Quantitative Chemistry — AQA Combined Science: Trilogy
Quantitative chemistry is about the amounts of substances in reactions — masses, moles and concentrations. It involves a lot of calculation, so practise is essential.
Conservation of mass
In a chemical reaction, no atoms are created or destroyed, so the total mass of the products equals the total mass of the reactants. This is why symbol equations must be balanced.
Sometimes mass appears to change:
- If a gas is given off (e.g. CO₂ from a carbonate), the mass appears to decrease because gas escapes.
- If a substance reacts with a gas from the air (e.g. a metal reacting with oxygen), the mass appears to increase.
In a sealed container, the mass stays constant.
Relative formula mass (Mr)
The relative formula mass (Mr) of a compound is the sum of the relative atomic masses (Ar) of all the atoms in its formula.
Example: water, H₂O. Ar(H) = 1, Ar(O) = 16. Mr = (2 × 1) + 16 = 18.
In a balanced equation, the sum of the Mr of the reactants equals the sum of the Mr of the products — a check on conservation of mass.
The mole (Higher Tier)
Chemical amounts are measured in moles (mol).
- One mole of any substance contains the same number of particles: 6.02 × 10²³ (the Avogadro constant).
- The mass of one mole of a substance (in grams) equals its relative formula mass.
$$\text{moles} = \frac{\text{mass (g)}}{\text{Mr}}$$
You can use this to convert between mass and moles in calculations.
Reacting masses
The balanced equation tells you the ratio in which substances react. Steps for a reacting-mass calculation:
- Write the balanced equation.
- Convert the known mass to moles.
- Use the equation ratio to find moles of the substance you want.
- Convert moles back to mass.
Example: in 2Mg + O₂ → 2MgO, 2 moles of magnesium make 2 moles of magnesium oxide — a 1:1 ratio between Mg and MgO.
Limiting reactants
When two reactants are mixed, one is often used up first — this is the limiting reactant. It limits the amount of product formed; the other reactant is in excess. The amount of product is directly proportional to the amount of the limiting reactant. To find which is limiting, work out the moles of each and compare with the equation ratio.
Concentration of solutions
The concentration of a solution tells you how much solute is dissolved in a given volume.
$$\text{concentration (g/dm}^3) = \frac{\text{mass of solute (g)}}{\text{volume (dm}^3)}$$
(Remember 1 dm³ = 1000 cm³, so divide a volume in cm³ by 1000.)
- The more solute dissolved in a given volume, the more concentrated the solution.
- The larger the volume for the same mass, the more dilute the solution.
At Higher Tier you may also use concentration in mol/dm³: $$\text{concentration (mol/dm}^3) = \frac{\text{moles}}{\text{volume (dm}^3)}$$
Worked example: concentration
Dissolve 20 g of sodium chloride in 250 cm³ of water.
- Volume = 250 ÷ 1000 = 0.25 dm³.
- Concentration = 20 ÷ 0.25 = 80 g/dm³.
Worked example: reacting mass (Higher)
What mass of magnesium oxide forms from 6 g of magnesium? (Ar: Mg = 24, O = 16)
- 2Mg + O₂ → 2MgO
- Moles of Mg = 6 ÷ 24 = 0.25 mol.
- Ratio Mg : MgO is 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol.
- Mr of MgO = 24 + 16 = 40.
- Mass = 0.25 × 40 = 10 g.
Exam tips
- Always balance the equation first in reacting-mass questions.
- Watch your units — convert cm³ to dm³ by dividing by 1000.
- Lay calculations out clearly, step by step, and show your working (method marks).
- For "mass appears to change" questions, identify whether a gas is lost or gained.
- Learn the three key equations: Mr, moles = mass ÷ Mr, and concentration = mass ÷ volume.