Atoms, Molecules and Stoichiometry

Physical Chemistry Weightage: AS · Papers 1, 2 and 3 (assumed in Papers 4 and 5) Topic 2

“Welcome! This is the topic where most Cambridge candidates quietly lose marks, and almost never because the chemistry is hard. The marks go on a wrong unit, a missing ratio, a number rounded too early or a line of working that was never written down. Stoichiometry also runs through every other paper: Paper 3 titrations, Paper 4 energetics and equilibria, Paper 5 planning and evaluation. Learn the moles routine below until it is automatic, write your working the way the mark scheme rewards it, and this topic becomes the most reliable source of marks in the whole syllabus.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in calculations

Show working. If a question says “show your working” or is worth two or more marks, a bare correct answer may still score, but a wrong answer with no working scores zero.
Error carried forward (ecf). If your first answer is wrong but you use it correctly in later steps, you can still earn the later marks. This only works when the steps are written down.
Significant figures. Give the final answer to the same number of s.f. as the least precise data, and to 3 s.f. if in doubt. One s.f. or a long calculator string can lose the answer mark. Never round in the middle of a calculation.
Units. A missing or wrong unit on the final answer usually loses the mark: mol, g, g mol⁻¹, dm³, mol dm⁻³.
Keywords in definitions. “Weighted average”, “one twelfth of carbon-12” and “specified particles” are the marking words.

How working earns marks. Examiners can only award a method mark if they can see it, so write down the amount in mol, the ratio, then the answer with its unit.

1. Relative Masses and Mass Spectra

Definitions to learn word for word:

Term	Exact wording
Unified atomic mass unit (u)	One twelfth of the mass of a carbon-12 atom
Relative isotopic mass	The mass of an atom of an isotope compared with one twelfth of the mass of a carbon-12 atom
Relative atomic mass, A_r	The weighted average mass of the atoms of an element compared with one twelfth of the mass of a carbon-12 atom
Relative molecular mass, M_r	The weighted average mass of a molecule compared with one twelfth of the mass of a carbon-12 atom
Relative formula mass	The same idea for ionic or giant compounds (use the formula unit)

The relative masses have no units. A_r is not a whole number because it is a weighted average of the isotopes.
To find M_r, add the A_r values, multiplying by the number of each atom. For a hydrated salt, include the water: M_r of MgSO₄·7H₂O = 24.3 + 32.1 + 64.0 + 7(18.0) = 246.4.

Mass spectrum of boron. A_r = (10 × 18.7 + 11 × 81.3) ÷ 100 = 10.8. Each peak is one isotope: its position gives the isotopic mass and its height gives the abundance.

Worked example: A_r from a mass spectrum. Boron has peaks at m/e 10 (18.7%) and 11 (81.3%).

Multiply each isotopic mass by its abundance: 10 × 18.7 = 187.0 and 11 × 81.3 = 894.3.
Add and divide by the total abundance (100): (187.0 + 894.3) ÷ 100 = 10.813.
Answer to 3 s.f.: A_r = 10.8 (no unit).

⚠️ Examiner Trap: The mean is weighted. Averaging the two masses as (10 + 11) ÷ 2 = 10.5 scores nothing. Also divide by the total of the abundances, not by the number of isotopes. If the abundances are given as heights (not percentages) divide by their sum.

Mass spectra of organic molecules (the M⁺ peak, the [M + 1] peak for carbon and the [M + 2] peak for Cl and Br) are covered in Topic 22.

2. The Mole and the Moles Routine

Definition: The mole is the amount of substance that contains the Avogadro number (6.022 × 10²³) of specified particles. The particles can be atoms, molecules, ions or electrons, so always say which particle.

Every stoichiometry question routes through moles. Convert what you are given into amount (mol), use the equation ratio, then convert to what is asked. Check the unit of volume before you use c × V.

The routine, for every calculation:

Write the balanced equation.
Turn the quantity you are given into moles.
Use the equation ratio to find moles of the other substance.
Turn moles into what the question asks (mass, volume, concentration).
Check the unit and s.f. on the final line.

⚠️ Examiner Trap: Particle counts. One mole of CaCl₂ contains 6.022 × 10²³ formula units, but 1.81 × 10²⁴ ions (three per formula unit). Always state the particle asked for.

3. Formulae and Ions

Empirical formula: the simplest whole-number ratio of atoms of each element in the compound. Molecular formula: the actual number of atoms of each element in one molecule.
Method for empirical formula: divide each percentage (or mass) by A_r, divide all by the smallest answer, and turn the result into whole numbers (1.5 → multiply all by 2; 1.33 → multiply by 3).
Molecular formula: find M_r of the empirical formula, then multiply by (M_r of compound ÷ M_r of empirical formula).

Worked example. A hydrocarbon contains 85.7% C and 14.3% H by mass and has M_r = 84.

C: 85.7 ÷ 12.0 = 7.14; H: 14.3 ÷ 1.0 = 14.3.
Divide by 7.14: C = 1, H = 2.0, so the empirical formula is CH₂.
M_r of CH₂ = 14.0; 84 ÷ 14.0 = 6, so the molecular formula is C₆H₁₂.

