AQA A-Level Chemistry: Atomic Structure and Amount of Substance — Complete Revision Guide (7405)

Atomic structure and amount of substance is the foundation of the entire AQA 7405 course. Every other topic — energetics, kinetics, equilibrium, acids and buffers, transition metals, organic mechanisms — relies on the tools introduced here. The mole concept turns macroscopic masses into countable particles. The ideal gas equation links pressure, volume and temperature to mole quantities. Titration arithmetic makes quantitative reaction analysis possible. If your fluency with these tools is shaky, every later topic will be slower than it should be, and marks will quietly leak away across all three papers.

This topic is profoundly synoptic. The mole calculations here are the same calculations that appear in energetics, in kinetics and equilibrium and in acids and buffers. The ideal gas equation reappears whenever a reaction produces gas. Electron configuration underpins the periodic-table trends in inorganic chemistry. Treat this course as the foundation of every paper, not just the opener.

This guide walks through the ten lessons in the LearningBro AQA Atomic Structure and Amount of Substance course: atomic models and subatomic particles; mass spectrometry and isotopic abundance; electron configurations; first and successive ionisation energies; the mole concept; empirical and molecular formulae; balanced equations and reacting masses; solution concentrations; titration calculations (the Required Practical 1 anchor); and the ideal gas equation. Each section gives the core idea, the key equation, one common pitfall and a link into the relevant lesson on LearningBro.

What the AQA 7405 Specification Covers

AQA A-Level Chemistry (7405) is examined through three papers. Paper 1 (Inorganic and Relevant Physical, 2h, 105 marks, 35 percent) covers physical alongside inorganic content. Paper 2 (Organic and Relevant Physical, 2h, 105 marks, 35 percent) covers physical alongside organic. Paper 3 (Practical Skills and Across-Specification, 2h, 90 marks, 30 percent) tests practical techniques, data analysis and synoptic understanding. Atomic structure and amount of substance is examined on all three papers.

Sub-topic	Spec area	Typical paper weight
Atomic models and subatomic particles	3.1.1	2-4 marks
Mass spectrometry	3.1.1	3-5 marks
Electron configuration	3.1.1	2-4 marks
Ionisation energies	3.1.1	3-6 marks
The mole concept	3.1.2	3-5 marks
Empirical and molecular formulae	3.1.2	3-5 marks
Balanced equations and reacting masses	3.1.2	4-6 marks
Solutions and concentrations	3.1.2	3-5 marks
Titration calculations (RP1)	3.1.2	4-8 marks
Ideal gas equation	3.1.2	3-5 marks

These weights are estimates based on the structure of recent 7405 papers. What is reliable is that a multi-step calculation tying together moles, concentration and either gas volume or titre — typically six to ten marks — appears on essentially every Paper 1 and Paper 3, with the same arithmetic embedded inside organic and inorganic questions on Paper 2.

Atomic Models and Subatomic Particles

An atom contains three subatomic particles: protons (relative mass 1, charge +1), neutrons (relative mass 1, charge 0) and electrons (relative mass 1/1836, charge -1). Protons and neutrons sit in the dense central nucleus; electrons occupy a much larger volume in quantised energy levels. The atomic number Z is the number of protons. The mass number A is the total of protons and neutrons. Isotopes are atoms of the same element (same Z) with different numbers of neutrons (different A).

The current model is the product of successive refinements. Dalton's hard-sphere picture gave way to Thomson's "plum pudding" once the electron was discovered; gold-foil scattering forced a nuclear model; Bohr quantisation rescued atomic spectra; quantum mechanics replaced fixed orbits with probability orbitals. You should describe the evidence at each step rather than just naming the scientists.

A common pitfall is to confuse mass number A with relative atomic mass Ar. A is an integer property of one isotope; Ar is the weighted average across the natural isotopic mixture. Another pitfall is to ignore the electron mass altogether — for most calculations this is fine, but the wording "approximately equal to" matters in definitions. See the atomic models and subatomic particles lesson.

Mass Spectrometry

The time-of-flight (TOF) mass spectrometer is the canonical AQA technique. A sample is ionised (either by electron impact or by electrospray), accelerated through a known potential difference so that all ions of the same charge gain the same kinetic energy, drifted through a field-free region, and detected when they arrive. Because KE = ½ mv^2 is fixed, lighter ions travel faster, so flight time encodes mass-to-charge ratio (m/z).

