GCSE · AQA Combined Science · Chemistry Paper 1 · C3 Quantitative Chemistry

Quantitative chemistry, made countable.

The whole of C3 — conservation of mass, relative formula mass, concentration, and (for Higher) the mole, reacting masses and limiting reactants. Built for both tiers.

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Both tiers in one booklet. Everything here is for Foundation and Higher. Anything that's Higher tier only sits in a purple HT box — Foundation students can skip those. Green boxes are required practicals (there are none in this topic). Do one topic at a time; each is about 10–15 minutes.

Topic 01 · 4.3.1 · Conservation & Mₕ

Conservation of mass & balanced equations

By the end of this topic you'll know why mass never changes in a reaction, you'll balance equations confidently, and you'll work out the relative formula mass of any compound.

Part 1Mass is conserved

In a chemical reaction, no atoms are created or destroyed — they're just rearranged into new substances. So the total mass of the products equals the total mass of the reactants. This is the conservation of mass.

Because the atoms are only swapped around, a balanced symbol equation must have the same number of each type of atom on both sides. That's exactly what "balanced" means.

SAME ATOMS, REARRANGED reactants 100 g = products 100 g Nothing is gained or lost — the atoms are just regrouped.
Total mass of reactants = total mass of products

Worked example — finding a missing mass

12 g of carbon reacts completely with oxygen to make 44 g of carbon dioxide. What mass of oxygen reacted?

Ideamass of reactants = mass of products
Set up12 + (oxygen) = 44
Answeroxygen = 44 − 12 = 32 g

⚠ Watch out — you can't change formulae to balance

You balance an equation by putting big numbers in front (the balancing numbers), never by changing a formula. H₂O is water; writing H₂O₂ to "make it balance" turns it into hydrogen peroxide — a different substance. Only the front numbers may change.

Part 2Relative formula mass, Mₕ

The relative atomic mass (Aₕ) of an element is the number under its symbol on the periodic table. The relative formula mass (Mₕ) of a compound is just the Aₕ values of all its atoms added together.

A useful check: the total Mₕ of the reactants equals the total Mₕ of the products in a balanced equation — another way of seeing conservation of mass.

What to add up

Mₕ = sum of all the Aₕ values in the formula recall
Count every atom, including those multiplied by a small subscript number. Remember a bracket multiplies everything inside it, e.g. Ca(OH)₂ has two O and two H.

Worked example — Mₕ of calcium hydroxide

Calculate the Mₕ of Ca(OH)₂. (Aₕ: Ca = 40, O = 16, H = 1)

Count1 × Ca, 2 × O, 2 × H (the bracket means ×2)
Add40 + (2 × 16) + (2 × 1) = 40 + 32 + 2
AnswerMₕ = 74
Quick check

What is the relative formula mass (Mₕ) of sodium carbonate, Na₂CO₃? (Aₕ: Na = 23, C = 12, O = 16)

  • A83
  • B106
  • C51
  • D96
Show answer
B — 106. (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106. Answer A forgets to double the sodium; D leaves out the carbon and one sodium.
Topic 1 — quick quiz
Click to reveal · 4 questions
  1. State the law of conservation of mass.
    No atoms are created or destroyed in a chemical reaction, so the total mass of the products equals the total mass of the reactants.
  2. Calculate the Mₕ of water, H₂O. (Aₕ: H = 1, O = 16)
    (2 × 1) + 16 = 18.
  3. When balancing an equation, what is the one thing you are allowed to change?
    Only the big balancing numbers in front of the formulae. You must never change a chemical formula (its subscripts).
  4. 4 g of hydrogen reacts with 32 g of oxygen to form water. What mass of water is made?
    Mass is conserved, so 4 + 32 = 36 g of water.
Topic 02 · 4.3.1.3 · Mass & gases

Mass changes when a gas is involved

Mass is always conserved — so why does a reaction on a balance seem to get heavier or lighter? The answer is always a gas entering or leaving.

