SPEC 4.3.1.1 • QUANTITATIVE CHEMISTRY • PAPER 1
Conservation of Mass & Balanced Equations
The mathematical backbone of all quantitative chemistry. Atoms are rearranged, never made or destroyed — and every mole calculation in Module 3 begins with a balanced equation.
The Law of Conservation of Mass
The single principle that underlies every calculation in this module. Learn the mark-scheme wording word-for-word.
Law of Conservation of Mass
No atoms are lost or made in a chemical reaction. The total mass of reactants equals the total mass of products.
Why it works
In a chemical reaction, the bonds between atoms are broken and new bonds form — but the atoms themselves are simply rearranged. The same atoms that started in the reactants end up in the products. So the total mass cannot change.
total mass of reactants = total mass of products
Visualising the Principle
Methane combustion: CH₄ + 2O₂ → CO₂ + 2H₂O. Count the atoms before and after — they match.
Same atoms, different arrangement. Same atoms means the same total mass — that's conservation of mass.
PEE Chain: Conservation Principle
The structured answer pattern for any "explain why the total mass does not change" or "why must the equation balance" question. Use this chain when asked to justify mass equality.
PPoint: No atoms are lost or made during a chemical reaction — atoms are only rearranged.
EEvidence: A balanced symbol equation has the same number of atoms of each element on both sides.
EExplain: Therefore the total mass of reactants equals the total mass of products — mass is conserved.
Exam Tip
Mark schemes accept "atoms are rearranged" or "no atoms are lost or made" — both score. They do not accept "mass is conserved" alone without saying why. Always link the principle to the atoms.
Reading a Balanced Equation
The spec asks you to "understand the use of the multipliers in equations in normal script before a formula and in subscript within a formula." Translation: you need to know what every number in the equation means.
4Fe + 3O₂ → 2Fe₂O₃
NumberTypeWhat it tells you
4 in 4FeCoefficient (multiplier in normal script)Four iron atoms involved in the reaction
3 in 3O₂CoefficientThree molecules of O₂ — that's 6 oxygen atoms (3 × 2)
2 (small) in O₂Subscript (multiplier within formula)Two oxygen atoms inside one O₂ molecule
2 in 2Fe₂O₃CoefficientTwo formula units of Fe₂O₃ — that's 4 Fe (2 × 2) and 6 O (2 × 3)
Atom Tally
Left: 4 Fe, 6 O. Right: 4 Fe, 6 O. ✓ Both sides match — the equation is balanced.
Coefficients vs Subscripts
The single most important rule for balancing — and the one most students break.
COEFFICIENTS — change these
The big numbers in front of a formula. They tell you how many molecules/units there are.
2H₂O
Two water molecules. Coefficient = 2.
SUBSCRIPTS — never change
The small numbers below the line. They are part of the formula — change them and the substance changes.
2H₂O
Two H atoms inside a water molecule. Subscript = 2.
⚠️ EXAMINER ALERT: Changing H₂O to H₂O₂ to "balance" oxygens turns water into hydrogen peroxide — a different substance. Only adjust coefficients in front of the formula.
Balancing Method — General
A reliable order that prevents you from getting stuck partway through.
1
Write correct formulae. Get the chemistry right first. Wrong formula = wrong everything.
2
Balance metals first. Metals usually appear in only one reactant and one product.
3
Then balance non-metals (except H and O).
4
Then balance H.
5
Balance O LAST. O usually appears in many compounds, so leave it until everything else is settled.
6
Check every element. Count atoms left and right — they must match.
Worked Example: Iron Rusting
Balance: Fe + O₂ → Fe₂O₃
1
Balance Fe. The right side has 2 Fe (in Fe₂O₃). Put a 2 in front of Fe on the left.
2Fe + O₂ → Fe₂O₃
2
Balance O. Right has 3 O; left has 2 O per O₂. Lowest common multiple is 6 → use 3 O₂ (6 O) on the left and 2 Fe₂O₃ (6 O) on the right. The doubled Fe₂O₃ now needs 4Fe on the left.
