Quantitative chemistry

Quantitative chemistry is the part of chemistry that works with amounts: masses, moles, concentrations and yields. It answers questions like how much product a reaction makes, how much of each reactant you need, and how concentrated a solution is. In AQA GCSE Chemistry it is topic 4.3, and much of it is also in Combined Science.

What quantitative chemistry covers

This is the list in AQA's GCSE Chemistry specification, with which parts are Higher tier (HT) only and which are in separate Chemistry only, not Combined Science.

The examples below use relative atomic masses H = 1, C = 12, O = 16, Mg = 24, S = 32, Ca = 40 and Fe = 56.

1. Conservation of mass

AQA puts it simply: no atoms are lost or made during a chemical reaction, so the mass of the products equals the mass of the reactants. That is why equations must balance. If 2.4 g of magnesium burns to make 4.0 g of magnesium oxide, the extra 4.0 − 2.4 = 1.6 g must be oxygen from the air.

In an open container, mass can seem to change when a gas is involved. Burning magnesium gains mass, because oxygen from the air joins on. Heating calcium carbonate loses mass, because carbon dioxide escapes. The atoms are all still there; some just entered or left as gas.

2. Relative formula mass

To find the relative formula mass, Mr, add up the relative atomic mass of every atom in the formula. For calcium carbonate, CaCO₃: 40 + 12 + 3 × 16 = 100. More examples are in the relative formula mass guide, and the molar mass calculator works it out for any formula.

3. Moles (Higher)

A mole is a fixed number of particles: the Avogadro constant, 6.02 × 10²³. The mass of one mole of a substance in grams is numerically equal to its relative formula mass, so:

For 22 g of carbon dioxide (Mr 44): 22 ÷ 44 = 0.5 mol, which is 0.5 × 6.02 × 10²³ = 3.01 × 10²³ molecules. See what a mole is for more.

4. Reacting masses (Higher)

To find how much product a reaction makes, go through moles: mass → moles → use the equation's ratio → moles → mass.

Example: what mass of calcium oxide comes from heating 25 g of calcium carbonate? CaCO₃ → CaO + CO₂.

1. Moles of CaCO₃ = 25 ÷ 100 = 0.25 mol.
2. The equation is 1 : 1, so 0.25 mol of CaO forms.
3. Mass of CaO = 0.25 × 56 = 14 g, since Mr of CaO is 40 + 16 = 56.

The stoichiometry calculator works through reacting mass problems step by step.

5. Using moles to balance equations (Higher)

Turn the masses that react into moles and simplify the ratio. 5.6 g of iron reacts exactly with 3.2 g of sulfur. Iron: 5.6 ÷ 56 = 0.1 mol. Sulfur: 3.2 ÷ 32 = 0.1 mol. The ratio is 1 : 1, so the equation is Fe + S → FeS.

6. Limiting reactants (Higher)

Whichever reactant gets used up first is the limiting reactant: once it has gone, the reaction stops, so it decides how much product forms. The other reactant is in excess.

Example: 6 g of magnesium burns in 8 g of oxygen. 2Mg + O₂ → 2MgO. Magnesium: 6 ÷ 24 = 0.25 mol, which needs 0.125 mol of O₂ (4 g). There are 8 g = 0.25 mol of O₂, more than enough, so oxygen is in excess and magnesium is limiting. It makes 0.25 mol of MgO, which is 0.25 × 40 = 10 g.

7. Concentration of solutions

Volumes in cm³ must be divided by 1000 to get dm³. 12 g dissolved to make 250 cm³ (0.250 dm³) of solution has a concentration of 12 ÷ 0.250 = 48 g/dm³. The concentration guide has more.

8. Percentage yield (Chemistry only)

In section 4, the theoretical mass of calcium oxide was 14 g. If only 11.2 g is collected, the yield is 11.2 ÷ 14 × 100 = 80%. Yields are below 100% because some product is lost when it is separated, the reaction may not go to completion, or other reactions happen.

9. Atom economy (Chemistry only)

If calcium oxide is the useful product of CaCO₃ → CaO + CO₂, the atom economy is 56 ÷ 100 × 100 = 56%. The rest of the atoms end up in the carbon dioxide, a waste product. See the atom economy guide.

10. Concentration in mol/dm³ (Chemistry, Higher)

Concentration in mol/dm³ = moles ÷ volume in dm³. 0.05 mol dissolved in 200 cm³ gives 0.05 ÷ 0.200 = 0.25 mol/dm³. This is the unit used in titration calculations, worked through in the neutralisation guide, and the molarity calculator checks it.

11. Volumes of gases (Chemistry, Higher)

AQA's specification states that one mole of any gas at room temperature and pressure (20 °C and 1 atmosphere) takes up 24 dm³. So volume in dm³ = moles × 24. The 0.25 mol of carbon dioxide made in section 4 would fill 0.25 × 24 = 6 dm³.

Common mistakes

• Multiplying Mr by the number in front. In 2MgO the 2 is a number of moles; the Mr of MgO is still 40.
• Ignoring the ratio. Always check the balanced equation before going from moles of one substance to another.
• Leaving volumes in cm³. Divide by 1000 to get dm³ before using concentration or gas volume equations.
• Mixing up yield and atom economy. Yield compares what you got with what you could get; atom economy compares useful product with all the reactants.
• Rounding too early. Keep full values until the last step.

To practise these calculations under time pressure, try today's GCSE Combined Science Higher or GCSE Chemistry Higher short timed paper. Each paper mixes topics.

Read next

Chemistry · Moles and quantitiesRelative formula mass (Mr): how to calculate itHow to calculate relative formula mass from a formula, with brackets and hydrates, the relative atomic mass formula for isotopes, examples and practice.