Atoms, Molecules & Stoichiometry

— Chemistry · AS Level (7405) · Crash Course · (Ruhab Iqbal) —

★ Exam Tip: The definitions here are boring & mostly “ratta” — learn the key ones, but they are rarely asked. What scores marks is the calculations: moles, titrations, gas volumes, empirical formulae and mass spectra. Your periodic table (“Data Booklet”) is given in the exam — it is your best friend for every A_r/M_r.

1. Key Definitions

Remember: the word “relative” always means relative to carbon-12. Every “relative …” definition must include: “on a scale where 1 atom of carbon-12 has a mass of 12 amu” (or “…where 12 g of carbon represents one mole”).

Relative atomic mass (A_r): the weighted average mass of all naturally occurring isotopes of an element, on a scale where 1 atom of carbon-12 has a mass of 12 amu.
Isotopes: atoms of the same element having the same number of protons but different numbers of neutrons (same atomic no., different mass no.).
Relative isotopic mass: mass of one particular isotope of an element, on a scale where 1 atom of carbon-12 has a mass of 12 amu.
Relative molecular mass (M_r): the mass of one molecule of an element or compound, on a scale where 1 atom of carbon-12 has a mass of 12 amu.
Relative formula mass: the mass of one formula unit of a compound, on a scale where 1 atom of carbon-12 has a mass of 12 amu.
Mole: amount of substance containing the same number of particles as 12 g of carbon-12 / or 6.02×10²³ particles.
Molar mass: the mass of 1 mole of a substance, in g mol⁻¹.
Solution concentration: the amount of solute (in mol) dissolved in 1 dm³ of solution.

Note: the red slash on the board in “Mole” separates two ways of saying it: “12 g of C-12” or “6.02×10²³ particles”.

Same idea, different “one thing”

Only the unit counted changes: isotopic → one isotope · molecular → one molecule · formula → one formula unit. Molecule vs formula unit rarely differs, but e.g. sulfur is usually written S but actually exists as S₈ (8 atoms bonded together).

A_r is the odd one out: it is a weighted average (each isotope × its abundance) — this is the only relative definition with those words.

Why the decimals? Ca = 40.1 does not mean 0.1 of a neutron — it is a weighted average of isotopes, quoted to 1 d.p. H has 3 isotopes (¹H, ²H deuterium, ³H tritium) but ¹H is by far the most abundant, so A_r ≈ 1.

What is a “particle”?

Metals (Na, Fe, Al) exist as single atoms → 1 mol = 6.02×10²³ atoms.
Cl₂ gas exists as molecules → 1 mol = 6.02×10²³ molecules of Cl₂ (each with 2 Cl atoms).
S₈ → 1 mole = 6.02×10²³ S₈ molecules.
Molar mass = the number in the periodic table with units g mol⁻¹, e.g. Mg = 24.3 g mol⁻¹. (Use the bottom number = A_r, not the proton number.)

2. Formulas & Mole Calculations

① Moles from massmol = mass (g)M_r / A_r

② Number of particlesNo. of particles (N) = mol × 6.02×10²³

Common mistakes:

A_r or M_r? Atom (Na, Fe, Al – monatomic) → A_r. Molecule (Cl₂, NH₃) → M_r. Cl₂ = 2×35.5 = 71 (not 70, not 35.5).
Mass must be in GRAMS. Physics habit = kg — in this formula kg gives wrong answers. If the question gives kg, convert first.
Don’t memorise 10,000 formulas — rearrange the two above (particles ÷ 6.02×10²³ = mol; mol × M_r = mass).

the mole is the hub — go into moles from what you are given, then out to what is asked

Worked questions — number of particles

Board Q1calculate the number of particles in …

(a) 35.5 g of Cl₂

Cl₂ is a molecule → use M_r = 2 × 35.5 = 71mol = 35.571mol = 0.5 molN = mol × 6.02×10²³N = 0.5 × 6.02×10²³N = 3.01×10²³ molecules

(b) 336 g of iron

Fe is a metal (monatomic) → use A_r = 55.8mol = 33655.8mol = 6.022 molN = 6.022 × 6.02×10²³N = 3.62×10²⁴ atoms(board line reads ×10²³ — arithmetic gives 36.2×10²³ = 3.62×10²⁴)

Board Q2what mass of aluminium contains … (going backwards)

Now particles are given → atoms → moles → mass (reverse working). A_r(Al) = 27.

(a) 2×10²³ atomsmol = N6.02×10²³mol = 2×10²³6.02×10²³ = 0.3322mass = mol × A_rmass = 0.3322 × 27 = 8.97 gmass ≈ 9 g

(b) 6×10²⁰ atomsmol = 6×10²⁰6.02×10²³ = 9.967×10⁻⁴mass = 9.967×10⁻⁴ × 27mass = 0.0269 gmass ≈ 0.03 g

Calculator habits (sir stressed these):

Put brackets around any term with an exponent — otherwise the calculator evaluates it in the wrong order (many lost marks on 6×10²⁰ ÷ 6.02×10²³). Or use the fraction button: top = 6×10²⁰, bottom = 6.02×10²³.
Put brackets around a sum of masses before multiplying: 0.011 × (63.5 + 32.1 + 4×16) — BODMAS!
Punch numbers in with the teacher in class — fast, accurate calculator use is a skill; practise it.
Memorise a few: C = 12, H = 1, O = 16 (others come with practice; periodic table is given).

3. Gas Volumes at r.t.p.

Molar gas volume: 1 mole of any gas occupies 24 dm³ = 24 000 cm³ at r.t.p. r.t.p. = 25 °C & 1 atm

③ Volume of gas at r.t.p.Volume (V) = mol × 24 dm³ or mol × 24 000 cm³

Reverse it: mol = V ÷ 24 (V in dm³) or mol = V ÷ 24 000 (V in cm³). e.g. 640 cm³ → 640 ÷ 24 000 mol.