Part 1Open systems and escaping gas

Mass is always conserved. But if you do a reaction in an open container — one a gas can get in or out of — the mass on the balance can appear to change. The atoms haven't vanished; a gas has simply crossed the boundary of what you're weighing.

Two patterns to recognise:

The mass decreases when a gas is given off and escapes into the air. Example: heating a metal carbonate releases carbon dioxide, which floats away, so the solid left behind weighs less.

The mass increases when a gas from the air is a reactant and gets added in. Example: a metal heated in air combines with oxygen, so the solid oxide formed weighs more than the metal you started with.

TWO OPEN SYSTEMS mass falls carbonate CO₂ escapes mass rises metal O₂ added from air
Left: a gas leaves, balance reads less. Right: a gas joins, balance reads more.

⚠ Watch out — mass is NOT really lost or gained

The total mass of everything, including the gas, is unchanged. The balance only shows what sits on the pan. To prove mass is conserved, do the reaction in a sealed (closed) container — then no gas can enter or leave and the reading stays exactly the same.

Worked example — explaining a mass decrease

A student heats copper carbonate in an open crucible. The mass on the balance falls. Explain why.

Reactioncopper carbonate → copper oxide + carbon dioxide
WhyCO₂ gas is produced and escapes into the air from the open crucible.
SoThe solid left weighs less, but total mass (solid + gas) is unchanged.
Quick check

A piece of magnesium ribbon is heated in an open crucible and the mass increases. Why?

  • AMass is created in the reaction
  • BOxygen from the air joins the magnesium to form magnesium oxide
  • CA gas escapes, taking mass with it
  • DThe heat adds energy, which has mass
Show answer
B. Magnesium reacts with oxygen from the air — that oxygen is now part of the solid weighed, so the mass rises. A breaks conservation of mass; C would make it fall, not rise.
Topic 2 — quick quiz
Click to reveal · 4 questions
  1. Why can the mass on a balance fall during a reaction in an open container?
    A gas is produced and escapes into the air, so it's no longer being weighed — but total mass is still conserved.
  2. Why can the mass on a balance rise during a reaction in an open container?
    A gas from the air (e.g. oxygen) reacts and is added to the substance, so the solid being weighed gains mass.
  3. How could you show that mass is actually conserved when a gas is given off?
    Carry out the reaction in a sealed/closed container so no gas can escape — the total mass then stays the same.
  4. Iron rusts slowly in air and the mass increases. Which gas has been added?
    Oxygen (and water) from the air combines with the iron to form iron oxide.
Topic 03 · 4.3.2 · % mass & concentration

Percentage by mass & concentration

Two everyday calculations — how much of a compound is a useful element, and how strong a solution is in g/dm³.

Part 1Percentage by mass of an element

The percentage by mass of an element in a compound tells you what fraction of the compound's mass comes from that element. Farmers use it to compare fertilisers; metallurgists use it to compare ores.

Equation

% mass = (Aₕ × number of atoms of that element) ÷ Mₕ × 100 recall
The "Aₕ × number of atoms" is the total mass of that element in one formula unit; Mₕ is the mass of the whole compound.

Worked example — % nitrogen in ammonium nitrate

Find the percentage by mass of nitrogen in NH₄NO₃. (Aₕ: N = 14, H = 1, O = 16)

Mₕ(2 × 14) + (4 × 1) + (3 × 16) = 28 + 4 + 48 = 80
N mass2 nitrogen atoms = 2 × 14 = 28
Divide% N = 28 ÷ 80 × 100
Answer= 35%

⚠ Watch out — count every atom of the element

If the element appears more than once in the formula, add all of it. NH₄NO₃ has nitrogen in two places — that's 2 × 14, not 14. Forgetting the second nitrogen is the classic slip here.

