OCR A-level Chemistry A (H432) · Module 2: Foundations in chemistry
Mini-Lesson
Foundations in Chemistry
Module 2 is the toolkit every other module leans on: atoms, ions and isotopes, the mole, acids, redox, electron structure, bonding and shapes of molecules.
Expect the heaviest calculation load of any module: n = m/M, n = cV, pV = nRT, titrations, percentage yield and atom economy — plus VSEPR shapes and intermolecular forces.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
2.1.1 Atomic structure and isotopes
Isotopes and relative atomic mass
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have identical chemical properties (same electron configuration) but slightly different physical properties, such as density and rate of diffusion.
Ar = Σ(isotopic mass × % abundance) ÷ 100relative atomic mass is the weighted mean of the isotopic masses, on the ¹²C = 12 scale
A mass spectrum gives you the abundances directly: each peak is one isotope, plotted at its m/z value.
Definitions OCR will ask for word-for-word:relative isotopic mass = the mass of an atom of an isotope compared with 1/12th the mass of an atom of carbon-12. Relative atomic mass = the weighted mean mass of an atom of an element compared with 1/12th the mass of an atom of carbon-12.
Calculate
Your turn — calculation 1
1Chlorine consists of 75.0% ³⁵Cl (isotopic mass 35.0) and 25.0% ³⁷Cl (isotopic mass 37.0). Calculate the relative atomic mass of chlorine to 1 decimal place.
One mole is 6.02 × 10²³ particles (the Avogadro constant, L).
n = m ÷ Mmoles = mass (g) ÷ molar mass (g mol⁻¹)
n = c × Vmoles = concentration (mol dm⁻³) × volume (dm³)
pV = nRTp in Pa · V in m³ · T in K · R = 8.314 J mol⁻¹ K⁻¹
Unit traps that cost marks: for pV = nRT, kPa → ×1000 to get Pa; dm³ → ÷1000 to get m³; cm³ → ÷1,000,000 to get m³; °C → +273 to get K. For n = cV, cm³ → ÷1000 to get dm³.
Calculate
Your turn — calculation 2
2Calculate the amount, in mol, of sodium hydrogencarbonate, NaHCO₃, in 4.20 g. (Ar: Na = 23.0, H = 1.0, C = 12.0, O = 16.0)
mol
Hint: M(NaHCO₃) = 23.0 + 1.0 + 12.0 + (3 × 16.0) = 84.0. Then n = m ÷ M.
Method
M = 23.0 + 1.0 + 12.0 + 48.0 = 84.0 g mol⁻¹. n = 4.20 ÷ 84.0 = 0.0500 mol.
Calculate
Your turn — calculation 3
3Calculate the volume, in dm³, occupied by 0.250 mol of an ideal gas at 100 kPa and 25 °C. (R = 8.314 J mol⁻¹ K⁻¹). Give your answer to 3 significant figures.
dm³
Hint: T = 298 K, p = 100 000 Pa. V = nRT ÷ p, giving V in m³ — then × 1000 for dm³.
Acids release H⁺ in aqueous solution. Strong acids (HCl, HNO₃, H₂SO₄) fully dissociate; weak acids (CH₃COOH) only partially dissociate.
Neutralisation always gives a salt + water. The ionic equation for any strong acid–strong base neutralisation is:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The titration routine, every time:
Step 1 — moles of the substance you know everything about: n = c × V (V in dm³).
Step 2 — use the balanced equation ratio to get the moles of the unknown.
Step 3 — divide by its volume in dm³ to get its concentration.
Calculate
Your turn — calculation 4
4A 25.0 cm³ sample of NaOH(aq) is exactly neutralised by 22.50 cm³ of 0.100 mol dm⁻³ HCl. NaOH + HCl → NaCl + H₂O. Calculate the concentration of the NaOH in mol dm⁻³, to 3 significant figures.
mol dm⁻³
Hint: n(HCl) = 0.02250 × 0.100. The ratio is 1 : 1. Then c = n ÷ 0.0250.