Careful: 24 dm³ is a constant only at r.t.p. If conditions are different, a different method (ideal gas equation) is used later in the course. Use the unit the examiner asks for — and don’t mix cm³ with dm³.

Worked questions

Board Q3volume occupied by 0.40 mol of NO₂ at r.t.p.

V = mol × 24V = 0.40 × 24V = 9.6 dm³

(= 9600 cm³ if asked in cm³.)

Board Q4mass of CO₂ present in 500 cm³ at r.t.p.

cm³ given → mol = V ÷ 24 000mol = 50024 000 = 148 molmass = mol × M_rM_r(CO₂) = 12 + 2(16) = 44mass = 148 × (12 + 32)mass = 0.91667 g ≈ 0.9 g

Board Q5volume occupied by 32 g of CO₂ at r.t.p.

mol = massM_rmol = 32(12 + 32) = 0.72727V = mol × 24V = 0.72727 × 24 = 17.45V ≈ 17 dm³

4. Concentration of Solutions

④ Concentrationconcentration (mol dm⁻³) = molesVolume (dm³)

rearrangedmoles = concentration × volume (dm³)

Unit conversion (do it as exponents!): cm³ × 10⁻³ = dm³ · dm³ × 10³ = cm³. Make it a habit: 250 cm³ → 250×10⁻³ dm³ (= ÷ 1000).

Other units of concentration (sometimes asked): g dm⁻³ → grams ÷ dm³; mol cm⁻³ → mol ÷ cm³. The unit tells you the formula. Normally it is mol dm⁻³.

Board Q6concentration of CH₃COOH solution containing 12 g in 250 cm³

M_r(CH₃COOH) = 2(12) + 4(1) + 2(16) = 60mol = 1260 = 0.2 molconc = molesvolume (dm³)conc = 0.2250 / 1000conc = 0.80 mol dm⁻³

Board Q7mass of anhydrous copper(II) sulfate in 55 cm³ of 0.20 mol dm⁻³ solution

Formula of copper(II) sulfate: Cu is 2+ (the “(II)”), sulfate is SO₄²⁻ → charges cancel → CuSO₄.

mol = conc × volumemol = 0.20 × 551000 = 0.011 molmass = mol × M_r (CuSO₄)mass = 0.011 × [63.5 + 32.1 + 4(16)]mass = 1.7556 g ≈ 1.8 g

Ions worth memorising: carbonate CO₃²⁻ · nitrate NO₃⁻ · sulfate SO₄²⁻ · H⁺ · Na⁺ (group 1 = +1) · group 2 = 2+ · halides (Cl, Br…) = −1. Make a formula by criss-crossing charges (cancel if equal): Cu²⁺ + SO₄²⁻ → CuSO₄.

5. Empirical & Molecular Formulae

Empirical formulasimplest ratio formula of atoms
e.g. C₄H₈ → CH₂

Molecular formulathe exact formula (actual no. of atoms)
e.g. C₄H₈

Molecular formula tells you more, so we find the empirical formula first and then convert to molecular.

empirical formula method: mass → moles → ÷ smallest → whole numbers

Board Q89.8 g of an oily liquid = 0.2 g H + 3.2 g S + 6.4 g O. Find the empirical formula.

	H	S	O
mass (g)	0.2	3.2	6.4
moles = mass ÷ A_r	0.21 = 0.2	3.232.1 = 0.09968	6.416 = 0.4
÷ smallest (0.09968)	0.20.09968 ≈ 2	0.099680.09968 = 1	0.40.09968 ≈ 4
simplest ratio	2	1	4

Empirical formula: H₂SO₄

Use the exact moles (0.09968) when dividing — not a rounded value. (Here the molecular formula is also H₂SO₄; sulfuric acid.)

Rounding trick: if the ratio is not whole — .5 → multiply everything by 2 (2.5 → 5). .33 → multiply by 3 (1.33 → 4). You may also divide/multiply by any number to reach whole numbers. Mostly you’ll meet .5 or .33 in AS.

Molecular formula from empirical

⑤ Molecular formulaMolecular formula = (Empirical formula)_n
n = Molar mass of molecule (M_r)Molar mass of empirical formula

(“× n” means multiply every subscript by n. e.g. if empirical is C₅H₁₀O₂ and n = 2 → C₁₀H₂₀O₄.)

Board Q9organic compound: 58.8 % C, 9.8 % H, 31.4 % O; molar mass 102 g mol⁻¹. Find empirical & molecular formulae.

Percentages given? Don’t panic — treat them as grams out of 100 g: 58.8 g C, 9.8 g H, 31.4 g O.

	C	H	O
mass	58.8	9.8	31.4
moles	58.812 = 4.9	9.81 = 9.8	31.416 = 1.9625
÷ smallest (1.9625)	2.5	5	1
× 2 (to remove .5)	5	10	2

Empirical formula: C₅H₁₀O₂ (empirical mass = 5(12)+10(1)+2(16) = 102)

n = M_r of moleculeM_r of empirical formulan = 102102 = 1Molecular formula = (C₅H₁₀O₂)₁ = C₅H₁₀O₂

Class extra — percentage composition

% by mass of an element% = (no. of atoms) × A_r of elementM_r of compound × 100

Class practice% by mass of carbon in tetraethyl lead, Pb(C₂H₅)₄

C: 8 atoms → 8 × 12 = 96M_r = 207.2 + 8(12) + 20(1) = 323.2% C = 8 × 12207.2 + 8(12) + 20(1) × 100 = 29.7 %

Class practicemass of urea, CO(NH₂)₂, containing the same mass of N as 101.1 g of KNO₃

Use the fraction directly (same as % × mass): fraction of N in compound × mass of compound.