Part 2Concentration of a solution

The concentration of a solution measures how much solute is dissolved in a given volume. For combined science you work in grams per cubic decimetre, g/dm³. The more solute in the same volume, or the smaller the volume for the same solute, the more concentrated the solution.

Two volume facts you must know: 1 dm³ = 1000 cm³ (one litre), so to change cm³ into dm³ you divide by 1000.

Equation

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³) recall
Volume must be in dm³. If you're given cm³, divide by 1000 first.
SAME SOLUTE, DIFFERENT VOLUME dilute spread out concentrated packed close
More solute particles in the same volume means a higher concentration

Worked example — concentration in g/dm³

25 g of sodium chloride is dissolved to make 500 cm³ of solution. Calculate the concentration.

Volume500 cm³ ÷ 1000 = 0.5 dm³
Equationconcentration = mass ÷ volume = 25 ÷ 0.5
Answer= 50 g/dm³
Quick check

20 g of solute is dissolved in 250 cm³ of solution. What is the concentration in g/dm³?

  • A0.08 g/dm³
  • B5 g/dm³
  • C80 g/dm³
  • D5000 g/dm³
Show answer
C — 80 g/dm³. Convert volume first: 250 ÷ 1000 = 0.25 dm³. Then 20 ÷ 0.25 = 80 g/dm³. Answer A divides by 250 cm³ without converting; B forgets to convert and divides by 4.
Topic 3 — quick quiz
Click to reveal · 4 questions
  1. Write the equation for percentage by mass of an element in a compound.
    % mass = (total Aₕ of that element ÷ Mₕ of the compound) × 100.
  2. Calculate the % by mass of oxygen in water, H₂O. (Aₕ: H = 1, O = 16)
    Mₕ = 18. % O = 16 ÷ 18 × 100 = 88.9% (to 3 s.f.).
  3. How many cm³ are there in 1 dm³, and how do you convert cm³ to dm³?
    1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.
  4. 40 g of sugar is dissolved to make 2 dm³ of solution. Find the concentration.
    concentration = mass ÷ volume = 40 ÷ 2 = 20 g/dm³.
Topic 04 · 4.3.2.1 · The mole (HT)

The mole

Higher tier only. A way to count atoms by weighing them — Avogadro's constant, and the one equation that links mass, moles and Mₕ.

Part 1Counting by weighing

Atoms are far too small and too many to count one by one, so chemists count them in batches called moles. One mole of any substance contains the same huge number of particles.

The clever part: the mass of one mole, in grams, is equal to the relative formula mass (Mₕ). So one mole of carbon (Aₕ = 12) is 12 g; one mole of water (Mₕ = 18) is 18 g. Weigh out the Mₕ in grams and you've counted out exactly one mole of particles.

Higher tier — the mole and Avogadro's constant

One mole of a substance contains 6.02 × 10²³ particles (atoms, molecules, ions or formula units). This number, 6.02 × 10²³ per mole, is the Avogadro constant.

So one mole of carbon atoms is 6.02 × 10²³ carbon atoms, and one mole of water molecules is 6.02 × 10²³ water molecules. Different substances, same number of particles per mole.

Equation

moles = mass (g) ÷ Mₕ recall
Rearranges to mass = Mₕ × moles. Mₕ is the relative formula mass; mass is in grams.
THE FORMULA TRIANGLE mass Mₕ mol cover what you want to find it
mass = Mₕ × moles, or moles = mass ÷ Mₕ

Worked example — mass to moles

How many moles are there in 36 g of water, H₂O? (Mₕ = 18)

Equationmoles = mass ÷ Mₕ
Sub in= 36 ÷ 18
Answer= 2.0 mol

⚠ Watch out — moles, mass and number aren't the same thing

"Moles" is a count of particles, not a mass. 1 mol of hydrogen and 1 mol of lead contain the same number of particles but very different masses. To turn a mass into a number of particles you go mass → moles → × Avogadro constant.

