HookThe roundest objects ever made exist to count atoms
Between 2007 and 2019, metrologists ground two one-kilogram spheres of isotopically enriched silicon-28 so close to perfectly round that, scaled up to the size of the Earth, the tallest bump on either would stand only a few metres high. The polish was not vanity. Silicon crystallises in a flawlessly regular lattice, so X-ray measurements of the spacing between atoms let the team count the atoms in each sphere — about \(2.15\times10^{25}\) of them — more precisely than any balance could ever weigh them. The result rewrote the SI itself: on 20 May 2019 the Avogadro constant stopped being a measured quantity and became a defined one, fixed for ever at exactly \(6.02214076\times10^{23}\ \text{mol}^{-1}\).
Chemistry needed that number nailed down because the mole is the unit the whole subject runs on. You cannot weigh one atom, but you can weigh \(6.02\times10^{23}\) of them — a mole of carbon-12 is exactly 12 g by construction. Every calculation in 3.1.2 is the same conversion run in a different direction: mass to moles (\(n = \dfrac{m}{M}\)), solution to moles (\(n = c \times V\)), gas to moles (\(pV = nRT\)), and balanced-equation ratios to carry moles from one substance to another. Master that triangle and this becomes the most reliable section on the paper — AQA threads amount-of-substance arithmetic through physical, inorganic, organic and practical questions alike.
ModelRelative mass: everything is weighed against carbon-12
Atoms are too light for grams to be a sensible everyday unit, so chemistry uses ratios. The relative atomic mass \(A_r\) is the weighted mean mass of an atom of an element divided by one-twelfth of the mass of an atom of carbon-12; the relative molecular mass \(M_r\) is the same comparison for a molecule, and both are pure numbers with no units. Carbon-12 is the anchor because a standard had to be chosen and fixed: one atom of \({}^{12}\text{C}\) is defined as exactly 12, and every other mass in the subject is measured against it — the weighted mean over isotopes coming, as section 3.1.1 showed, straight off a mass spectrometer.
In practice you build \(M_r\) by adding \(A_r\) values from the periodic table: ethanoic acid \(\text{CH}_3\text{COOH}\) is \(12.0\times2 + 1.0\times4 + 16.0\times2 = 60.0\). For ionic compounds, which have no molecules, the honest term is relative formula mass — same arithmetic, applied to the formula unit. Hydrated salts fold in their water of crystallisation: washing soda, \(\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}\), is \(106.0 + 180.0 = 286.0\), and forgetting the ten waters is the single most common slip in RP1-style calculations.
ModelThe mole triangle: n = m/M and n = cV
One mole is the amount of substance containing \(6.02\times10^{23}\) particles — atoms, molecules, ions or electrons, so always name the particle. Its power is that one mole of anything has a mass in grams equal to its relative mass, giving the workhorse equation \(n = \dfrac{m}{M}\) (moles = mass in g ÷ molar mass in g mol\(^{-1}\)). To count particles, multiply moles by the Avogadro constant: \(N = n \times N_A\).
For solutions, concentration in mol dm\(^{-3}\) links moles to volume: \(n = c \times V\), with \(V\) in dm\(^{3}\) — divide cm\(^{3}\) by 1000 first, every time. Converting a concentration from mol dm\(^{-3}\) to g dm\(^{-3}\) is one multiplication by \(M\). Dilution questions are moles-bookkeeping: adding water changes the volume but not the moles, so \(c_1V_1 = c_2V_2\) falls straight out of \(n\) being conserved.
Two quick conversions, both single lines of the triangle. (1) How many molecules are in a 250 cm\(^{3}\) glass of water (density 1.00 g cm\(^{-3}\), so 250 g)? \(n = \dfrac{250}{18.0} = 13.9\) mol, and \(N = 13.9 \times 6.02\times10^{23} = 8.36\times10^{24}\) molecules — several hundred times more molecules in the glass than there are stars in the observable universe (around \(10^{22}\)). (2) You dissolve 5.30 g of anhydrous sodium carbonate, \(M = 106.0\) g mol\(^{-1}\), and make the solution up to 250 cm\(^{3}\): \(n = \dfrac{5.30}{106.0} = 0.0500\) mol, so \(c = \dfrac{0.0500}{0.250} = 0.200\ \text{mol dm}^{-3}\). That second calculation is, line for line, the first half of Required Practical 1.