N in KNO₃: 1439.1 + 14 + 3(16) × 101.1 = 14 gurea: N is 2 × 14 = 28 in M_r = 12 + 16 + 28 + 4 = 602860 × mass = 14mass = 14 × 6028 = 30 g

Class extra — “peak” empirical-formula question (combustion)

Class practicecompound X (C, H, O only): 2 g burns completely → 4 g CO₂ + 1.63 g H₂O. Empirical formula?

Idea: all the C in CO₂ and all the H in H₂O came from X (O₂ supplies extra O, so oxygen must be found by subtraction).

Percentage composition trick (above): mass of element = element’s mass in formulaM_r of compound × mass of compound.

C in X = 1244 × 4 = 1.09 gH in X = 218 × 1.63 = 0.18 gO in X = 2 − 1.09 − 0.18 = 0.73 gmoles: C = 1.09/12 = 0.09 H = 0.18/1 = 0.18 O = 0.73/16 = 0.0456÷ 0.0456 → 1.97 : 3.94 : 1 ≈ 2 : 4 : 1empirical formula = C₂H₄O

Sir: this is the hardest an empirical-formula question gets — you know it if you can do this.

6. Stoichiometry & Titrations

Stoichiometry = all calculations that involve moles and balanced equations (reacting amounts). Your first step is always the balanced equation — without it you most probably can’t do the question. Old papers guide you (“construct an equation…”); newer ones won’t.

every stoichiometry question = this chain. Trace the “story” of the experiment first, then work backwards

Reaction types to memorise:

acid + alkali → salt + water (neutralisation). Acid, alkali & salt are all aqueous (aq); water is liquid.
carbonate + acid → salt + water + CO₂ (e.g. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂).
State symbols: write them only if you are sure of all of them — unless the question says “include state symbols”.

Mole-ratio method (“unitary method”)

Write the two substances, their ratio under them, the known moles under the one you know, and x under the unknown; then cross-multiply. e.g. ratio 1 : 2, known 2.16×10⁻³ for the “2”, unknown x for the “1” → 2x = 2.16×10⁻³ → x = 1.08×10⁻³. Sir: many students just “halve it” — fine, but when the ratio is 9 : 2 or 9 : 5 the unitary method saves you.

Board Q4 — titration of a weak diprotic acid R

Vocabulary: diprotic (dibasic) = loses two H⁺ in water (e.g. H₂SO₄; formula H₂X). Monoprotic = loses one H⁺ (e.g. HNO₃). “Weak”, “very soluble” — just information here.

the “story”: write these steps down first — then work backwards from the titre to the original mass

Q4(a)(i)balanced equation for H₂X + NaOH

Acid + alkali → salt + water. H → H⁺ (1+), X is 2−; Na → Na⁺ (1+). Criss-cross: Na⁺ + X²⁻ → Na₂X (charges don’t cancel, so swap them).

H₂X + 2NaOH → Na₂X + 2H₂O

Balancing: 2 Na on the right → 2 NaOH; then 4 H on the left → 2 H₂O on the right; O balances.

Q4(a)(ii)moles of OH⁻ ions used in the titration

OH⁻ comes from NaOH: 1 mol NaOH gives 1 mol OH⁻ → find moles of NaOH.

mol = conc × volumemol = 0.100 × 21.61000mol = 2.16×10⁻³ mol

Q4(a)(iii)moles of R in 25.0 cm³ of solution

ratio H₂X : NaOH = 1 : 2n(R) = 2.16×10⁻³2n(R) = 1.08×10⁻³ mol

Q4(a)(iv)moles of R in the whole 250 cm³

25 cm³ was only a portion of the 250 cm³ → scale up (10×).

25 cm³ : 1.08×10⁻³250 cm³ : xx = 1.08×10⁻³ × 25025x = 0.0108 mol

Q4(a)(v)calculate M_r of R [5 marks in total]

All 0.0108 mol came from the 1.25 g weighed at the start.

mol = massM_r ⇒ M_r = massmolM_r = 1.250.0108 = 115.7M_r ≈ 116

(Board writes “116 g” — M_r has no units; as a molar mass it is 116 g mol⁻¹.)

Board Q6 — MCQ titration (no guidance this time)

MCQ25.0 cm³ of H₂SO₄ reacts with 36.2 cm³ of 0.225 mol dm⁻³ NaOH. Concentration of the acid?

A 36.2 × 0.22525.0B 2 × 36.2 × 0.22525.0C ✓ 36.2 × 0.2252 × 25.0D 25.02 × 36.2 × 0.225

Titration recipe: known conc & volume → moles of one → equation ratio → moles of other → ÷ its volume → concentration.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂On(NaOH) = conc × vol = 0.225 × 36.21000 = 8.145×10⁻³n(H₂SO₄) = ½ × 8.145×10⁻³ = 4.0725×10⁻³conc = 4.0725×10⁻³25 / 1000 = 0.1629 mol dm⁻³

Algebra: 36.2×10⁻³ × 0.2252 ÷ 25×10⁻³ → the 10⁻³ cancel → 36.2 × 0.2252 × 25 = option C. Salt: SO₄²⁻ + Na⁺ → Na₂SO₄ (not NaSO₄).

Answer: C

Class extra — “which carbonate releases the most CO₂?” (shortcut MCQ)

Class MCQ1 g samples of CaCO₃, Li₂CO₃, MgCO₃, Na₂CO₃ + excess dilute acid. Which releases the most CO₂?

Each equation: 1 mol carbonate → 1 mol CO₂ (salt differs only). Group 1 metal (Li, Na) → M₂CO₃; group 2 (Ca, Mg) → MCO₃. So Mg and Ca use the same equation as each other; Li and Na the same as each other — write only two.