Quick check

How many moles are in 80 g of sodium hydroxide, NaOH? (Mₕ = 40)

  • A0.5 mol
  • B2 mol
  • C3200 mol
  • D40 mol
Show answer
B — 2 mol. moles = mass ÷ Mₕ = 80 ÷ 40 = 2. Answer A divides the wrong way round (40 ÷ 80); C multiplies instead of dividing.
Topic 4 — quick quiz
Click to reveal · 4 questions
  1. [HT] How many particles are in one mole, and what is this number called?
    6.02 × 10²³ particles — the Avogadro constant.
  2. [HT] Write the equation linking moles, mass and Mₕ.
    moles = mass ÷ Mₕ (so mass = Mₕ × moles).
  3. [HT] What is the mass of 0.5 mol of carbon dioxide, CO₂? (Mₕ = 44)
    mass = Mₕ × moles = 44 × 0.5 = 22 g.
  4. [HT] How many moles are in 56 g of nitrogen gas, N₂? (Mₕ = 28)
    moles = 56 ÷ 28 = 2 mol.
Topic 05 · 4.3.2.3 · Reacting masses (HT)

Moles, equations & reacting masses

Higher tier only. Use moles to read a balanced equation as a recipe — and to work out exactly how much product a mass of reactant will make.

Part 1The balancing numbers are mole ratios

The big numbers in front of a balanced equation aren't just there to make the atoms match — they tell you the ratio of moles that react. The equation is a recipe written in moles.

For example, in 2H₂ + O₂ → 2H₂O, two moles of hydrogen react with one mole of oxygen to make two moles of water. If you have the moles of one substance, the ratio gives you the moles of any other.

Higher tier — using moles to balance equations

You can work the other way too: if an experiment tells you the masses that reacted, convert each to moles (moles = mass ÷ Mₕ), then simplify the mole numbers to the smallest whole-number ratio. Those whole numbers are the balancing numbers of the equation.

READ THE EQUATION AS A RATIO 2H₂ + O₂ → 2H₂O 2 mol 1 mol 2 mol the front numbers are the mole ratio
Balancing numbers = the ratio in which substances react, in moles

Part 2Calculating reacting masses

Putting it together gives the most important Higher-tier calculation in C3. Whenever you're given the mass of one substance and asked for the mass of another, follow the same four steps every time: mass → moles → ratio → mass.

Higher tier — calculating reacting masses

The reliable method:

1. Write the balanced equation. 2. Turn the known mass into moles (÷ its Mₕ). 3. Use the equation's ratio to find the moles of the substance you want. 4. Turn those moles back into a mass (× its Mₕ).

Worked example — mass of product

What mass of magnesium oxide forms when 6 g of magnesium burns? 2Mg + O₂ → 2MgO. (Aₕ: Mg = 24, O = 16)

Moles Mg= mass ÷ Aₕ = 6 ÷ 24 = 0.25 mol
Ratio2 Mg : 2 MgO is 1 : 1, so moles MgO = 0.25 mol
Mₕ MgO24 + 16 = 40
Answermass = Mₕ × moles = 40 × 0.25 = 10 g

⚠ Watch out — don't skip the ratio step

It's tempting to convert mass to moles and straight back to mass — but you must apply the balancing ratio in between, or you'll get the wrong answer whenever the ratio isn't 1 : 1. Always check the front numbers of both substances.

Quick check

For 2H₂ + O₂ → 2H₂O, how many moles of water are made from 4 mol of hydrogen (with plenty of oxygen)?