% yield = (actual moles ÷ theoretical moles) × 100equivalently (actual mass ÷ theoretical mass) × 100 for the same product
atom economy = (M of desired product ÷ Σ M of all products) × 100
The distinction OCR test:yield is about how much of the possible product you actually isolated (losses, side reactions, incomplete reaction). Atom economy is fixed by the equation itself — it tells you what fraction of the reactant atoms end up in the product you want. An addition reaction has 100% atom economy; a substitution never does.
Calculate
Your turn — calculation 5
55.00 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂. The mass of calcium oxide obtained is 2.24 g. Calculate the percentage yield. (M: CaCO₃ = 100.0, CaO = 56.0)
%
Hint: n(CaCO₃) = 5.00 ÷ 100.0 = 0.0500 mol → theoretical CaO = 0.0500 × 56.0 g. Then (2.24 ÷ theoretical) × 100.
Total mass of products = 56.0 + 44.0 = 100.0. Atom economy = (56.0 ÷ 100.0) × 100 = 56.0%. The CO₂ is waste.
2.1.5 Redox
Oxidation numbers and redox
Oxidation is loss of electrons (oxidation number increases); reduction is gain (oxidation number decreases). The oxidising agent is itself reduced; the reducing agent is itself oxidised.
Uncombined element = 0. Simple ion = its charge.
O is usually −2 (but −1 in peroxides); H is usually +1 (but −1 in metal hydrides); F is always −1.
The oxidation numbers in a neutral compound sum to 0; in an ion they sum to the charge.
Worked example — Mn in MnO₄⁻
Let x be the oxidation number of Mn. x + 4(−2) = −1 → x − 8 = −1 → x = +7.
Quick check
Think it through
?What is the oxidation number of chromium in the dichromate(VI) ion, Cr₂O₇²⁻?
2.2.1 Electron structure
Orbitals, sub-shells and ionisation energy
Electrons occupy orbitals (s holds 2, p holds 6, d holds 10, f holds 14). Fill in order of increasing energy — note 4s fills before 3d.
Cu is an exception: [Ar] 3d¹⁰ 4s¹, and Cr is [Ar] 3d⁵ 4s¹
The first ionisation energy is the energy to remove one electron from each atom in one mole of gaseous atoms: X(g) → X⁺(g) + e⁻. It depends on nuclear charge, atomic radius and shielding.
The two dips across Period 3: Al is lower than Mg because Al's outer electron is in a 3p orbital (higher energy, more shielded). S is lower than P because in S two electrons are paired in the same 3p orbital and repel each other.
Quick check
Think it through
?What is the electron configuration of the Fe²⁺ ion? (Fe is element 26)
2.2.2 Bonding and structure
Electronegativity, polarity and intermolecular forces
Electronegativity is the ability of an atom to attract the bonding electron pair in a covalent bond. A difference in electronegativity makes a bond polar (a permanent dipole).
Three intermolecular forces, weakest to strongest:
London (induced dipole–dipole) forces — present between all molecules; caused by instantaneous dipoles from moving electrons. Stronger for more electrons.
Permanent dipole–dipole — between polar molecules (e.g. HCl).
Hydrogen bonding — only where H is bonded directly to N, O or F, and a lone pair on N/O/F of another molecule attracts it.
Why ice floats: hydrogen bonds are relatively long and hold the H₂O molecules in an open lattice with large gaps. Ice is therefore less dense than liquid water. Water's high boiling point and high surface tension come from the same hydrogen bonding.
A molecule can be polar in its bonds but non-polar overall. CO₂ has two polar C=O bonds, but it is linear so the dipoles cancel. CCl₄ is tetrahedral and symmetrical, so it also cancels.
Sort it
What is the strongest intermolecular force present?
Tap a molecule, then tap the strongest intermolecular force it has.
💨 London forces only
🧲 Permanent dipole–dipole
💧 Hydrogen bonding
Quick check
Think it through
?Why does ice float on water?
2.2.2 Shapes
Electron pair repulsion (VSEPR)
Electron pairs around a central atom repel and get as far apart as possible. Lone pairs repel more strongly than bonding pairs, so each lone pair squeezes the bond angle by roughly 2.5°.