AS Chemistry: Atoms, Molecules and Stoichiometry — Revision Notes

Condensed for the final weeks. For the full explanation, use the Atoms, Molecules and Stoichiometry study guide. Related: Atoms, Molecules and Stoichiometry practice questions for further worked calculations.

Every equation

moles from mass         n = m / M
moles in solution       n = c V           (V in dm3)
moles of gas at rtp     n = V / 24        (V in dm3)
ideal gas equation      pV = nRT          R = 8.31 J K-1 mol-1
number of particles     N = n x 6.02 x 10^23

percentage yield        (actual / theoretical) x 100
atom economy            (sum of Mr x coefficient for desired product) /
                        (sum of Mr x coefficient for ALL reactants) x 100

Atom economy is background only, not a 9701 term.

Unit discipline: for pV = nRT, pressure in Pa, volume in m³, temperature in K. Convert: kPa × 1000, cm³ ÷ 10⁶, °C + 273.

Ionic formulas and equations

Build an ionic formula from charge, using the lowest common multiple to balance:

Iron(III) oxide: Fe3+ and O2-
LCM of 3 and 2 = 6  ->  2 Fe3+ and 3 O2-  ->  Fe2O3

Seven polyatomic ions you’re expected to recall without being given them: NO₃⁻, CO₃²⁻, SO₄²⁻, OH⁻, NH₄⁺, HCO₃⁻, PO₄³⁻. Also worth memorising: Zn²⁺ and Ag⁺, which are monatomic metal ions with charges that aren’t obvious from the Periodic Table group number. All nine come up constantly in ionic-formula and ionic-equation questions, so it is worth being able to write each one’s charge from memory rather than working it out each time.

An ionic equation shows only the species that actually change, omitting spectator ions that appear unchanged on both sides:

Full:   Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq)
Ionic:  Pb2+(aq) + 2I-(aq) -> PbI2(s)

K⁺ and NO₃⁻ are omitted because they appear on both sides in identical form; PbI₂ stays fully written out because it is a solid, not a free ion in solution. The same logic applies whenever you’re asked to write an ionic equation from a full symbol equation: identify which ions genuinely change (form a precipitate, a gas, or a molecule), keep those, and drop anything left unchanged on both sides.

The universal four-step method

1. Balance the equation.
2. Convert the known substance to moles.
3. Apply the mole ratio.
4. Convert back to the quantity asked for.

Step 1 is skipped surprisingly often, and it invalidates step 3.

Limiting reagent

1  Convert BOTH reactants to moles
2  Divide each by its coefficient in the balanced equation
3  The SMALLER value is limiting
4  Base ALL product calculations on the limiting reagent

Dividing by the coefficient is the step candidates omit — comparing raw moles gives the wrong answer whenever the ratio is not 1:1.

Empirical and molecular formula

1  Mass or % of each element
2  Divide by Ar          -> moles
3  Divide by the smallest -> ratio
4  Multiply to whole numbers if needed

molecular formula = empirical formula x n,
where n = Mr(molecular) / Mr(empirical)

Percentage yield vs atom economy

(“Atom economy” is not itself a 9701 syllabus term — only percentage yield is named in the syllabus learning outcomes — but the underlying idea is useful background for understanding why some reactions waste more reactant mass than others.)

• Yield measures how much of the theoretical product was actually obtained — losses come from incomplete reaction, side reactions and transfer losses.
• Atom economy measures how much of the reactant mass ends up in the desired product — calculated as (coefficient × M_r of desired product) ÷ Σ(coefficient × M_r of every reactant) × 100, using the coefficients of the balanced equation (by conservation of mass, Σ(coefficient × M_r) of all reactants equals that of all products, so either total can be used) — a reaction can have 100% yield and poor atom economy if it produces a lot of by-product.

C6H12O6 -> 2C2H5OH + 2CO2        Mr: glucose 180, ethanol 46.0, CO2 44.0
desired product:   2 x 46.0 = 92.0
all products:      2 x 46.0 + 2 x 44.0 = 180.0   (= 1 x 180 for the reactant)
atom economy = 92.0 / 180.0 x 100 = 51.1%
(leaving out the coefficient 2 would give 46.0 / 180 x 100 = 25.6%, which is wrong)

Addition reactions have 100% atom economy; substitution and elimination do not.

Exam traps

• Failing to balance before using a mole ratio.
• Comparing raw moles instead of moles ÷ coefficient for the limiting reagent.
• Mixing units in pV = nRT — Pa, m³ and K, always.
• Using 24 dm³ mol⁻¹ for a solid or liquid; it applies to gases at rtp only.
• Percentage yield above 100% means impure or wet product, or an arithmetic error.
• Rounding partway through instead of at the end.
• Splitting a solid or gas product into separate ions in an ionic equation — only species that are genuinely free ions in solution are written that way; precipitates, gases, water and other covalent molecules stay as their full formulae.
• Forgetting the lowest-common-multiple step when balancing an ionic formula, especially for charges like 3+ and 2− that don’t cancel directly.

Self-test

1. Convert 25 cm³ of 0.200 mol dm⁻³ solution to moles.
2. 6.0 g of Mg reacts with 4.0 g of O₂ (2Mg + O₂ → 2MgO). Which is limiting?
3. State the ideal gas equation and the units required.
4. A compound is 52.2% C, 13.0% H, 34.8% O. Find the empirical formula.
5. Distinguish percentage yield from atom economy (atom economy is background, not a 9701 syllabus term).
6. Find the formula of aluminium oxide from Al³⁺ and O²⁻.
7. Write the full and ionic equations for silver nitrate solution reacting with sodium chloride solution to form a silver chloride precipitate.

Answers: 1. V = 0.025 dm³; n = 0.200 × 0.025 = 5.00 × 10⁻³ mol. 2. Mg: 6.0/24.3 = 0.247 mol ÷ 2 = 0.123. O₂: 4.0/32 = 0.125 mol ÷ 1 = 0.125. Mg gives the smaller value, so Mg is limiting (O₂ is in excess). 3. pV = nRT, with p in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹. 4. 52.2/12 = 4.35; 13.0/1 = 13.0; 34.8/16 = 2.175. Divide by 2.175 → 2 : 6 : 1 → C₂H₆O. 5. Yield compares actual product obtained with the theoretical maximum; atom economy compares the mass of the desired product with the total mass of all reactants, each M_r multiplied by its coefficient in the balanced equation. 6. LCM of 3 and 2 is 6, so two Al³⁺ and three O²⁻ → Al₂O₃. 7. Full: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq); ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s), with Na⁺ and NO₃⁻ omitted as spectator ions.