Stoichiometry: IGCSE Chemistry 0620

Stoichiometry is the most expensive topic in 0620 to be weak at. It is not one question. It is the calculation engine inside titrations, electrolysis, energetics and organic yields, worth 8-12 marks across Papers 2 and 4 in a typical series. Examiner reports say the same thing every year: students who set out moles step by step score full marks; students who jump straight to an answer with no working lose everything when one number slips. Method marks exist here. Claim them.

Chemical formulae and equations (Core)

Core candidates must write formulae from ion charges and balance symbol equations. The ion charges to memorise: Group I +1, Group II +2, Group III +3, Group V −3, Group VI −2, Group VII −1, plus the compound ions ammonium NH4+, hydroxide OH−, nitrate NO3−, carbonate CO3 2−, sulfate SO4 2−. Swap-and-drop the charges: aluminium oxide = Al3+ and O2− → Al2O3.

Balancing changes only the large numbers in front of formulae, never the small subscripts. CH4 + 2O2 → CO2 + 2H2O balances; rewriting O2 as O4 destroys the chemistry. State symbols, (s), (l), (g), (aq), are asked by name, and ionic equations get their own treatment in our ionic equations and balancing guide.

Relative atomic and molecular mass (Core)

Relative atomic mass, Ar, is the average mass of the atoms of an element compared to 1/12 of the mass of a carbon-12 atom. The “average” matters because of isotopes: chlorine’s 35.5 comes from the 3:1 mix of chlorine-35 and chlorine-37, which connects back to atomic structure.

Relative molecular mass (or relative formula mass), Mr, is the sum of the Ar values in the formula. CuSO4·5H2O: 64 + 32 + 64 + 90 = 250. Two rules prevent the standard errors: multiply through brackets, so Ca(OH)2 is 40 + 2(16+1) = 74, and include the water of crystallisation when the dot is there.

The mole and the Avogadro constant (S)

Everything from here on is Supplement. Core candidates stop at Mr.

A mole is the amount of substance containing 6.02 × 10²³ particles; that number per mole is the Avogadro constant. One mole of carbon atoms weighs 12 g, one mole of water molecules 18 g. The molar mass in g/mol is numerically equal to Ar or Mr.

The three conversions cover every question on the topic:

Quantity	Formula
Mass	moles = mass (g) ÷ Mr
Gas volume at r.t.p.	moles = volume (dm³) ÷ 24
Solution	moles = concentration (mol/dm³) × volume (dm³)

Moles and reacting-mass calculations (S)

Every reacting-quantity question is the same four steps. Step 1: write (or copy down) the balanced equation. Step 2: convert the quantity you are given into moles. Step 3: use the ratio from the equation to find moles of the substance asked about. Step 4: convert those moles into the unit requested. This is the same four-step method used in every guide on this site.

Example: what mass of magnesium oxide forms when 6.0 g of magnesium burns? 2Mg + O2 → 2MgO. Moles of Mg = 6.0/24 = 0.25. Ratio Mg:MgO is 2:2, so 0.25 mol MgO. Mass = 0.25 × 40 = 10 g.

Limiting reactant questions add one check before step 2: divide each reactant’s moles by its equation coefficient; the smaller result is the limiting reactant, and all later steps use it. Examiners flag this as the discriminator between A and A* candidates: the question says “excess” or gives two masses, and that word is the instruction to check.

Concentration of solutions (S)

Concentration is measured in g/dm³ or mol/dm³, and the two convert through Mr: mol/dm³ × Mr = g/dm³. The working unit is dm³, so divide cm³ by 1000 first: a 25.0 cm³ pipette delivers 0.0250 dm³.

Titration calculations chain two solutions together: moles of the known solution, equation ratio, then concentration of the unknown. The practical side (pipette, burette, indicator choice) sits in acids, bases and salts; the maths is pure stoichiometry and follows the same four-step method as reacting masses.

Percentage yield and percentage purity (S)

Percentage yield = (actual yield ÷ theoretical yield) × 100. The theoretical yield comes from a mole calculation; the actual yield is the measured mass. Yields below 100% are normal: the reaction may be incomplete, product is lost on transfer, or side reactions occur.

Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100. The mole calculation tells you the pure mass; the question gives the sample mass. Read which mass goes on top. Inverting the fraction is a whole-mark error that the calculator will not catch because the answer still looks plausible.