Ions you must recall: nitrate NO₃⁻, carbonate CO₃²⁻, sulfate SO₄²⁻, hydroxide OH⁻, ammonium NH₄⁺, zinc Zn²⁺, silver Ag⁺, hydrogencarbonate HCO₃⁻, phosphate PO₄³⁻. Predict other ionic charges from the group number: Group 1 is 1+, Group 2 is 2+, Group 13 is 3+, Group 16 is 2−, Group 17 is 1−.

To write an ionic formula, balance the charges: aluminium sulfate is Al₂(SO₄)₃ (two Al³⁺ = 6+, three SO₄²⁻ = 6−). Use brackets around a polyatomic ion when you need more than one.

4. Equations and State Symbols

Balance atoms and charges. Do not change a formula to balance; only change the coefficients.
State symbols: (s), (l), (g), (aq). Include them whenever the question asks. For ionisation-energy or enthalpy equations they carry a mark.
Ionic equations show only the particles that change. Cancel the spectator ions, and check that the total charge is the same on both sides.

Example: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) becomes Ag⁺(aq) + Cl⁻(aq) → AgCl(s). The Na⁺ and NO₃⁻ are spectators.

5. Reacting Masses, Limiting Reagent, Yield

Worked example. 15.96 g of iron(III) oxide is reduced by an excess of carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. The mass of iron collected is 9.05 g. Calculate the percentage yield.

M_r of Fe₂O₃ = 2(55.8) + 3(16.0) = 159.6. Moles = 15.96 ÷ 159.6 = 0.100 mol.
Ratio: 1 mol Fe₂O₃ gives 2 mol Fe, so 0.200 mol Fe.
Theoretical mass = 0.200 × 55.8 = 11.16 g.
Percentage yield = (actual ÷ theoretical) × 100 = 9.05 ÷ 11.16 × 100 = 81.1%.

Limiting reagent: turn each reactant into moles, divide by its coefficient, and the smaller value is limiting. Percentage yield is always calculated from the limiting reagent.
Percentage composition by mass = (mass of element in the formula ÷ M_r) × 100.

⚠️ Examiner Trap: Excess reactant. To show which reagent is in excess, do not compare the two masses directly. Convert both to moles, apply the equation ratio, and state clearly which one is left over.

6. Solutions and Titrations

Concentration c (mol dm⁻³) = n ÷ V (dm³). To convert g dm⁻³ to mol dm⁻³, divide by M_r.

The four-step route for any titration calculation. The equation ratio (step 2) is where most marks are lost: check whether the acid and alkali react 1 : 1 or 1 : 2 before you continue.

Worked example. 1.20 g of impure NaOH is made up to 250 cm³. A 25.0 cm³ portion needs 22.40 cm³ of 0.100 mol dm⁻³ HCl.

n(HCl) = 0.100 × 22.40 ÷ 1000 = 2.24 × 10⁻³ mol.
NaOH + HCl → NaCl + H₂O is 1 : 1, so n(NaOH) in 25.0 cm³ = 2.24 × 10⁻³ mol.
In 250 cm³ (ten times more): 2.24 × 10⁻² mol.
Mass = 2.24 × 10⁻² × 40.0 = 0.896 g. Purity = 0.896 ÷ 1.20 × 100 = 74.7%.

Paper 3 practical rules: read the burette to 0.05 cm³, do a rough titration first, and use the mean of the accurate titres that are within 0.10 cm³ of each other.
Back titration: add a known excess of reagent, titrate the excess, then subtract to find what reacted (used in Paper 3, for example with a carbonate or hydroxide and excess acid).

⚠️ Examiner Trap: The ratio and the dilution. H₂SO₄ reacts with NaOH 1 : 2, not 1 : 1. If the flask contained only a portion of the original solution, multiply back up by the dilution factor before you go on.

7. Gas Volumes

At r.t.p., 1 mol of any gas occupies 24.0 dm³ (24 000 cm³); at s.t.p., 22.4 dm³. Both values, and the conditions, are printed in the Data Booklet.
Avogadro: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So gas volumes react in the same ratio as the equation coefficients.
Water formed at room temperature is a liquid, so it does not count as gas volume.

Worked example. 0.010 mol of calcium nitrate is heated strongly. 2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂. What volume of gas forms at r.t.p.?

2 mol Ca(NO₃)₂ gives 4 + 1 = 5 mol gas, so 0.010 mol gives 0.025 mol gas.
Volume = 0.025 × 24.0 = 0.600 dm³ = 600 cm³.

8. Quick Sheet and Checklist

Definitions: u, A_r, isotopic mass, M_r, mole.
A_r from abundances: weighted mean, divide by 100 (or by the total).
n = m ÷ M_r; n = V ÷ 24.0 (r.t.p.) or 22.4 (s.t.p.); n = cV ÷ 1000 for cm³.
Empirical formula method; molecular formula from M_r.
Nine ions to recall; ionic equations without spectators; state symbols.
Limiting reagent and percentage yield; titration route with the ratio and dilution.
Working shown, unit on the answer, 3 s.f., no rounding in the middle.

Before you leave this topic, can you:

Define the mole and A_r exactly as above?
Work out the volume of 0.0250 mol of CO₂ at r.t.p. in cm³?
Write the ionic equation for silver nitrate with sodium chloride?
Explain why a titre is only valid when it is within 0.10 cm³ of another?

If yes to all four, attempt the ten exam-style questions with the mark schemes covered, then open the flashcards.