Most moles of carbonate ⇒ most CO₂. Moles = 1 ÷ M_r; CO₃ is in all four, so just compare the metal part of M_r: smaller M_r ⇒ more moles (inverse proportion).

compound	metal part of M_r
CaCO₃	40.1
Li₂CO₃	2 × 6.9 = 13.8 ← smallest
MgCO₃	24.3
Na₂CO₃	2 × 23 = 46

Li₂CO₃ releases the most CO₂.

7. Stoichiometry: Gas Volumes

For ONLY-gas reactions, the mole ratio = the volume ratio

Then no conversion to moles is needed: read the ratio from the balanced equation and apply it to the volumes directly (dm³ or cm³). Both substances involved must be gases — if one is a solid or liquid, use the full mole method.

Proof (why it works): mol = V ÷ 24 000 for every gas, so the “÷ 24 000” cancels in every ratio. That’s also why you don’t need r.t.p. — only that all gases are measured under the same conditions.

Board Q1050 cm³ propane mixed with 600 cm³ O₂ (excess) and ignited. Volume of CO₂ produced? O₂ unused?

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Balancing: 3 C → 3 CO₂; 8 H → 4 H₂O; O = 6 + 4 = 10 → 5 O₂. Volume ratio C₃H₈ : O₂ : CO₂ = 1 : 5 : 3.

50 cm³ : 250 cm³ (O₂ used) : 150 cm³ (CO₂)(1) CO₂ produced = 150 cm³(2) O₂ unused = 600 − 250 = 350 cm³

propane is limiting (O₂ is in excess) → calculate from propane

Why not 24 000? In the ethane question (below) the conditions were 101 kPa, 200 °C — not r.t.p. We never used 24 dm³, because we used the volume ratio only. Also note: at 200 °C, water is a gas (boils at 100 °C) so count it in the total volume!

Board Q11 (MCQ)10 cm³ ethane burned in 45 cm³ O₂ at 101 kPa, 200 °C (complete combustion). Total volume of gas after reaction, same conditions? A 30 B 50 C 55 D 60

C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O

Balancing: 2 C → 2 CO₂; 6 H → 3 H₂O; O = 4 + 3 = 7 → 3.5 O₂. (Fraction/decimal coefficients are fine at A-level — or double the whole equation.)

ratio → C₂H₆ 10 : O₂ 35 : CO₂ 20 : H₂O 30unused O₂ = 45 − 35 = 10 cm³total = 20 + 30 + 10 = 60 cm³

Answer: D

Board Q12 (MCQ)Assume air is 21 % oxygen. Minimum volume of air for complete combustion of 10 cm³ butane (room conditions)? A 14 B 27 C 65 D 310

C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O

Balancing: 4 C → 4 CO₂; 10 H → 5 H₂O; O = 8 + 5 = 13 → 6.5 O₂. Ratio 1 : 6.5 : 4 : 5.

butane 10 : O₂ 65 : CO₂ 40 : H₂O 50O₂ needed = 65 cm³ — but it comes from AIR (21 %)x × 21100 = 65x = 65 × 10021 = 309.5x ≈ 310 cm³

Trap: 65 cm³ is only the oxygen volume (option C) — the question asks for air, so divide by 21 %.

Answer: D

8. Stoichiometry: Tougher Questions

Recent papers give no step-by-step guidance (“Show your working” + [3] marks). You must trace the story yourself: what was taken, what was diluted, what reacted, what was measured — then work backwards.

Board Q13 · [3]bleach S: NaClO conc. = x g dm⁻³. H₂O₂(aq) + NaClO(aq) → H₂O(l) + NaCl(aq) + O₂(g). 5.00 cm³ of S reacts completely and gives 24.0 cm³ O₂ at room conditions. Calculate x.

Equation already balanced (check! Na 1, Cl 1, O 3 = 3, H 2 = 2). Plan: O₂ volume → mol O₂ → (1 : 1) mol NaClO → mass (g/dm³ asked, not mol/dm³) → ÷ volume.

n(O₂) = 2424 000 = 1×10⁻³ molratio NaClO : O₂ = 1 : 1n(NaClO) = 1×10⁻³ molmass = mol × M_r (M_r = 23 + 35.5 + 16 = 74.5)mass = 1×10⁻³ × (23 + 35.5 + 16)mass = 0.0745 gconc = 0.07455×10⁻³x = 14.9 g dm⁻³

Sir’s calculator warning: put 5×10⁻³ in brackets → 0.0745 ÷ (5×10⁻³). If you get 1.49 you forgot brackets. Since gas volume was in cm³, use 24 000 cm³ (no need to convert to dm³).

Board Q14 — Chile saltpetre (back-titration) · [9]

Chile saltpetre (mainly NaNO₃) is purified for fertiliser/fireworks. To find purity, a sample is heated with NaOH(aq) + Devarda’s alloy (contains Al), which reduces the nitrate to ammonia:

3NaNO₃ + 8Al + 5NaOH + 18H₂O → 3NH₃ + 8NaAl(OH)₄
2NH₃ + H₂SO₄ → (NH₄)₂SO₄
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

back-titration story: acid in EXCESS ⇒ (total acid) − (acid left over, found by NaOH) = acid that reacted with NH₃

Q14(a) (i)–(iv)acid amounts

(i) n(H₂SO₄) added = c × V = 1 × 25×10⁻³ = 0.025 mol(ii) n(NaOH) = c × V = 2 × 16.2×10⁻³ = 0.0324 mol(iii) ratio H₂SO₄ : NaOH = 1 : 2n(H₂SO₄ left over) = 0.03242 = 0.0162 mol(iv) H₂SO₄ that reacted with NH₃ = original − leftover= 0.025 − 0.0162 = 8.80×10⁻³ mol

Q14(a) (v)–(viii)from acid to % purity

(v) ratio NH₃ : H₂SO₄ = 2 : 1n(NH₃) = 2 × 8.80×10⁻³ = 0.0176 mol(vi) ratio NaNO₃ : NH₃ = 3 : 3 = 1 : 1n(NaNO₃) = 0.0176 mol(vii) mass = mol × M_r = 0.0176 × (23 + 14 + 3(16))mass of NaNO₃ = 1.496 g(viii) % = pure masssample mass × 100 = 1.4961.64 × 100% NaNO₃ = 91.2 %

1.64 g was impure — only 1.496 g was NaNO₃, the rest impurities. “Suitable s.f.” = same as data (3 s.f.). Percentage = what you want ÷ total × 100.