A mass spectrum is a plot of relative abundance against m/z. For an element, peaks correspond to isotopes; their positions give isotopic masses and their heights give isotopic abundances. The relative atomic mass is then a weighted average: Ar = Σ(isotopic mass × percentage abundance) / 100.

Worked example. Chlorine shows peaks at m/z 35 (75 percent) and m/z 37 (25 percent). Ar(Cl) = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 35.5. That matches the periodic table value.

A common pitfall is to confuse molecular peaks with isotope peaks. For Cl2 you would also see peaks at m/z 70, 72 and 74 in a 9:6:1 ratio. Another pitfall is to forget that electrospray ionisation produces (M+H)+, so the molecular peak appears at Mr + 1, not Mr. See the mass spectrometry lesson.

Electron Configuration

Electrons fill atomic orbitals in order of increasing energy. The relevant orbitals at A-Level are s (holds 2 electrons), p (6), d (10) and f (14). The standard filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p — note that 4s fills before 3d because it is lower in energy in neutral atoms.

The full configuration of iron (Z = 26) is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6, which can be written in shorthand as [Ar] 4s^2 3d^6. For positive transition-metal ions, 4s electrons are removed before 3d electrons, so Fe^2+ is [Ar] 3d^6 and Fe^3+ is [Ar] 3d^5.

Two exceptions matter at A-Level: chromium is [Ar] 4s^1 3d^5 and copper is [Ar] 4s^1 3d^10, because half-filled and fully-filled d subshells are particularly stable. You should know these by name.

A common pitfall is to fill 3d before 4s in neutral atoms — wrong for filling, right for ionisation. Another is to forget the chromium and copper exceptions. A third is to miscount p- and d-subshell capacities. See the electron configuration lesson.

Ionisation Energies

The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms: X(g) → X+(g) + e-. Successive ionisation energies remove successive electrons from the resulting cation. All ionisation energies are endothermic and positive.

The trend across a period is generally increasing (rising nuclear charge, similar shielding, smaller atomic radius), with two characteristic dips: between group 2 and group 13 (the new p-electron is at higher energy than the filled s) and between group 15 and group 16 (the first paired p-electron experiences extra electron-electron repulsion). Down a group, first ionisation energy decreases (more shells, greater shielding, larger radius).

Successive ionisation energies for a single element rise smoothly until a new inner shell is reached, then jump sharply. Plotting log(IE) against electron number reveals shell structure directly — for sodium you see one low value, then a jump, then eight intermediate values, then another jump, then two high values, mirroring 2,8,1.

A common pitfall is to invoke "shielding" loosely without distinguishing it from nuclear charge and orbital energy. Another is to forget the two characteristic dips across period 3. See the ionisation energies lesson.

The Mole Concept

The mole is the SI unit of amount of substance. One mole contains Avogadro's constant of entities: N_A = 6.022 × 10^23 mol^-1. The mass of one mole of a substance in grams is numerically equal to its relative formula mass: M(H2O) = 18.0 g mol^-1.

Three relationships drive most mole calculations:

• n = m / M (moles from mass and molar mass)
• n = c × V (moles from concentration and volume of solution)
• n = pV / RT (moles from gas pressure, volume and temperature)

Worked example. How many molecules are in 9.0 g of water? n = 9.0 / 18.0 = 0.50 mol. Molecules = 0.50 × 6.022 × 10^23 = 3.01 × 10^23.

A common pitfall is to confuse molar mass M (g mol^-1) with Avogadro's number (a count). Another is to forget that for solutions, V must be in dm^3 not cm^3 when c is in mol dm^-3. A third is to use the formula unit instead of the atom count when asked for "atoms" rather than "molecules" — there are three atoms per H2O molecule. See the mole concept lesson.

Empirical and Molecular Formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms in one molecule, which is always an integer multiple of the empirical formula.

The standard procedure is to take a sample (often 100 g, so percentages become masses), divide each element's mass by its atomic mass to get moles, divide by the smallest mole quantity to get a ratio, multiply if necessary to get whole numbers, and read off the empirical formula. To convert to molecular formula, divide the molecular mass by the empirical formula mass to find the multiplier.