ModelpV = nRT and the unit conversions that decide the mark
Gases refuse to be weighed conveniently, but the ideal gas equation counts them anyway: \(pV = nRT\), with the gas constant \(R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}\). Because \(R\) is fixed in SI units, everything else must arrive in SI too: pressure in pascals (multiply kPa by \(10^{3}\)), volume in cubic metres (multiply cm\(^{3}\) by \(10^{-6}\), dm\(^{3}\) by \(10^{-3}\)), temperature in kelvin (add 273 to °C). The model treats molecules as point particles with no intermolecular forces — an idealisation that real gases approach closely at the temperatures and pressures exams use.
The equation's classic A-level job is finding the relative molecular mass of a volatile liquid: vaporise a weighed sample, measure the volume, temperature and pressure of the vapour, use \(n = \dfrac{pV}{RT}\) to count the moles, then \(M = \dfrac{m}{n}\). Two measurable numbers and two ambient conditions identify an unknown compound — no spectrometer required.
A 0.116 g sample of a volatile liquid is injected into a gas syringe at 373 K and 100 kPa, where it vaporises to occupy 62.0 cm\(^{3}\). Convert first: \(p = 1.00\times10^{5}\) Pa, \(V = 6.20\times10^{-5}\) m\(^{3}\). Then \[ n = \dfrac{pV}{RT} = \dfrac{1.00\times10^{5} \times 6.20\times10^{-5}}{8.31 \times 373} = 2.00\times10^{-3}\ \text{mol} \] so \(M = \dfrac{0.116}{2.00\times10^{-3}} = 58.0\ \text{g mol}^{-1}\), matching propanone, \(\text{CH}_3\text{COCH}_3\) (\(M_r\) 58.1). Almost every dropped mark on this question type is a unit: 62.0 substituted as cm\(^{3}\), or 100 °C used instead of 373 K. Convert before you substitute, as a written line — examiners credit it.
MechanismEmpirical and molecular formulae from data
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound; the molecular formula is the actual number in one molecule, always a whole-number multiple of the empirical one. From percentage composition (or combustion masses) the routine is mechanical: divide each element's mass by its \(A_r\) to get moles, divide through by the smallest, then scale away any stubborn decimals — a ratio ending in .5 doubles, one ending in .33 triples.
To climb from empirical to molecular you need one extra fact: the relative molecular mass, which in modern practice arrives from the molecular-ion peak of a mass spectrum. Divide \(M_r\) by the empirical-formula mass to find the multiplier. The same logic finds the formula of a hydrated salt, with the mole ratio running between anhydrous salt and water driven off on heating.
An organic acid is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. In 100 g: C \(= \dfrac{40.0}{12.0} = 3.33\), H \(= \dfrac{6.7}{1.0} = 6.7\), O \(= \dfrac{53.3}{16.0} = 3.33\). Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is \(\text{CH}_2\text{O}\), formula mass 30.0. A TOF spectrum of the compound shows its molecular ion at m/z = 60, so the multiplier is \(\dfrac{60}{30} = 2\) and the molecular formula is \(\text{C}_2\text{H}_4\text{O}_2\) — ethanoic acid. The step matters because \(\text{CH}_2\text{O}\) is also the empirical formula of methanal (\(M_r\) 30) and glucose (\(M_r\) 180): the ratio narrows the field, but only the molecular mass names the compound.
CaseYield versus atom economy: how ibuprofen got greener
Balanced equations are exchange rates between substances, and reacting-mass questions just run the triangle twice: convert the given mass to moles, cross the equation on the mole ratio, convert back to mass. Real reactions then under-deliver, and percentage yield \(= \dfrac{\text{actual mass}}{\text{theoretical mass}}\times100\) measures the shortfall — lost to reversible reactions that never complete, side reactions, and product left behind in purification and transfer.
Atom economy asks a different question: of all the reactant mass fed in, how much could ever end up in the useful product? \(\text{Atom economy} = \dfrac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all reactants}}\times100\) — a design property you can compute before entering the lab. The textbook case is real. Ibuprofen was made from the 1960s by a six-step Boots synthesis with an atom economy of about 40%: most of the atoms bought as reactants left as waste. In the 1990s the BHC Company replaced it with a three-step catalytic route at roughly 77% atom economy — and its main by-product, ethanoic acid, is recovered and sold, pushing effective atom use towards 99%. The process collected a US Presidential Green Chemistry Challenge Award in 1997. Industry cares because waste is paid for twice: once as raw material coming in, again as disposal going out.
