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CCEA GCSE Chemistry · Quantitative Chemistry
Mini-Lesson

Quantitative Chemistry

This mini-lesson walks you through the whole of CCEA Quantitative Chemistry: relative formula mass, the mole, reacting masses, percentage composition and yield, empirical formulae, water of crystallisation, concentration, titration, gas volumes and atom economy.

mass (g) weigh it moles amount particles 6.02×10²³ / mol ÷ Mr × Nᴀ

Work through each screen, answer the questions as you go (some are concepts, most are calculations) and collect ⭐ stars. Use the periodic table values H=1, C=12, N=14, O=16, Na=23, Mg=24, S=32, Cl=35.5, Ca=40, Fe=56. Press Start when you're ready.

Relative formula mass, Mr

Adding up the atoms

The relative formula mass (Mr) — called relative molecular mass for covalent molecules — is the sum of the relative atomic masses (Ar) of every atom in the formula.

Mr = sum of all Ar valuesMultiply the Ar by the number of each atom, then add. Brackets multiply everything inside.
Worked example — calcium nitrate, Ca(NO₃)₂

Ca: 1 × 40 = 40

N: 2 × 14 = 28   (two NO₃ groups → two N)

O: 6 × 16 = 96   (two NO₃ groups → 2 × 3 = 6 O)

Mr = 40 + 28 + 96 = 164

Misconception: Mr has no units — it is a relative mass (a ratio to ¹²C), not a mass in grams. Only when you weigh out one mole does that number become a mass in grams.

Calculate

Your turn — relative formula mass

1Work out the relative formula mass (Mr) of sulfuric acid, H₂SO₄. (Ar: H=1, S=32, O=16)
Hint: (2 × 1) + 32 + (4 × 16).
The mole & Avogadro's constant

Counting by weighing

Chemists count particles in moles. One mole always contains the same number of particles — Avogadro's constant, 6.02 × 10²³ per mole. The clever part: the mass of one mole in grams is numerically equal to the Mr.

mass (g) mol Mr cover what you want → the rest is the sum
The mole triangle: mass = mol × Mr,  mol = mass ÷ Mr,  Mr = mass ÷ mol.
moles = mass (g) ÷ Mre.g. 80 g of NaOH (Mr = 40) is 80 ÷ 40 = 2 mol.
Calculate

Your turn — moles from mass

2How many moles are there in 19.6 g of sulfuric acid, H₂SO₄? (You found Mr = 98 a moment ago.)
mol
Hint: moles = mass ÷ Mr = 19.6 ÷ 98.
Reacting masses

The equation gives the ratio

The big numbers in a balanced equation are the mole ratio in which substances react. To find a reacting mass: mass → moles → ratio → moles → mass.

2Mg + O₂ → 2MgO 2 mol Mg = 48 g 1 mol O₂ = 32 g 2 mol MgO = 80 g + 48 + 32 = 80 g — mass is conserved
The 2 : 1 : 2 ratio is in moles, not grams. Notice the total mass in equals the total mass out.

Misconception: mass is conserved — atoms are only rearranged, never created or destroyed. If a product looks lighter (e.g. a gas escapes) the "missing" mass left as gas.

Calculate

Your turn — reacting masses

3In 2Mg + O₂ → 2MgO, what mass of magnesium oxide forms when 6 g of magnesium burns completely? (Ar: Mg=24, O=16)
g
Hint: mol Mg = 6 ÷ 24 = 0.25; ratio Mg : MgO is 2 : 2, so 0.25 mol MgO; mass = 0.25 × 40.
Percentage composition by mass

How much of the mass is one element?

The percentage by mass of an element tells you what fraction of a compound's mass comes from that element — vital for choosing the best fertiliser or ore.

