This mini-lesson walks you through the main ideas 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.
Make a guess: how can you count atoms just by weighing?
Mission 1How do you count atoms by weighing?
Mission 2How can you find an unknown concentration?
Mission 3Why does a mole of any gas fill the same space?
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.
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.
Mission 1 check
How do you count atoms by weighing?
?A chemist weighs out 24 g of magnesium (Aᵣ = 24) and 12 g of carbon (Aᵣ = 12). Which statement about the number of atoms in each pile is correct?
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.
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.
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)
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)
Mission 2 of 3
How can you find an unknown concentration?
Make a guess: how can measuring a volume reveal a concentration?
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 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.
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.
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.
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.
?25.0 cm³ of a mystery hydrochloric acid is exactly neutralised by 15.0 cm³ of 0.200 mol/dm³ sodium hydroxide. HCl + NaOH → NaCl + H₂O. What is the concentration of the acid?
Mission 3 of 3
Why does a mole of any gas fill the same space?
Make a guess: why does a mole of any gas fill the same space?
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.
Mission 3 check
Why does a mole of any gas fill the same space?
?At room temperature and pressure, a flask holds 24 dm³ of oxygen. It is emptied and filled with 24 dm³ of chlorine, which has much heavier molecules. Which statement is correct?
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)
Atom economy: (Mr desired ÷ total Mr products) × 100
You've covered the main ideas 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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Mini-lesson complete!
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You've worked through Quantitative Chemistry for CCEA GCSE Chemistry. 🎉