Class extras — more “story” questions

Class practice · sample vs dilutionbleach W: 10 cm³ of W is diluted with water to 100 cm³. 25 cm³ of the diluted solution added to excess H₂O₂ gives 42 cm³ O₂ (r.t.p.). Conc. of NaClO in W?

Key idea: adding water (dilution) never changes the moles; but taking a portion does — a smaller portion contains fewer moles. So: work back through the portion (25 → 100), then straight across the dilution (100 → 10, same moles).

moles: 10 cm³ = 100 cm³ (just water added) ≠ 25 cm³ (a portion taken)

n(O₂) = 4224 000 = 1.75×10⁻³ = n(NaClO) in 25 cm³ (1 : 1)in 100 cm³: 25 : 1.75×10⁻³ = 100 : xx = 1.75×10⁻³ × 10025 = 7×10⁻³ mol (= moles in the original 10 cm³ too)mass = 7×10⁻³ × 74.5 = 0.5215 gconc = 0.521510×10⁻³ = 52.2 g dm⁻³

Class practice · identify the metal0.250 g of group 2 bromide XBr₂ needs 33.65 cm³ of 0.05 mol dm⁻³ AgNO₃: XBr₂ + 2AgNO₃ → 2AgBr + X(NO₃)₂. Find X.

Aim: find M_r of XBr₂ = mass ÷ moles, subtract 2 Br, then match A_r to the periodic table (group 2).

n(AgNO₃) = 0.05 × 33.65×10⁻³ = 1.68×10⁻³ratio XBr₂ : AgNO₃ = 1 : 2 → n(XBr₂) = ½ × 1.68×10⁻³ = 8.4×10⁻⁴M_r = massmol = 0.2508.4×10⁻⁴ = 297.6A_r(X) = 297.6 − 2(79.9) = 137.8closest group 2 element: barium (Ba)

Don’t worry if it isn’t exact — small experimental error. Remember to subtract Br₂: the 297.6 was for the whole XBr₂.

Class practice · gas left over0.198 g As₂S₃ heated in 0.1 dm³ O₂ (excess). Ratios from the equation: As₂S₃ : O₂ = 2 : 9 and As₂S₃ : SO₂ = 2 : 6. Total volume of gas at the end? (1 mol = 24 dm³)

“Excess” O₂ ⇒ As₂S₃ is limiting → calculate from it. End gas = O₂ left over + SO₂ made.

M_r(As₂S₃) = 2(74.9) + 3(32.1) → n = 8.05×10⁻⁴ molO₂ used: x = 9 × 8.05×10⁻⁴2 = 3.62×10⁻³ mol → × 24 = 0.0869 dm³O₂ left = 0.1 − 0.08688 = 0.01312 dm³SO₂: x = 6 × 8.05×10⁻⁴2 = 2.415×10⁻³ mol → × 24 = 0.058 dm³total = 0.058 + 0.01312 = 0.0711 dm³

Use A_r for each atom (As, S) then add them up to make the M_r of the compound.

9. Percentage Yield & Limiting Reagent

⑥ Percentage yield% yield = experimental (actual) masstheoretical mass × 100

Theoretical = what the balanced equation says should form (maths/chemistry). Experimental = what you actually collected (impurities, losses, incomplete reaction…). e.g. expected 100 g, got 50 g → 50 % yield — half of it was “lost”.

Board Q15C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O. A student used 10.0 g benzene and obtained 9.0 g nitrobenzene. % yield?

9.0 g = experimental. Find theoretical from 10.0 g benzene (1 : 1); keep exact fractions.

M_r(C₆H₆) = 6(12) + 6(1) = 78n(C₆H₆) = 1078 = 539 molratio 1 : 1 → n(C₆H₅NO₂) = 539 moltheoretical mass = 539 × 123 = 15.77 g (M_r = 72 + 5 + 14 + 32 = 123)% yield = 915.77 × 100 = 57 %

Class extra — limiting reagent

If the question says “excess”, the other reagent is limiting and you calculate from that one. If it gives amounts for both reactants (no “excess” stated), you must work out which is limiting: find moles of both → use the balanced-equation ratio → compare.

Class practice3.7 g CuCO₃ reacts with 25 cm³ HCl (0.05 mol HCl). Volume of CO₂ at r.t.p.?

CuCO₃ + 2HCl → CuCl₂ + H₂O + CO₂

Salt: Cu²⁺ + 2Cl⁻ → CuCl₂. Cu(II) = 2+; CO₃²⁻ → CuCO₃.

n(CuCO₃) = 3.763.5 + 12 + 3(16) = 0.03 moln(HCl) = 0.05 molratio 1 : 2 → 0.03 mol CuCO₃ needs 0.06 mol HCl — only 0.05 availableor: 0.05 mol HCl needs 0.025 mol CuCO₃ (< 0.03 ✓)⇒ HCl is limiting, CuCO₃ in excessn(CO₂) = ½ × 0.05 = 0.025 molV = 0.025 × 24 = 0.6 dm³

Don’t just pick the smaller number of moles (0.03 < 0.05 would wrongly say CuCO₃) — always check against the ratio.

