H₂O into H₂O₂.
Formulas are fixed, so only the numbers in front can change. For H₂ + O₂ → H₂O, put a 2 in front of H₂O and then another in front of H₂: 2H₂ + O₂ → 2H₂O.
Saying mass was not conserved because the contents of an open flask got lighter or heavier.
Look for a gas: heating a metal carbonate lets carbon dioxide escape, so the solid gets lighter, and burning magnesium takes in oxygen from the air, so the product is heavier. Either way, mass is still conserved.
Using a volume in cm³ in a concentration calculation without converting it to dm³.
Divide by 1000 to turn cm³ into dm³. Dissolving 5 g in 250 cm³ gives 5 ÷ 0.25 = 20 g/dm³, not 0.02 g/dm³.
Ignoring the balancing numbers in a reacting-mass calculation, as if every substance had the same number of moles.Higher only
Convert mass to moles (mass ÷ Mr), use the ratio from the balanced equation, then convert back. In 2Mg + O₂ → 2MgO, 0.2 mol of magnesium reacts with only 0.1 mol of oxygen.
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.