  • A2 mol
  • B8 mol
  • C4 mol
  • D1 mol
Show answer
C — 4 mol. The ratio of H₂ to H₂O is 2 : 2, i.e. 1 : 1. So 4 mol of hydrogen makes 4 mol of water. Answer B wrongly doubles it.
Topic 5 — quick quiz
Click to reveal · 4 questions
  1. [HT] What do the big balancing numbers in an equation tell you?
    The ratio of moles in which the substances react and are produced.
  2. [HT] List the four steps for a reacting-mass calculation.
    Mass → moles (÷ Mₕ) → use the equation ratiomoles → mass (× Mₕ).
  3. [HT] In 2Mg + O₂ → 2MgO, how many moles of MgO come from 0.5 mol of Mg?
    The ratio Mg : MgO is 2 : 2 = 1 : 1, so 0.5 mol of MgO.
  4. [HT] What mass of MgO forms from 0.5 mol of MgO? (Mₕ MgO = 40)
    mass = Mₕ × moles = 40 × 0.5 = 20 g.
Topic 06 · 4.3.2.5 · Limiting reactants (HT)

Limiting reactants

Higher tier only. When two reactants are mixed, one usually runs out first — and that one decides how much product you get.

Part 1The reactant that runs out

In most reactions one reactant is used up before the others. That reactant is the limiting reactant — once it's gone, the reaction stops, no matter how much of the other reactants is left over. The reactant left spare is said to be in excess.

The key consequence: the amount of product is controlled by the limiting reactant. Add more of the limiting reactant and you make more product; adding more of the excess reactant does nothing once the limiting one is gone.

ONE REACTANT RUNS OUT FIRST reactant A only 2 — runs out LIMITING + reactant B spare left over product set by A A is used up; the leftover B is "in excess".
The limiting reactant (A) is used up first and fixes the amount of product

Higher tier — limiting reactants and moles

To find the limiting reactant by calculation: convert the mass of each reactant to moles, then compare those moles against the ratio in the balanced equation. The reactant you have too few moles of, relative to the ratio, is the limiting reactant — and you base the product calculation on it.

⚠ Watch out — limiting isn't just "the smaller mass"

The limiting reactant is not simply whichever has the smaller mass or the fewer moles. You must compare the moles against the equation's ratio. A reactant present in a larger amount can still be limiting if the equation demands even more of it.

Part 2Working it out

Worked example — which reactant is limiting?

0.3 mol of magnesium is added to 0.1 mol of oxygen molecules: 2Mg + O₂ → 2MgO. Which is limiting, and how many moles of MgO form?

Ratio2 Mg react with 1 O₂ (a 2 : 1 ratio)
Mg needs0.3 mol Mg needs 0.15 mol O₂ — but only 0.1 mol O₂ is present
SoO₂ runs out first → oxygen is the limiting reactant (Mg is in excess)
Answer0.1 mol O₂ makes 0.2 mol MgO (ratio O₂ : MgO = 1 : 2)
Quick check

Hydrogen and chlorine react: H₂ + Cl₂ → 2HCl. You mix 5 mol of H₂ with 3 mol of Cl₂. Which is the limiting reactant?

  • AHydrogen — there's more of it
  • BChlorine — there's less of it (a 1 : 1 ratio)
  • CNeither — they react fully
  • DYou can't tell without the masses
Show answer
B — chlorine. The ratio is 1 : 1, so 5 mol H₂ would need 5 mol Cl₂, but only 3 mol is present. Chlorine runs out first; 2 mol of hydrogen is left in excess. (Here the smaller amount is limiting because the ratio is 1 : 1 — but always check the ratio.)
Topic 6 — quick quiz
Click to reveal · 4 questions
  1. [HT] Define the limiting reactant.
    The reactant that is completely used up first, stopping the reaction. It controls the amount of product formed.
  2. [HT] What name is given to the reactant left over at the end?
    It is said to be in excess.
  3. [HT] How do you find the limiting reactant by calculation?
    Convert each reactant's mass to moles, then compare against the equation's mole ratio; the one you have too few moles of (relative to the ratio) is limiting.
  4. [HT] Why does adding more of the excess reactant not make more product?
    Once the limiting reactant is used up the reaction stops, so extra excess reactant has nothing left to react with.
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