% element = (total Ar of that element ÷ Mr) × 100Use the total mass of that element if there is more than one atom of it.
N 35% O 60% H 5%
Ammonium nitrate, NH₄NO₃ (Mr = 80): nitrogen makes up 28 ÷ 80 = 35% of the mass.
Calculate

Your turn — percentage by mass

4Calculate the percentage by mass of oxygen in calcium carbonate, CaCO₃. (Ar: Ca=40, C=12, O=16)
%
Hint: Mr = 40 + 12 + 48 = 100; oxygen mass = 3 × 16 = 48; % = (48 ÷ 100) × 100.
Empirical formula

The simplest whole-number ratio

The empirical formula is the simplest whole-number ratio of atoms of each element. The molecular formula shows the actual numbers — always a simple multiple of the empirical formula.

Method — from masses

1. Divide each element's mass by its Ar → moles.

2. Divide every answer by the smallest of them.

3. Scale up to whole numbers if needed.

Worked example

An oxide contains 11.2 g Fe and 4.8 g O.

Fe: 11.2 ÷ 56 = 0.2  |  O: 4.8 ÷ 16 = 0.3

Divide by 0.2 → Fe : O = 1 : 1.5 → ×2 → 2 : 3

Empirical formula = Fe₂O₃

Misconception: empirical is not always the molecular formula. Glucose is C₆H₁₂O₆ but its empirical formula is just CH₂O — the simplest ratio.

Quick check

Find the empirical formula

?A compound contains 2.4 g of carbon and 0.6 g of hydrogen. What is its empirical formula? (Ar: C=12, H=1)
Water of crystallisation

Water locked in the crystal

Water of crystallisation is water chemically bonded into a crystal structure. A salt with it is hydrated; with it removed (by heating to constant mass) it is anhydrous. The dot in a formula shows how many water molecules — the degree of hydration.

CuSO₄·5H₂O  →  CuSO₄ + 5H₂OBlue hydrated copper(II) sulfate turns white when heated; the degree of hydration here is 5.
Finding the degree of hydration, x

1. moles of anhydrous salt = mass ÷ Mr(salt)

2. moles of water lost = mass of water ÷ 18

3. x = moles of water ÷ moles of salt

Tip: Mr of H₂O = 18. Heating "to constant mass" means you keep heating until the mass stops changing — all the water has gone.

Calculate

Your turn — degree of hydration

5Heating hydrated magnesium sulfate, MgSO₄·xH₂O, leaves 0.05 mol of MgSO₄ and drives off 0.35 mol of water. Find x.
Hint: x = moles of water ÷ moles of salt = 0.35 ÷ 0.05.
Percentage yield

How much you actually got

The theoretical yield is the maximum mass the equation predicts. The actual yield is what you really collected. The percentage yield compares the two.

% yield = (actual yield ÷ theoretical yield) × 100e.g. theoretical 8.0 g, actual 6.0 g → (6.0 ÷ 8.0) × 100 = 75%.

Yield is below 100% because some product is lost on separation/transfer, side reactions make other products, or the reaction is reversible and never goes to completion.

Misconception: % yield is not % composition. Yield compares mass got to mass possible for a reaction; composition is the fixed mass make-up of one compound. And neither is atom economy, which is about how much of the product mass is the one you want.

Calculate

Your turn — percentage yield

6A reaction should make 40 g of product (theoretical yield) but a student collects only 30 g. Calculate the percentage yield.
%
Hint: (30 ÷ 40) × 100.
Concentration of solutions

How crowded the solution is

Concentration measures how much solute is dissolved in a given volume. CCEA uses two units: mol/dm³ and g/dm³. Remember 1 dm³ = 1000 cm³, so always convert cm³ → dm³ first.

concentration (mol/dm³) = moles ÷ volume (dm³)Triangle: moles = conc × volume; volume = moles ÷ conc.
g/dm³ = mol/dm³ × MrMultiply a concentration in mol/dm³ by the Mr to get g/dm³.

Misconception: mol/dm³ ≠ g/dm³. One counts particles per litre, the other counts grams per litre. They are only equal if Mr happened to be 1 — which never happens. Convert with the Mr.

