Overview
What you’ll learn
- Explain conservation of mass, including “lost” or “gained” mass in gas reactions.
- Balance equations by changing only the big numbers, and add state symbols.
- Calculate Mr from Ar values.
- Compute concentrations in g/dm³ with correct volume conversions.
- (Higher) Convert mass ↔ moles and chain moles through a balanced equation to a reacting mass.
- (Higher) Spot the limiting reactant and predict the maximum yield.
- Estimate uncertainty as half the range of repeats.
Specification checklist (AQA Trilogy 8464 · Edexcel 1SC0 · OCR Gateway J250)
Statements are paraphrased from the shared DfE combined science subject content and grouped by tier. Foundation content is studied by every student; Higher statements are examined on Higher-tier papers only. This chapter covers only what Combined Science (double award) examines — separate-science extras live on the GCSE Biology, Chemistry and Physics hubs. QuizQuip is not affiliated with AQA, Pearson Edexcel or OCR.
Foundation — every student
- 13.1Explain conservation of mass in chemical reactions — atoms are rearranged, never created or destroyed — and account for apparent mass changes when gases enter or escape.
- 13.2Balance symbol equations by adjusting the large balancing numbers, and write balanced equations with state symbols (s), (l), (g) and (aq).
- 13.3Calculate the relative formula mass (Mr) of a compound from relative atomic masses.
- 13.4Calculate the concentration of a solution in g/dm³ from the mass of solute and the volume of solution, converting cm³ to dm³.
- 13.7Estimate the uncertainty in a mean from repeated measurements as half the range.
Higher — Higher tier only
- 13.5(Higher) Use the Avogadro constant, convert between mass and moles with moles = mass ÷ Mr, and calculate reacting masses from balanced equations.
- 13.6(Higher) Identify the limiting reactant from amounts in moles and calculate the maximum mass of product.
Key points
- 1Conservation of mass: atoms are only rearranged — total mass of reactants = total mass of products.
- 2Apparent losses: a gas escaping an open flask. Apparent gains: burning magnesium picking up oxygen from the air.
- 3Balance equations by changing big numbers only — never the small subscript numbers.
- 4Mr = all the Ar values in a formula added up: H₂SO₄ = 2(1) + 32 + 4(16) = 98.
- 5Concentration (g/dm³) = mass (g) ÷ volume (dm³); 1000 cm³ = 1 dm³.
- 6(HT) One mole = 6.02 × 10²³ particles; moles = mass ÷ Mr.
- 7(HT) Reacting masses: moles of what you know → ratio from the equation → moles of what you want → mass.
- 8(HT) The limiting reactant runs out first and fixes the maximum product.
- 9Uncertainty of a mean ≈ half the range of the repeat measurements.
Notes · 1 of 6
Conservation of mass
No atom vanishes — the books always balance.
In any chemical reaction the atoms of the reactants are rearranged into new substances — none are destroyed, none appear from nowhere. So the total mass of reactants equals the total mass of products, every time. Apparent exceptions come from gases: magnesium burning in an open crucible gains mass (oxygen joins from the air), and a fizzing tablet in an open flask loses mass (carbon dioxide escapes). Seal the system and the balance never moves.
Worked example — the escaping gas
In a sealed flask, 5.0 g of calcium carbonate reacts completely with excess dilute acid, releasing carbon dioxide. Afterwards the gas is allowed to escape, and the flask plus its remaining contents lose 2.2 g of mass. Since the only thing that left the flask was the gas, 2.2 g of CO₂ must have been produced — conservation of mass lets you weigh what you cannot see.
Notes · 2 of 6
Balancing equations
Fix the big numbers, never the little ones.
A balanced equation has equal numbers of every atom on both sides. Balance by putting large numbers in front of formulas — changing a subscript (like H₂O → H₂O₂) invents a different substance and loses all the marks. Method: tally each element, adjust the biggest molecule first, then sweep back through to re-check everything.
Worked example — balancing propane combustion
C₃H₈ + O₂ → CO₂ + H₂O. Three carbons → 3CO₂; eight hydrogens → 4H₂O. That demands 3×2 + 4 = 10 oxygens, so O₂ needs a 5: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check: C 3 = 3 ✓, H 8 = 8 ✓, O 10 = 10 ✓. State symbols finish it: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).
Notes · 3 of 6
Relative formula mass
Add up the atomic masses named in the formula.
The relative formula mass (Mr) is the sum of the relative atomic masses (Ar) of every atom shown in the formula. Subscripts multiply; brackets multiply everything inside.