Aspirin is made from salicylic acid (\(M_r\) 138.1) and excess ethanoic anhydride; one mole of acid gives one mole of aspirin (\(M_r\) 180.2). Starting from 5.00 g of salicylic acid: \(n = \dfrac{5.00}{138.1} = 0.0362\) mol, so the theoretical mass of aspirin is \(0.0362 \times 180.2 = 6.52\) g. A student isolates 4.90 g of purified product, so \[ \text{percentage yield} = \dfrac{4.90}{6.52}\times100 = 75.1\% \] Notice what did and did not vary: the atom economy of this reaction was fixed the moment the equation was written; the 75.1% is the student's, lost in the flask, the filter paper and the recrystallisation. Yield is a performance; atom economy is a blueprint.
DataRequired Practical 1: the standard solution and the titration
RP1 has two halves. First, making a standard solution — one whose concentration is known exactly. Weigh the solid by difference (weigh the boat full, tip, reweigh the boat, subtract), dissolve in a beaker with less distilled water than the final volume, then transfer to a 250 cm\(^{3}\) volumetric flask with rinsings — every rinse of the beaker and funnel carries stray moles that belong in the flask. Fill to within a centimetre of the line, then add water dropwise until the bottom of the meniscus sits on the calibration mark at eye level. Stopper and invert slowly about ten times; swirling is not mixing.
Second, the titration. Rinse the burette with the solution it will hold (water left inside would dilute it), but rinse the conical flask only with distilled water — extra water changes nothing, extra moles change everything. Pipette a 25.0 cm\(^{3}\) aliquot, add a few drops of a single indicator (methyl orange or phenolphthalein for a strong acid–strong base pair), and titrate over a white tile: a rough run first, then accurate runs adding dropwise near the end point with constant swirling, until two titres are concordant — within 0.10 cm\(^{3}\) of each other. Average only the concordant titres. The error analysis is where AQA aims: a burette is read to ±0.05 cm\(^{3}\), and a titre is the difference of two readings, so its uncertainty is ±0.10 cm\(^{3}\); percentage uncertainty \(= \dfrac{0.10}{\text{titre}}\times100\), and the way to shrink it is a larger titre — a more dilute burette solution or a bigger aliquot. 'Human error' earns nothing; name the apparatus limit or the procedural fault.
A 25.0 cm\(^{3}\) aliquot of sodium hydroxide of unknown concentration is neutralised by a mean titre of 27.30 cm\(^{3}\) of 0.100 mol dm\(^{-3}\) hydrochloric acid. Moles of acid: \(n = 0.100 \times \dfrac{27.30}{1000} = 2.73\times10^{-3}\) mol. The equation \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\) is 1 : 1, so the aliquot held \(2.73\times10^{-3}\) mol of NaOH and \[ c = \dfrac{2.73\times10^{-3}}{0.0250} = 0.109\ \text{mol dm}^{-3} \] Now the apparatus check examiners reward: the titre's uncertainty is \(2\times0.05 = \pm0.10\) cm\(^{3}\), so its percentage uncertainty is \(\dfrac{0.10}{27.30}\times100 = 0.37\%\) — comfortably the least precise step, and the number you would quote when asked what limits the accuracy of the whole determination.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Amount-of-substance marks are method marks. Write the mole statement for every step — \(n = \dfrac{m}{M}\), \(n = cV\), \(n = \dfrac{pV}{RT}\) — substitute with units visible, and only then evaluate: a slipped final digit loses one mark, a missing method line loses them all. Give answers to three significant figures unless the data says otherwise, never fewer than the question's own precision, and put a unit on every final answer.
Structure titration calculations the way mark schemes are written: balanced equation, moles of the known solution, mole ratio, moles of the unknown, then concentration — and if the question began with a 250 cm\(^{3}\) volumetric flask and a 25.0 cm\(^{3}\) aliquot, remember the ×10 scaling step, the most commonly forgotten line in the whole topic. For gas questions the examiner is watching the conversions: kPa to Pa, cm\(^{3}\) to m\(^{3}\), °C to K, each worth showing explicitly.
Practical evaluation is AO3 territory and has fixed vocabulary: uncertainty of a burette titre is ±0.10 cm\(^{3}\) because two readings each carry ±0.05; percentage uncertainty falls when the titre is larger; results are reliable when concordant within 0.10 cm\(^{3}\); and 'human error' scores zero — name the apparatus limit or the specific procedural fault. Section 3.1.2 is examined on Papers 1, 2 and 3, and at least 20% of A-level Chemistry marks are Level 2 mathematics: this section is where most of them live, which is exactly why it repays over-practice.