Calculate

Your turn — concentration

75.85 g of sodium chloride, NaCl (Mr = 58.5), is dissolved to make 250 cm³ of solution. Find the concentration in mol/dm³.
mol/dm³
Hint: moles = 5.85 ÷ 58.5 = 0.1; volume = 250 ÷ 1000 = 0.25 dm³; conc = 0.1 ÷ 0.25.
Titration

Finding an unknown concentration

A titration finds the volume of one solution that exactly reacts with a measured volume of another. CCEA uses methyl orange or phenolphthalein to spot the end point. You pipette a fixed volume into the flask, then run acid/alkali from the burette until the indicator just changes colour.

burette (acid) conical flask (alkali + indicator)
Repeat until concordant titres (within 0.1 cm³) are obtained, then average only the accurate titres.
Calculate

Your turn — titration

825.0 cm³ of sodium hydroxide is exactly neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. For HCl + NaOH → NaCl + H₂O, find the concentration of the NaOH in mol/dm³.
mol/dm³
Hint: mol HCl = (20.0 × 0.100) ÷ 1000 = 0.00200; ratio 1 : 1 so mol NaOH = 0.00200; conc = (0.00200 × 1000) ÷ 25.0.
Gas volumes & Avogadro's Law

One mole of any gas = 24 dm³

At room temperature and pressure (20 °C, 1 atm), one mole of any gas occupies 24 dm³ (24 000 cm³) — its molar volume.

moles of gas = volume (dm³) ÷ 24or volume (cm³) ÷ 24 000. Rearranged: volume (dm³) = moles × 24.

Avogadro's Law: equal volumes of gases at the same temperature and pressure contain the same number of molecules. So gases react in simple volume ratios — you can use the balancing numbers directly on the volumes.

Example: in N₂ + 3H₂ → 2NH₃, 20 cm³ of N₂ reacts with 3 × 20 = 60 cm³ of H₂. No need to find moles for a gas-only ratio.
Calculate

Your turn — gas volume

9What volume, at room temperature and pressure, is occupied by 0.5 mol of carbon dioxide gas?
dm³
Hint: volume = moles × 24 = 0.5 × 24.
Atom economy

How much ends up as useful product

Atom economy measures what fraction of the total mass of products is the desired product. A high atom economy means less waste — important for sustainability and lower costs.

atom economy = (Mr of desired product ÷ total Mr of all products) × 100
Worked example — making quicklime

CaCO₃ → CaO + CO₂

Desired CaO: Mr = 40 + 16 = 56

All products: 56 (CaO) + 44 (CO₂) = 100

Atom economy = (56 ÷ 100) × 100 = 56%

Calculate

Your turn — atom economy

10Hydrogen is made by Zn + 2HCl → ZnCl₂ + H₂. Calculate the atom economy for making H₂. (Mr: ZnCl₂ = 136, H₂ = 2)
%
Hint: total products Mr = 136 + 2 = 138; atom economy = (2 ÷ 138) × 100.
Sort it

Which unit goes where?

Tap a quantity, then tap the box for the unit CCEA uses for it.

📦 mol/dm³ or dm³

⚖️ g or g/dm³

Recap

The equations to know

Relative formula mass: Mr = sum of all Ar (no units)

Mole: moles = mass ÷ Mr  (NA = 6.02×10²³ /mol)

% by mass: (total Ar of element ÷ Mr) × 100

% yield: (actual ÷ theoretical) × 100

Concentration: mol/dm³ = moles ÷ volume(dm³);  g/dm³ = mol/dm³ × Mr

Titration: moles = (vol cm³ × conc) ÷ 1000

Gas volume: moles = volume(dm³) ÷ 24 (at RTP)

Atom economy: (Mr desired ÷ total Mr products) × 100

You've covered all of CCEA Quantitative Chemistry — from Mr and the mole to titrations, gas volumes and atom economy. Press Finish to see your score.

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