Empirical and molecular formulae (S)

The empirical formula is the simplest whole-number ratio of atoms. The routine: write the mass or percentage of each element, divide each by its Ar, then divide all results by the smallest. A compound with 80% carbon and 20% hydrogen: 80/12 = 6.67 and 20/1 = 20; divide by 6.67 to get 1 : 3, CH3.

The molecular formula is the actual number of atoms, found by comparing the empirical formula mass to the given Mr. CH3 has mass 15; if Mr = 30, the molecular formula is C2H6. Set the working out as a table. Cambridge mark schemes award the divide-by-Ar line and the ratio line separately.

Worked exam question

A student heats 12.4 g of copper(II) carbonate, CuCO3, until it fully decomposes: CuCO3 → CuO + CO2. (a) Calculate the mass of copper(II) oxide formed. (b) Calculate the volume of carbon dioxide produced at r.t.p. [Ar: Cu 64, C 12, O 16] [4]

Model answer: (a) Mr of CuCO3 = 64 + 12 + 48 = 124. Moles = 12.4 ÷ 124 = 0.10 mol (1). Ratio 1:1, so 0.10 mol CuO; mass = 0.10 × 80 = 8.0 g (1). (b) 0.10 mol CO2 (1); volume = 0.10 × 24 = 2.4 dm³ (1).

Mark-by-mark: mark 1 is the correct moles of CuCO3. The Mr can even be wrong and later marks survive on error carried forward, but only if working is shown. Mark 2 needs the Mr of CuO (80) applied to the right mole number. Mark 3 is recognising the 1:1 ratio gives 0.10 mol of gas. Mark 4 is moles × 24 with the unit dm³. A bare “8 g” with no working risks all four if any digit is off.

The mistakes that cost marks

1. No working shown. Cambridge awards method marks and error-carried-forward. One line per step protects 3 marks even when the final answer is wrong.
2. Skipping the mole-ratio step. Students convert mass to moles, then convert straight back without applying the 2:1 or 1:2 ratio from the equation. Underline the coefficients before calculating.
3. Unit slips: cm³ used as dm³ in concentration and gas-volume work. Divide by 1000 first, every time, and write the converted value down.
4. Brackets ignored in Mr: Mg(NO3)2 is 24 + 2(14 + 48) = 148, not 86. Water of crystallisation dropped from hydrated salts causes the same damage.
5. Rounding mid-calculation. Carry at least three significant figures through and round only the final answer: 0.0417 rounded early to 0.04 shifts a titration answer outside the allowed range.

How to phrase it for full marks

Student wording	Mark-scheme wording
”A mole is a big number of particles"	"The amount of substance containing 6.02 × 10²³ particles"
"I times-ed it by the equation"	"Moles of MgO = moles of Mg × 2/2 = 0.25 mol (ratio from equation)"
"Some product got lost"	"Yield is below 100% because the reaction is incomplete / product is lost during transfer"
"25 cm³ of acid” (used as 25 in formula)	“25.0 cm³ = 0.0250 dm³"
"The answer is 8"	"Mass = 0.10 mol × 80 g/mol = 8.0 g”

Calculations are not prose, but labels are still phrasing: write what each number is (“moles of CuCO3 = …”), keep units on every line, and the examiner can award each marking point even when arithmetic stumbles.

The Malaysia note

Malaysian students sitting 0620 in the May/June or Oct/Nov series consistently report stoichiometry as the topic that decides their grade, and school feedback matches: classes move through the mole in two or three weeks of Year 10, then assume it for two years. SPM-stream friends and siblings learn the same concept as “konsep mol”, so the ideas transfer, but 0620’s insistence on shown working and its r.t.p. value of 24 dm³ (SPM references use 22.4 at s.t.p.) catch transfer students out. Mole calculations are the single most-requested fix in our free 1-hour trial lesson, and most students lock in the four-step method within 2-3 sessions.

Every sub-topic in Stoichiometry

• Chemical Formulae
• Relative Atomic and Molecular Mass (Ar and Mr)
• The Mole and the Avogadro Constant
• Moles and Reacting-Mass Calculations
• Concentration of Solutions
• Percentage Yield and Percentage Purity (Supplement)
• Empirical and Molecular Formulae (Supplement)