Topic 3 of 10

Quantitative Chemistry

Master the maths of chemistry — from calculating moles and reacting masses to titrations and percentage yield.

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Conservation of Mass

The law of conservation of mass states that no atoms are lost or made during a chemical reaction. The total mass of the products equals the total mass of reactants.

This is because the same atoms are present before and after the reaction — they have just been rearranged. This is why we must balance chemical equations.

Atoms are neither created nor destroyed in a chemical reaction — only rearranged into different substances.

Apparent Changes in Mass

Sometimes, the mass of a reaction vessel appears to change:

  • If a gas escapes (e.g., CO₂ from a thermal decomposition), mass appears to decrease.
  • If a gas is gained from the air (e.g., oxygen during oxidation of metals), mass appears to increase.

In reality, if the system were sealed, the total mass would remain unchanged.

Relative Formula Mass (Mr)

The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula.

Calculate the Mr of calcium carbonate (CaCO₃)

Ca = 40, C = 12, O = 16

Mr = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100

Calculate the Mr of magnesium hydroxide (Mg(OH)₂)

Mg = 24, O = 16, H = 1

Mr = 24 + 2 × (16 + 1) = 24 + 34 = 58

Moles & Avogadro's Constant

A mole is simply a number: 6.02 × 10²³ (Avogadro's constant). One mole of any substance contains exactly this number of particles (atoms, molecules, or ions).

moles = mass ÷ Mr

How many moles in 11 g of CO₂?

Mr of CO₂ = 12 + (2 × 16) = 44

Moles = 11 ÷ 44 = 0.25 mol

The moles equation triangle is your best friend. Cover what you want to find: mass = moles × Mr, moles = mass ÷ Mr, Mr = mass ÷ moles.

Reacting Masses

Use a balanced equation plus the moles formula to predict masses used or produced in reactions.

What mass of magnesium oxide is produced from 6 g of magnesium?

Equation: 2Mg + O₂ → 2MgO

Step 1: Moles of Mg = 6 ÷ 24 = 0.25 mol

Step 2: Ratio — 2Mg : 2MgO → 1:1 so moles of MgO = 0.25 mol

Step 3: Mass of MgO = 0.25 × 40 = 10 g

Limiting Reactants

In many reactions, one reactant is used up before the others. This reactant is called the limiting reactant — it limits the amount of product that can be formed.

The other reactant(s) are said to be in excess.

The amount of product formed is always determined by the limiting reactant, not the one in excess.
To identify the limiting reactant: (1) calculate moles of each reactant, (2) use the balanced equation ratios to compare, (3) the one that runs out first is the limiting reactant.

Concentration

Concentration tells you how much solute is dissolved in a given volume of solution.

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³)
Remember: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.

2.5 g of NaOH dissolved in 500 cm³. Find concentration in g/dm³.

Volume = 500 ÷ 1000 = 0.5 dm³

Concentration = 2.5 ÷ 0.5 = 5 g/dm³

Percentage Yield

The percentage yield compares the actual amount of product obtained to the theoretical maximum.

% yield = (actual yield ÷ theoretical yield) × 100

Three main reasons why yields are always less than 100%:

  • The reaction is reversible and doesn't go to completion.
  • Some product is lost during transfer (e.g., filtration, evaporation).
  • Side reactions produce unwanted by-products.

Atom Economy

Atom economy measures the proportion of reactant atoms that become useful product.

atom economy = (Mr of desired product ÷ total Mr of all products) × 100

High atom economy is desirable. It means less waste, lower costs, and a more sustainable process.

Addition reactions have 100% atom economy (all atoms end up in one product). Substitution reactions usually have low atom economy because they produce by-products.

Titrations

A titration is a technique used to find the concentration of an unknown acid or alkali by reacting it with one of known concentration.

The Method

  1. Use a pipette to measure a fixed volume of alkali into a conical flask.
  2. Add a few drops of indicator (e.g., phenolphthalein or methyl orange).
  3. Fill a burette with acid of known concentration.
  4. Add acid gradually, swirling, until the indicator permanently changes colour — the end point.
  5. Record the titre (volume of acid added). Repeat until concordant results are achieved (within 0.10 cm³).
When describing the titration method, always mention: using a pipette for the fixed volume, a burette for the variable volume, and an indicator for the end point. Repeat to get concordant results.

Molar Volume of Gases (HT)

At room temperature and pressure (RTP: 20°C, 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).

volume (dm³) = moles × 24
moles = volume (dm³) ÷ 24

What volume does 0.5 mol of oxygen gas occupy at RTP?

Volume = 0.5 × 24 = 12 dm³