Worked example. A hydrocarbon is 85.7 percent carbon and 14.3 percent hydrogen by mass with Mr = 56. In 100 g: 85.7 g C = 7.14 mol; 14.3 g H = 14.3 mol. Ratio C:H = 1 : 2.00. Empirical formula CH2. Empirical mass = 14. Molecular mass / empirical mass = 56 / 14 = 4. Molecular formula = C4H8.

A common pitfall is to round mole ratios prematurely — a ratio of 1 : 1.49 should suggest 2 : 3, not 1 : 1. Another is to forget that combustion analysis gives masses of CO2 and H2O, from which you must back-calculate the C and H content. See the empirical and molecular formulae lesson.

Balanced Equations and Reacting Masses

A balanced equation specifies the mole ratio in which reactants combine and products form. Given moles of any one species, you can find moles of any other from the stoichiometric coefficients.

The mass-to-mass calculation has a fixed five-step rhythm: write the balanced equation; calculate moles of the given reactant from its mass; use the mole ratio to find moles of the target; convert moles of target to mass; report to an appropriate number of significant figures.

Worked example. Calculate the mass of MgO produced from 4.86 g of Mg burning in excess O2. Balanced equation: 2Mg + O2 → 2MgO. n(Mg) = 4.86 / 24.3 = 0.200 mol. Mole ratio Mg : MgO = 1 : 1, so n(MgO) = 0.200 mol. m(MgO) = 0.200 × 40.3 = 8.06 g.

Percentage yield = (actual yield / theoretical yield) × 100. Atom economy = (Mr of desired product / Σ Mr of all products) × 100. Atom economy is a property of the equation; yield is a property of the procedure. AQA expects you to distinguish them carefully.

A common pitfall is to use mass ratios from coefficients instead of mole ratios. Another is to identify the wrong limiting reagent — compute moles of each reactant divided by its coefficient and take the smallest. See the balanced equations and reacting masses lesson.

Solutions and Concentrations

Concentration is moles of solute per unit volume of solution: c = n / V. The standard unit at A-Level is mol dm^-3 (equivalent to mol L^-1, sometimes written as M). For mass-based concentrations, g dm^-3 is also used; to convert, divide by molar mass.

To make a standard solution: weigh a known mass of solute accurately, dissolve in a small volume of solvent in a beaker, transfer quantitatively to a volumetric flask (with rinsings), and make up to the mark. The accuracy of the resulting concentration is limited by the accuracy of the mass weighing and the volumetric flask calibration.

Worked example. What mass of NaOH is needed to make 250 cm^3 of a 0.100 mol dm^-3 solution? V = 0.250 dm^3. n = c × V = 0.100 × 0.250 = 0.0250 mol. m = n × M = 0.0250 × 40.0 = 1.00 g.

A common pitfall is to mix cm^3 and dm^3 carelessly — 1 dm^3 = 1000 cm^3, and forgetting the factor of 1000 destroys an entire calculation. Another is to dissolve solute directly in a volumetric flask, where you cannot stir effectively without splashing. See the solutions and concentrations lesson.

Titration Calculations

A titration measures the volume of one solution (the titrant, in the burette) required to react exactly with a known volume of another (the analyte, in the conical flask). The end point is detected with an indicator (or a pH meter for non-coloured systems). Average concordant titres — typically agreeing within 0.10 cm^3 — give the volume used to four significant figures.

The standard calculation is again a mole-ratio problem. From the known concentration and volume of the standard solution, find moles. Use the mole ratio of the balanced equation to find moles of the unknown. Divide by its volume to get its concentration.

Worked example. 25.0 cm^3 of 0.100 mol dm^-3 NaOH neutralises 24.50 cm^3 of HCl. Find [HCl]. n(NaOH) = 0.100 × 0.0250 = 2.50 × 10^-3 mol. NaOH : HCl = 1 : 1, so n(HCl) = 2.50 × 10^-3 mol. [HCl] = 2.50 × 10^-3 / 0.02450 = 0.102 mol dm^-3.

A common pitfall is to ignore the mole ratio for diprotic systems — 25.0 cm^3 of 0.100 mol dm^-3 H2SO4 neutralises 50.0 cm^3 of 0.100 mol dm^-3 NaOH, not 25.0 cm^3. Another is to quote answers to too many significant figures; titre volumes are reliable to four, so concentrations should be reported to three. See the titration calculations lesson.