10. Water of Crystallisation

Some salts are hydrated: their crystals hold water molecules, e.g. CuSO₄·yH₂O. The question gives the salt but not the number of H₂O. Find the mole ratio salt : water = 1 : y, then divide so that the salt = 1. (No formula needed — the unitary method does it.)

value of y (or x)y = moles of H₂Omoles of anhydrous salt

Board Q16 · [4]17.43 g CuSO₄·yH₂O; titration shows 0.0982 mol CuSO₄. Find integer y. (Data Booklet = periodic table.)

mass of CuSO₄ = mol × M_r= 0.0982 × [63.5 + 32.1 + 4(16)] = 15.672 ≈ 15.67 gmass of H₂O = 17.43 − 15.67 ≈ 1.757 g (the rest of the crystal)mol = 1.75718 = 0.0978 mol H₂Oy = 0.09780.0982 ≈ 1y = 1

M_r(H₂O) = 18. Subtracting the anhydrous mass from the total leaves the water — “definitely, obviously”. [Total: 9 with part (a).]

Class practice · struvitehydrated NH₄MgPO₄·xH₂O has M_r 245.3; anhydrous NH₄MgPO₄ has M_r 137.3. Find x.

M_r(anhydrous) = 14 + 4(1) + 24.3 + 31.0 + 4(16) = 137.3M_r due to water = 245.3 − 137.3 = 108x = 10818 = 6 so the crystal is NH₄MgPO₄·6H₂O

Board Q17 — hydrated CuSO₄·xH₂O by iodine–thiosulfate titration

A hydrated copper(II) sulfate sample is dissolved in water; excess KI is added; total volume made up to 250.0 cm³: 2CuSO₄ + 4KI → 2CuI + I₂ + K₂SO₄. The I₂ is found by titrating 25.0 cm³ of this solution with 20.0 cm³ of 0.10 mol dm⁻³ thiosulfate: I₂ + 2S₂O₃²⁻ → S₄O₆²⁻ + 2I⁻.

go backwards: thiosulfate → I₂ (in 25 cm³) → I₂ (in 250 cm³) → CuSO₄

Q17(d)(i) · [2]moles of copper(II) sulfate in the original sample

n(S₂O₃²⁻) = c × V = 0.10 × 20×10⁻³ = 2×10⁻³ratio I₂ : S₂O₃²⁻ = 1 : 2 → n(I₂) = 1×10⁻³ (in 25 cm³)in 250 cm³: I₂ = 1×10⁻³ × 25025 = 0.01ratio CuSO₄ : I₂ = 2 : 1n(CuSO₄) = 2 × 0.01 = 0.02 mol

Q17(d)(ii) · [3]7.98 g CuSO₄ in 10.68 g CuSO₄·xH₂O (M_r of CuSO₄ = 159.6). Find x.

row	working	answer
amount of CuSO₄	7.9863.5 + 32.1 + 4(16)	0.05 mol
amount of H₂O	10.68 − 7.98 = 2.7 g → 2.718	0.15 mol
value of x	0.150.05	x = 3

Don’t use the whole-hydrate M_r for CuSO₄’s moles — 7.98 g goes with the 159.6 of CuSO₄ only; the rest of the hydrate is H₂O.

Board Q18 — Mohr’s salt (NH₄)₂Fe(SO₄)₂·xH₂O by KMnO₄ titration

0.784 g of the hydrated salt is dissolved in water and the solution acidified. All of it is titrated with 0.0200 mol dm⁻³ KMnO₄; 20.0 cm³ is needed to react completely with the Fe²⁺.

Q18(d)(i)balance using oxidation-number changes [1]

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Mn goes +7 → +2 (gains 5 e⁻); each Fe goes +2 → +3 (loses 1 e⁻) → need 5 Fe²⁺ per MnO₄⁻; then balance O with 4 H₂O and H with 8 H⁺. Part (ii) “state the role of Fe²⁺ and explain” was crossed out / skipped on the board (redox chapter comes later: Fe²⁺ is oxidised, so it is the reducing agent).

Q18(d)(iii)–(vi)find x

(iii) n(MnO₄⁻) = c × V = 0.02 × 20×10⁻³ = 4.00×10⁻⁴ mol(iv) ratio Fe²⁺ : MnO₄⁻ = 5 : 1n(Fe²⁺) = 4.00×10⁻⁴ × 5 = 2.00×10⁻³ mol(v) 1 Fe per formula unit → n(salt) = 2.00×10⁻³M_r = massmol = 0.7842.00×10⁻³ = 392(vi) 392 = 28 + 8 + 55.8 + 64.2 + 8(16) + 2x + 16x392 = 284 + 18x ⇒ 108 = 18xx = 6

(NH₄)₂: 2(14) + 8(1) = 36 = “28 + 8” · Fe = 55.8 · (SO₄)₂: 2(32.1) + 8(16) = 64.2 + 128 · then each H₂O is 18 (= 2x + 16x).

11. Mass Spectrometry & Isotopes

Overview only here — the full topic comes in organic chemistry. All A_r values in the periodic table were calculated from mass spectra (weighted average of isotopes).

Ne atoms → +1 ions → deflected by the field; the detector records how much of each mass arrives

Must-know rules:

Every particle is given +1 charge, so name peaks with the +: ²⁰Ne⁺ not ²⁰Ne — no + = no mark.
x-axis: m/e (or m/z, or M_r/mass number) — all mean the same because charge = +1, so the x-value = mass number.
y-axis: relative abundance (peak height = how much of that isotope).