Worked example — Mr of H₂SO₄
Using Ar: H = 1, S = 32, O = 16. Mr = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98. For Ca(OH)₂: 40 + 2 × (16 + 1) = 40 + 34 = 74 — the bracket doubles the whole OH group, not just the oxygen.
Notes · 4 of 6
Concentration
How much stuff per cubic decimetre of solution.
Concentration = mass of solute ÷ volume of solution, quoted in g/dm³. Volumes measured in cm³ must be divided by 1000 first: 250 cm³ = 0.25 dm³. A bigger mass in the same volume means a more concentrated solution; adding water dilutes it.
Worked example — concentration from crystals
25 g of salt is dissolved in water and made up to 500 cm³ of solution. Volume = 500 ÷ 1000 = 0.5 dm³, so concentration = 25 ÷ 0.5 = 50 g/dm³. If the same salt were made up to 1 dm³, the concentration would fall to 25 g/dm³ — same solute, more water, weaker solution.
Notes · 5 of 6
Moles and reacting masses (Higher)
The chemistry accountant’s unit — and the three-step mass prediction.
One mole of a substance is 6.02 × 10²³ particles (the Avogadro constant) and weighs exactly its Mr in grams. The converting equation is moles = mass ÷ Mr — rearrange to mass = moles × Mr. From there, reacting-mass questions are a three-step chain: convert the known mass to moles, apply the balancing-number ratio, convert the answer back to mass.
Exam tips
- Worked example — a full reacting-mass chain. What mass of magnesium oxide forms from 12 g of magnesium? 2Mg + O₂ → 2MgO; Ar Mg = 24, Mr MgO = 40. Step 1: moles Mg = 12 ÷ 24 = 0.5 mol. Step 2: the ratio Mg : MgO is 2 : 2 = 1 : 1, so moles MgO = 0.5 mol. Step 3: mass MgO = 0.5 × 40 = 20 g. Write the ratio line explicitly — it carries the method mark.
- Worked example — the limiting reactant. 2H₂ + O₂ → 2H₂O. A vessel holds 4 g of H₂ (Mr 2 → 2 mol) and 64 g of O₂ (Mr 32 → 2 mol). Each mole of O₂ needs 2 mol of H₂, so 2 mol O₂ would demand 4 mol H₂ — but only 2 mol is present. Hydrogen is limiting: it runs out first, and the yield is fixed by it. Moles of water = 2 mol (2:2 ratio with H₂) → mass = 2 × 18 = 36 g, whatever the oxygen surplus.
Notes · 6 of 6
Uncertainty
Honest numbers: half the range tells you the wobble.
Repeated measurements rarely agree exactly. The uncertainty of a mean is estimated as half the range (the largest result minus the smallest, divided by 2). Quote results as mean ± uncertainty — for readings of 24.1, 24.5 and 24.3 cm³, the mean is 24.3 cm³, the range is 0.4 cm³, and the result is 24.3 ± 0.2 cm³.
Revise
Key terms
The words examiners expect you to use precisely.
- Conservation of mass
- The total mass of reactants equals the total mass of products — atoms are only rearranged.
- Balanced equation
- A symbol equation with equal numbers of every atom on both sides.
- State symbol
- (s), (l), (g) or (aq) showing the physical state of each substance.
- Relative formula mass (Mr)
- The sum of the relative atomic masses in a formula.
- Concentration
- Mass of solute per dm³ of solution, in g/dm³.
- Mole
- 6.02 × 10²³ particles of a substance — its Mr in grams.
Show 3 more key termsShow fewer key terms
- Avogadro constant
- 6.02 × 10²³ — the number of particles in one mole.
- Limiting reactant
- The reactant that is completely used up first, fixing the maximum product.
- Uncertainty
- Half the range of repeated measurements, showing the spread around the mean.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Claiming mass is not conserved when a reaction loses mass
The atoms left — they were not destroyed.
Instead: Gas escaping an open system explains the loss; seal it and mass stays constant.
Mistake: Balancing by changing subscripts
H₂O → H₂O₂ invents hydrogen peroxide.
Instead: Adjust only the big balancing numbers in front of formulas.
Show 10 more mistakesShow fewer mistakes
Mistake: Forgetting to convert cm³ to dm³
100 cm³ is not 100 dm³ — a ×1000 error.
Instead: Divide cm³ by 1000 before dividing by the volume.