The Ideal Gas Equation and Gas Calculations

The ideal gas equation is pV = nRT, where p is pressure in pascals (Pa), V is volume in cubic metres (m^3), n is moles, R is the gas constant (8.31 J K^-1 mol^-1) and T is temperature in kelvin (K). The equation assumes negligible particle volume and no intermolecular forces — a good approximation for most gases at moderate temperatures and pressures.

Unit hygiene is the dominant source of error. Convert kPa to Pa by multiplying by 1000. Convert dm^3 to m^3 by dividing by 1000 (and cm^3 to m^3 by dividing by 10^6). Convert Celsius to kelvin by adding 273 (or 273.15 if extreme precision is asked).

Worked example. What volume does 0.500 mol of an ideal gas occupy at 100 kPa and 25 °C? p = 100 000 Pa; n = 0.500 mol; T = 298 K. V = nRT / p = 0.500 × 8.31 × 298 / 100 000 = 0.0124 m^3 = 12.4 dm^3.

The ideal gas equation can be combined with mole-from-mass to give Mr from gas measurements: Mr = mRT / (pV). This is the standard route for finding the molecular mass of a volatile liquid from vapour density data.

A common pitfall is to leave pressure in kPa or volume in dm^3. Another is to use 273 instead of 298 K for "room temperature" without thinking. See the ideal gas equation lesson.

Required Practical 1: Volumetric Solution and Acid-Base Titration

The AQA Required Practical 1 — "Making a volumetric solution and carrying out a simple acid-base titration" — is anchored on this course. RP1 combines two of the practical skills introduced here: preparing a standard solution of known concentration from a primary standard (often anhydrous sodium carbonate or potassium hydrogen phthalate), and titrating it against an acid or alkali solution of unknown concentration.

The procedure tests careful technique: rinsing all glassware with the appropriate solution; reading the burette to 0.05 cm^3; using a white tile under the conical flask for end-point sharpness; performing a rough titration followed by accurate titrations to concordance; calculating from the average concordant titre rather than including outlier rough values; and reporting concentration to three significant figures.

RP1 also generates data for several Paper 3 questions. Expect to be asked to identify the limiting source of uncertainty (usually burette readings, since two are required per run), to calculate percentage uncertainty in the final concentration, and to suggest improvements.

How to Use This Guide for Exam Preparation

• Work in specification order. Each step builds on the previous one — subatomic particles, mass spectrometry, electron configuration, ionisation energies, mole, formulae, equations, solutions, titrations, gases.
• Drill calculations to fluency. Twenty mole-mass-volume conversions per day for two weeks until the arithmetic is automatic.
• Always report to three significant figures. Two is too few; four is overclaiming.
• Mind the units. kPa to Pa, cm^3 to dm^3 to m^3, Celsius to kelvin. Write units after every number, including intermediates.
• Memorise the period-3 ionisation-energy trend with its two dips and the chromium and copper exceptions — both appear on Paper 1 in essentially every cycle.
• Use the LearningBro AI tutor on calculation lessons and time yourself on past-paper-style questions at the AQA pacing target of roughly two minutes per mark.

Linking to Other Topics

The mole concept reappears in energetics every time you convert mass of fuel into moles before multiplying by enthalpy of combustion. Titration arithmetic carries forward into acids and buffers. Concentration in mol dm^-3 is the unit of every equilibrium constant in kinetics and equilibrium. Electron configuration sets up the trends seen in inorganic chemistry and the d-block behaviour of transition metals. Time spent here pays back across the specification.

Final Word

Atomic structure and amount of substance is foundational, not preliminary. Drill the mole-mass-volume arithmetic until it is automatic. Memorise the structural details — period-3 ionisation dips, chromium and copper exceptions, RP1 procedure — cold. The full LearningBro Atomic Structure and Amount of Substance course walks through every sub-topic with worked examples and AI tutor feedback.

Related Reading

• AQA A-Level Chemistry: Energetics — uses the mole concept in every enthalpy calculation.
• AQA A-Level Chemistry: Kinetics and Equilibrium — concentration-based rate laws and equilibrium constants build directly on mol dm^-3 units.
• AQA A-Level Chemistry: Acids, Bases and Buffers — titration arithmetic and pH reasoning extend the work from this course.
• AQA A-Level Chemistry: Period 3 Inorganic Chemistry — electron configuration and ionisation energies explain the periodic trends.
• AQA A-Level Chemistry: Transition Metals — d-block configurations and oxidation states rest on the foundations laid here.