Board — neon: three isotopes

²⁰Ne (90.9 %) & ²¹Ne (0.20 %) & ²²Ne (8.9 %) — no need to memorise these. The spectrum has one peak per isotope, at its mass number:

each peak = one isotope as an ion (X⁺); height = its % abundance (small peaks not to scale)

Ar from a mass spectrum = weighted averageA_r = Σ (% abundance100 × mass number)

A_r(Ne) = (90.9100 × 20) + (0.20100 × 21) + (8.9100 × 22)= 20.2

“Weighted average mass” is what makes A_r different from every other “relative” definition — quote it in the definition!

Board — potassium with no scale on the y-axis

no numbers given? MEASURE the peak heights with a ruler (4.7 cm and 0.3 cm here)

Q. The mass spectrum of potassium (scale missing) is shown. (a) relative abundances of ³⁹K and ⁴¹K? (b) A_r of potassium?

Examiners aren’t silly: if the scale is missing they expect you to measure the lengths and use the ratio. Total = 4.7 + 0.3 = 5.0.

(a) % ³⁹K = 4.74.7 + 0.3 × 100 = 94 %% ⁴¹K = 0.34.7 + 0.3 × 100 = 6 %(b) A_r = (94100 × 39) + (6100 × 41)= 39.12 ≈ 39.1 (board wrote 4.7/(3 + 4.7) by slip — the total must be 4.7 + 0.3)

Periodic-table A_r is always quoted to 1 d.p. → 39.1 ✓.

Mass spectrum of Cl and Cl₂ (LEARN)

Chlorine has 2 natural isotopes: ³⁵Cl (75 %) and ³⁷Cl (25 %). For Cl atoms (individual gaseous atoms, not Cl₂): one line per isotope, peaks ³⁵Cl⁺ and ³⁷Cl⁺ in ratio 3 : 1.

Cl atoms: ³⁵Cl⁺ = 75 %, ³⁷Cl⁺ = 25 %

For Cl₂ molecules the possibilities multiply. A molecule can be made from:

³⁵Cl–³⁵Cl (M_r 70)
³⁵Cl–³⁷Cl or ³⁷Cl–³⁵Cl (M_r 72) counts twice
³⁷Cl–³⁷Cl (M_r 74)

1) [³⁵Cl–³⁵Cl]⁺ = 34 × 34 = 9162) [³⁵Cl–³⁷Cl]⁺ or [³⁷Cl–³⁵Cl]⁺ = (34 × 14) + (14 × 34) = 6163) [³⁷Cl–³⁷Cl]⁺ = 14 × 14 = 116

“And” between the two atoms = multiply; “either/or” (two orders) = add. Multiply all by 16 → molecule peaks in the ratio 9 : 6 : 1.

Cl₂: molecular ions 70 : 72 : 74 = 9 : 6 : 1 · fragment atoms 35 : 37 = 3 : 1

Fragmentation: giving the molecule a + charge makes it unstable, so it can break into atoms → extra peaks at 35 (³⁵Cl⁺) and 37 (³⁷Cl⁺). The molecules are in ratio 9 : 6 : 1; the fragments are in ratio 3 : 1 (same as 75 : 25). There is NO link between the two ratios — don’t compare a fragment height with a molecule height.

Mass spectrum of Br and Br₂ (LEARN)

Bromine has 2 isotopes, equally abundant: ⁷⁹Br (50 %) and ⁸¹Br (50 %) (the board’s x-axis says 80 by mistake — it is 79). Br atoms → two equal lines at 79 and 81.

Br atoms: two equal peaks (50 % each)

1) [⁷⁹Br–⁷⁹Br]⁺ = 12 × 12 = 14 (M_r 158)2) [⁷⁹Br–⁸¹Br]⁺ or [⁸¹Br–⁷⁹Br]⁺ = (12 × 12) + (12 × 12) = 12 (M_r 160)3) [⁸¹Br–⁸¹Br]⁺ = 12 × 12 = 14 (M_r 162)

Br₂: molecules 158 : 160 : 162 = 1 : 2 : 1 · fragments 79 : 81 = 1 : 1

Br₂ summary: molecules in the ratio 1 : 2 : 1 (the middle peak is 2× — two arrangements); fragments ⁷⁹Br⁺ and ⁸¹Br⁺ in ratio 1 : 1; again no link between fragments and molecules — 1 : 1 does not mean equal heights across the two groups.

Board Q19 — tumbaga (Cu + Au alloy) & HCl spectra

y-axis = percentage abundance: copper-63 = 56.36 %, copper-65 = 25.14 %, gold-197 = x

Q19(d)(i) · [1]% abundance of gold, x, in the tumbaga sample

total = 100 %100 = 56.36 + 25.14 + xx = 18.50 %

Q19(d)(ii) · [2]A_r of the copper in the sample (2 d.p.)

Only the copper isotopes count: their combined abundance is 56.36 + 25.14 = 81.50, not 100 — so use each as a fraction of 81.50.

A_r(Cu) = (56.3656.36 + 25.14 × 63) + (25.1456.36 + 25.14 × 65)A_r(Cu) = 63.62

HCl mass spectrum

Some HCl molecules split into atoms in the spectrometer. Cl has two isotopes; H is ¹H only. Fragments ³⁵Cl⁺, ³⁷Cl⁺ give the lower peaks (35, 37); the molecular ions [H–Cl]⁺ give the taller ones (36, 38).

green = molecular ions; blue = chlorine fragments (heights 30 and 10 read off the y-axis)

Q19(b)(i)–(ii) · [2]particle responsible for the peak at mass 35? at mass 38?

(i) ³⁵Cl⁺ (ii) [¹H–³⁷Cl]⁺ (1 + 37 = 38)

Q19(c) · [2]use peak heights to find the proportions of the two Cl isotopes; explain simply

use the fragment peaks (Cl⁺): heights 30 (m = 35) and 10 (m = 37)% ³⁵Cl = 3030 + 10 × 100 = 75 %% ³⁷Cl = 1030 + 10 × 100 = 25 %

Explanation to write: “relative height of each isotope’s peak ÷ total height of the two peaks × 100”.