4Fe + 3O₂ → 2Fe₂O₃
4Fe + 3O₂ → 2Fe₂O₃
Check: Fe: 4 = 4 ✓ O: 6 = 6 ✓
Combustion: The C → H → O Order
For combustion of a hydrocarbon (CxHy + O₂ → CO₂ + H₂O), follow this order strictly. It is the single most repeated balancing question on Paper 2.
CCarbon first. Each C in the hydrocarbon → one CO₂. So the CO₂ coefficient = number of C atoms.
HHydrogen next. Each pair of H atoms → one H₂O. So the H₂O coefficient = (number of H atoms) ÷ 2.
OOxygen LAST. Count total O atoms on the right (in CO₂ AND H₂O), then divide by 2 for the O₂ coefficient.
⚠️ EXAMINER ALERT (P2 Jun22 Q04.3): The error "19 H₂O" is flagged as extremely common. It comes from balancing oxygen first and getting tangled. Always C → H → O.
Worked Example: Combustion of Propane
Balance: C₃H₈ + O₂ → CO₂ + H₂O
C
Carbon first. 3 carbons in propane → 3 CO₂.
C₃H₈ + O₂ → 3CO₂ + H₂O
H
Hydrogen next. 8 hydrogens in propane ÷ 2 = 4 H₂O.
C₃H₈ + O₂ → 3CO₂ + 4H₂O
O
Oxygen last. Right side: (3 × 2) + (4 × 1) = 10 O atoms. So O₂ coefficient = 10 ÷ 2 = 5.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Check: C: 3 = 3 ✓ H: 8 = 8 ✓ O: 10 = 10 ✓
Presentation Matters
An equation can be chemically correct but score zero if the examiner can't read it.
⚠️ EXAMINER ALERT (P2 Jun22 Q04.3): Badly written digits lose marks. Examiners reported "14" being read as "19", costing the mark even though the chemistry was right.
3 Things To Get Right
Coefficients clearly larger than subscripts. A coefficient is normal-sized; a subscript should be visibly smaller and lower.
Capital letters look capital. The examiner report flagged a small "O" being mistaken for "0" (zero) — costing the mark. Element symbols always start with a clear capital.
Numbers unambiguous. If your 4 looks like a 9, that's a mark lost — re-write it.
Exam Tip
This is a free mark you can give away or keep simply by writing slowly and clearly. In a Paper 2 balancing question, slow down for the final answer.
Worked Example: Metal + Acid
Combustion has its own C → H → O order. For most other reactions, use the general method: metals → other non-metals → H → O.
Balance: Mg + HCl → MgCl₂ + H₂
1
Mg. 1 Mg on each side already. ✓ Skip for now.
2
Cl. Right has 2 Cl (in MgCl₂); left has 1 per HCl. Put a 2 in front of HCl.
Mg + 2HCl → MgCl₂ + H₂
3
H. Left now has 2 H (in 2HCl); right has 2 H (in H₂). ✓ Already balanced.
Mg + 2HCl → MgCl₂ + H₂
Check: Mg: 1 = 1 ✓ H: 2 = 2 ✓ Cl: 2 = 2 ✓
Why this works
Acid + metal reactions always produce a salt + hydrogen gas. The salt formula is fixed by the metal's charge (Mg²⁺ needs 2 Cl⁻ → MgCl₂) — never adjust it. You balance by adjusting the acid coefficient.
Using Conservation to Find a Missing Mass
Because total mass on both sides is equal, you can calculate the mass of any one substance if you know the masses of all the others. This is a direct application of 4.3.1.1.
Worked Example
4.8 g of magnesium reacts completely with chlorine gas to form 19.0 g of magnesium chloride.
Mg + Cl₂ → MgCl₂
Calculate the mass of chlorine that reacted.