Mistake: Reversing moles = mass ÷ Mr
Multiplying by Mr inflates everything.
Instead: Moles = mass ÷ Mr; mass = moles × Mr.
Mistake: Using the balancing ratio in reverse
A 2:1 equation does not let 1 mol of A make 2 mol of B.
Instead: moles wanted = moles known × (wanted coefficient ÷ known coefficient).
Mistake: Calling the reactant with the smallest mass “limiting”
Masses must be converted to moles first.
Instead: The limiting reactant is whichever provides fewer moles relative to the ratio.
Mistake: Forgetting diatomic elements
Writing O instead of O₂ for oxygen gas.
Instead: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂ travel as molecules.
Mistake: Quoting Mr without units awareness
Mr is a relative number with no units; grams attach only after multiplying by moles.
Instead: Mr is unitless; moles × Mr gives grams.
Mistake: Rounding Mr values too early
Early rounding compounds through the calculation.
Instead: Carry full values through; round only the final answer.
Mistake: Mixing up concentration in g/dm³ with mol/dm³
One is mass per volume, the other moles per volume.
Instead: g/dm³ comes from mass ÷ volume; (triple only) mol/dm³ from moles ÷ volume.
Mistake: Computing uncertainty as the full range
The range overstates the wobble of the mean.
Instead: Uncertainty = range ÷ 2.
Mistake: Saying a crucible experiment’s mass gain breaks the conservation law
Oxygen from the air is a reactant.
Instead: The magnesium has combined with atmospheric oxygen — mass is conserved including the air’s contribution.
Check yourself
Quick quiz
20 questions with instant explanations. About 5 minutes.
Question 1 of 3· free preview of 20
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Explain conservation of mass
Mass is conserved in a chemical reaction because
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Practise
Topical questions
12 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this chapter, in three tiers, each with a model answer, mark scheme and examiner tip. Higher-tier questions are tagged.
Practise topical questionsExam practice
Practice paper
A 30-mark paper in the style of a Combined Science paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 13 Practice Paper
30 marks · 40 min · 5 pages
6 multiple-choice questions and structured questions with answer lines. Conservation of mass, balancing equations and state symbols, relative formula mass, concentration in g/dm³, Higher-tier moles, reacting masses and limiting reactants, uncertainty.
Mark scheme
PROChapter 13 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
How does this chapter fit the real exams?
Chemistry Paper 1
1 h 15 · 70 marks (Trilogy)
Atomic structure, bonding, quantitative chemistry, chemical and energy changes (chapters 11–16)
Chemistry Paper 2
1 h 15 · 70 marks (Trilogy)
Rates, organic chemistry, analysis, atmosphere and resources (chapters 17–20)
Tiers
Foundation (grades 1-1 to 5-5) · Higher (4-3 to 9-9)
Two grades are awarded together; the 21 required practicals are examined on the written papers
Want a full paper? Try the Biology Paper 1-style (Foundation), Biology Paper 1-style (Higher), Biology Paper 2-style (Foundation), Biology Paper 2-style (Higher), Chemistry Paper 1-style (Foundation), Chemistry Paper 1-style (Higher), Chemistry Paper 2-style (Foundation), Chemistry Paper 2-style (Higher), Physics Paper 1-style (Foundation), Physics Paper 1-style (Higher), Physics Paper 2-style (Foundation), Physics Paper 2-style (Higher), Required practicals practice sets.
Questions
Frequently asked questions
How is Combined Science quantitative chemistry different from triple Chemistry?
Combined stops at concentration in g/dm³ plus Higher-tier moles, reacting masses and limiting reactants. Triple adds titration calculations, yields, atom economy and mol/dm³ work in depth.
Which maths skills appear in this chapter?
Arithmetic with Mr, unit conversion (cm³ → dm³), moles = mass ÷ Mr, three-step reacting-mass chains, ratios and half-range uncertainty.
What is the three-step reacting mass method?
1) Moles = mass ÷ Mr for what you know. 2) Apply the balancing-number ratio. 3) Mass = moles × Mr for what you want. Show every step — the ratio line carries a mark.
How do I spot the limiting reactant?
Convert both reactants to moles, then check which one runs out first against the balancing numbers. Smallest MASS is not the test — smallest MOLES relative to the ratio is.
Why does mass “disappear” in some reactions?
It never does. Gases escaping an open system take their mass with them; oxygen from the air can add mass. Seal the system and the balance never changes.
Up next
Chapter 14: Reactivity and redox