Q19(d) · [1]explain why A_r of chlorine is 35.5

A_r = (75100 × 35) + (25100 × 37)A_r = 35.5

i.e. it is the weighted average of the isotopes — not a whole number because there is more ³⁵Cl than ³⁷Cl.

Board Q20 (MCQ) — how many peaks for O₂⁺?

MCQO has three stable isotopes ¹⁶O, ¹⁷O, ¹⁸O; all present in O₂. How many peaks for the O₂⁺ ion? A 3 B 5 C 6 D 9

List every possible pair of isotopes (order doesn’t matter) and add the masses:

[¹⁶O–¹⁶O]⁺	32	[¹⁷O–¹⁷O]⁺	34
[¹⁶O–¹⁷O]⁺	33	[¹⁷O–¹⁸O]⁺	35
[¹⁶O–¹⁸O]⁺	34	[¹⁸O–¹⁸O]⁺	36

6 combinations, but ¹⁶O–¹⁸O and ¹⁷O–¹⁷O both = 34 → they overlap into one peak ⇒ masses 32, 33, 34, 35, 36 = 5 peaks. C ✗ (6) counts 34 twice.

Answer: B

5 different masses ⇒ 5 peaks (heights schematic)

12. Full Chapter Cheat-Sheet

✓ Formulas (all in GRAMS, dm³, mol)

mol = mass ÷ M_r/A_r · N = mol × 6.02×10²³
gas at r.t.p.: V = mol × 24 dm³ (24 000 cm³)
conc (mol dm⁻³) = mol ÷ V(dm³) · mol = c × V · cm³ × 10⁻³ = dm³
n = M_r of molecule ÷ mass of empirical formula → molecular = (empirical)_n
% yield = experimental ÷ theoretical × 100 · y (or x) = mol H₂O ÷ mol salt
A_r = Σ (% abundance ÷ 100 × mass number) · % of element = (atoms × A_r) ÷ M_r × 100

✓ Method reminders

Balanced equation first; trace the story (dissolved? diluted? portion taken? excess?).
Atom → A_r; molecule → M_r (Cl₂ = 71). Data booklet = periodic table.
Unitary method: write both species, their ratio, known moles, x → cross-multiply.
Gas-only reactions: mole ratio = volume ratio (water is a gas at 200 °C!).
Dilution: same moles. Taking a portion: fewer moles. Back-titration: original − left-over.
Empirical: ÷ Ar → ÷ smallest → ×2 for .5, ×3 for .33. % given → take as grams.
Brackets on exponents & sums in the calculator; never kg in mol = mass/M_r.

✓ Learn: mass spectra

Always +1 ions: write X⁺ (no + = no mark); x-axis m/e = mass number.
Cl: ³⁵Cl 75 % / ³⁷Cl 25 %. Cl₂ molecules 70 : 72 : 74 = 9 : 6 : 1; fragments 35 : 37 = 3 : 1.
Br: ⁷⁹Br 50 % / ⁸¹Br 50 %. Br₂ molecules 158 : 160 : 162 = 1 : 2 : 1; fragments 79 : 81 = 1 : 1.
No link between the fragment ratio and the molecule ratio.
No y-scale? Measure peak lengths with a ruler. Count peaks by listing all isotope pairs (equal masses merge).

★ Rounding: sir promised the significant-figure rules next class; for now, answers here are quoted to about 3 s.f. — or as the question’s data suggests.

Atoms, Molecules & Stoichiometry

1. Key Definitions

Same idea, different “one thing”

What is a “particle”?

2. Formulas & Mole Calculations

Worked questions — number of particles

3. Gas Volumes at r.t.p.

Worked questions

4. Concentration of Solutions

5. Empirical & Molecular Formulae

Molecular formula from empirical

Class extra — percentage composition

Class extra — “peak” empirical-formula question (combustion)

6. Stoichiometry & Titrations

Mole-ratio method (“unitary method”)

Board Q4 — titration of a weak diprotic acid R

Board Q6 — MCQ titration (no guidance this time)

Class extra — “which carbonate releases the most CO2?” (shortcut MCQ)

7. Stoichiometry: Gas Volumes

8. Stoichiometry: Tougher Questions

Board Q14 — Chile saltpetre (back-titration) · [9]

Class extras — more “story” questions

9. Percentage Yield & Limiting Reagent

Class extra — limiting reagent

10. Water of Crystallisation

Board Q17 — hydrated CuSO4·xH2O by iodine–thiosulfate titration

Board Q18 — Mohr’s salt (NH4)2Fe(SO4)2·xH2O by KMnO4 titration

11. Mass Spectrometry & Isotopes

Board — neon: three isotopes

Board — potassium with no scale on the y-axis

Mass spectrum of Cl and Cl2 (LEARN)

Mass spectrum of Br and Br2 (LEARN)

Board Q19 — tumbaga (Cu + Au alloy) & HCl spectra

HCl mass spectrum

Board Q20 (MCQ) — how many peaks for O2+?

12. Full Chapter Cheat-Sheet

Class extra — “which carbonate releases the most CO₂?” (shortcut MCQ)

Board Q17 — hydrated CuSO₄·xH₂O by iodine–thiosulfate titration

Board Q18 — Mohr’s salt (NH₄)₂Fe(SO₄)₂·xH₂O by KMnO₄ titration

Mass spectrum of Cl and Cl₂ (LEARN)

Mass spectrum of Br and Br₂ (LEARN)

Board Q20 (MCQ) — how many peaks for O₂⁺?