PPoint: Conservation of mass — total mass of reactants = total mass of products.
EEvidence: mass(Mg) + mass(Cl₂) = mass(MgCl₂) → 4.8 + mass(Cl₂) = 19.0
EExplain (calculate): mass(Cl₂) = 19.0 − 4.8 = 14.2 g
Exam Tip
You don't need M_r values for this — just subtraction. The conservation principle alone is enough. Always quote the principle in your working: it earns the method mark even if the arithmetic slips.
Practice Question 1
Balance the equation for the complete combustion of ethane:
__ C₂H₆ + __ O₂ → __ CO₂ + __ H₂O2 marks
P
Plan: This is a combustion equation, so use the C → H → O order.
E
Execute: C: 2 carbons → 2 CO₂. H: 6 hydrogens ÷ 2 → 3 H₂O. O: right side has (2×2) + (3×1) = 7 atoms, halving gives 3.5 O₂. Multiply through by 2 to clear the fraction.
E
Evaluate: Final equation: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Check: C: 4=4 ✓ H: 12=12 ✓ O: 14=14 ✓
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Practice Question 2
3.00 g of magnesium reacts completely with oxygen to form magnesium oxide. The mass of magnesium oxide produced is 5.00 g.
2Mg + O₂ → 2MgO
Calculate the mass of oxygen that reacted with the magnesium. 2 marks
P
Point: By conservation of mass, total mass of reactants = total mass of products.
E
Evidence: mass(Mg) + mass(O₂) = mass(MgO) → 3.00 + mass(O₂) = 5.00 1 mark — method
E
Evaluate: mass(O₂) = 5.00 − 3.00 = 2.00 g1 mark — answer
Common Errors — Mark-Scheme Verbatim
Sit these in your memory exactly as written. Examiners are looking for the right-hand column.
Students often writeCorrect versionWhy it matters
"Atoms made or lost in the reaction""Atoms are rearranged ; no atoms are lost or made"Conservation principle wording (spec verbatim)
"Mass is conserved" — alone, no mechanism"No atoms are lost or made, so mass is conserved"Mark scheme requires the reasoning, not just the conclusion
"19 H₂O" in a combustion equationRe-balance in C → H → O order — answer will be much smallerDouble-counting oxygen is the trap (P2 Jun22 Q04.3)
Changing subscripts to "balance"Change only coefficients (numbers in front of formulae)Subscripts are part of the formula — change them and the substance changes
Final equation contains a fraction (e.g. 3.5 O₂)Multiply every coefficient by the denominator (×2 here) to clear itBalanced equations use whole-number coefficients only
Forgetting to check one elementTally every element on both sides at the endOne element off = whole equation wrong
Summary
Key Definitions
Conservation of mass: no atoms are lost or made during a chemical reaction; total mass of reactants = total mass of products
Balanced symbol equation: the number of atoms of each element is the same on both sides
Multipliers
Coefficients (normal script, in front) — adjust to balance. Subscripts (small, within formula) — never change. Both are multipliers, but only coefficients are tools for balancing.
Balancing — General
Order: metals → other non-metals → H → O. Always check every element at the end. Clear any fractions by multiplying through.
Balancing — Combustion
Hydrocarbon + O₂ → CO₂ + H₂O. Order: C → H → O. Avoids the "19 H₂O" trap (P2 Jun22 Q04.3).
Mass Calculations
Conservation gives you any one missing mass: total mass in = total mass out. Just subtract — no M_r needed.
Exam Tip
Write coefficients clearly and noticeably larger than subscripts. A correctly balanced equation in unreadable handwriting still loses the mark — examiners cannot credit what they cannot read.
Ready for 3.02
Next up: Relative Formula Mass, Mass Changes & Chemical Measurements — calculating M_r, percentage by mass, explaining apparent mass changes when a reactant or product is a gas, and the random/systematic error distinction.
Next: The Law of Conservation of Mass