HookThe reaction that has to finish in 40 milliseconds — or you go through the windscreen
When a car hits a wall at 30 mph, your head is thrown forward faster than you can blink. The airbag has about 40 milliseconds to be fully inflated before your face arrives — and it inflates not with a pump but with a chemical reaction. A pellet of sodium azide is detonated and decomposes, \(2\text{NaN}_3 \rightarrow 2\text{Na} + 3\text{N}_2\), flooding the bag with nitrogen gas. Here is the part that keeps engineers up at night: the mass of that pellet has to be exactly right. Weigh out too little solid and the bag inflates only halfway; too much and it bursts or hits you like a fist. The whole safety system rests on a calculation — a known mass of solid must produce a known volume of gas — done long before the crash, on a balance, using nothing more than the ideas in this section.
That is what quantitative chemistry is: chemistry you can count. It all hangs off one law — conservation of mass — because atoms are only ever rearranged in a reaction, never made or destroyed. From that single fact flow all the tools in C3: balancing equations, relative formula mass, working out when mass only appears to change, handling real measurements and their uncertainty, and — on the Higher tier — the mole, reacting masses, limiting reactants and concentration. It is also the most method-hungry section on the paper: the marks live in the working, not the final number, so you write every line.
ModelConservation of mass and balancing equations
Atoms are neither created nor destroyed in a chemical reaction — they are just rearranged — so the total mass of the products always equals the total mass of the reactants. Written as an equation, that means every element must have the same number of atoms on each side. You balance by putting large multiplying numbers in front of the formulae, and you must never change a subscript inside a formula to force a balance, because that turns the substance into something else entirely.
The same bookkeeping, done for electrons rather than atoms, gives a half equation — a shorthand for what happens at an electrode. When an aluminium atom forms its ion it loses three electrons, \(\text{Al} \rightarrow \text{Al}^{3+} + 3e^-\), and the charges as well as the atoms must balance across the arrow. Whether you are counting atoms or electrons, the rule is the same: what goes in must come out.
Balance the complete combustion of propane, which starts unbalanced as \(\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). Work through the elements in a fixed order. Carbon first: there are 3 carbon atoms on the left, so you need \(3\text{CO}_2\). Hydrogen next: 8 hydrogen atoms means \(4\text{H}_2\text{O}\). Now count the oxygen on the right — \((3\times2) + (4\times1) = 10\) atoms — so you need \(5\text{O}_2\) on the left. The balanced equation is \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \] Final check: 3 carbon, 8 hydrogen and 10 oxygen on each side. Notice you balanced by adding numbers in front — never by rewriting \(\text{H}_2\text{O}\) as \(\text{H}_2\text{O}_2\), which would be hydrogen peroxide, a completely different chemical.
ModelRelative formula mass
The relative formula mass \(M_r\) of a substance is the sum of the relative atomic masses \(A_r\) of every atom shown in its formula. The one trap that catches people is brackets: a subscript sitting outside a bracket multiplies everything inside it, so you have to expand the bracket before you add. Because mass is conserved, the total \(M_r\) of the reactants in a balanced equation equals the total \(M_r\) of the products — a fast sanity check that both your equation and your arithmetic are right.
Find the relative formula mass of ammonium sulfate, \((\text{NH}_4)_2\text{SO}_4\), using \(A_r\) values N \(= 14\), H \(= 1\), S \(= 32\), O \(= 16\). The subscript \(2\) applies to the whole \(\text{NH}_4\) bracket, so that part is \(2\times(14 + 4\times1) = 2\times18 = 36\). Then add the sulfate: \(S = 32\) and \(4\times O = 64\). Altogether \[ M_r = 2\times(14 + 4) + 32 + (4\times16) = 36 + 32 + 64 = 132 \] The classic error is to forget the bracket and count only one nitrogen and four hydrogens, giving \(14 + 4 + 32 + 64 = 114\). Doubling the whole \(\text{NH}_4\) group is exactly what separates the right answer from the wrong one.
MechanismWhen mass seems to change — reactions with a gas
If a reaction is open to the air, its measured mass can appear to change, and students wrongly conclude that mass has been made or lost. There are two cases, and in both mass is really conserved. Mass appears to increase when a substance combines with a gas from the air — a strip of magnesium or a wad of iron wool gains mass as it reacts with oxygen, because that oxygen was never sitting on the balance to start with.
Mass appears to decrease when a reaction gives off a gas that escapes — a metal carbonate decomposing, or a fizzing mixture of acid and carbonate on an open balance, leaves less behind because carbon dioxide has drifted away. If you could trap every gas that crossed the boundary of the apparatus, the total mass would not move at all. The apparent change is simply the mass of gas entering or leaving the system.
Burn \(2.4\) g of magnesium ribbon completely in air: \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\), with \(A_r\) Mg \(= 24\), O \(= 16\). The moles of magnesium are \(\frac{2.4}{24} = 0.10\) mol, so \(0.10\) mol of magnesium oxide forms, of mass \(0.10 \times 40 = 4.0\) g. The solid in the crucible has gained \(4.0 - 2.4 = 1.6\) g. Nothing was created: that \(1.6\) g is the mass of oxygen from the air (\(0.05\) mol of \(\text{O}_2\)) that has joined the magnesium. Weigh the crucible before and after and it looks as though mass appeared from nowhere — until you account for the invisible gas.
DataChemical measurements — repeats, means and uncertainty
Real readings scatter, so good practice is to repeat a measurement and take the mean of the concordant results — the ones that agree closely — after discarding any anomaly that clearly breaks the pattern. The resolution of an instrument is the smallest change it can register; a balance reading to \(0.01\) g has a finer resolution than one reading to \(0.1\) g. The uncertainty in a mean is estimated as plus or minus half the range of the repeats, and you should quote answers to a sensible number of significant figures — never more precise than your least precise measurement allows.
A carbonate reacts with acid on a top-pan balance and you record the mass of carbon dioxide lost four times: \(0.44\), \(0.45\), \(0.43\) and \(0.52\) g. The reading of \(0.52\) g is an anomaly — it sits well outside the cluster — so you leave it out. The mean of the three concordant values is \[ \frac{0.44 + 0.45 + 0.43}{3} = \frac{1.32}{3} = 0.44\ \text{g} \] Their range is \(0.45 - 0.43 = 0.02\) g, so the uncertainty is \(\pm 0.01\) g and you report \(0.44 \pm 0.01\) g. Quoting \(0.4400\) g would be dishonest — the balance simply cannot resolve that many figures.
ModelThe mole — counting atoms by weighing (Higher tier)
Atoms are far too small and too many to count one at a time, so chemists count them in moles. One mole is \(6.02\times10^{23}\) particles — the Avogadro constant — and, crucially, one mole of any substance has a mass in grams equal to its relative formula mass. That gives the single most useful relationship in the whole section, \(\text{mass} = M_r \times \text{moles}\), which rearranges to \(\text{moles} = \dfrac{\text{mass}}{M_r}\). It is the bridge between the grams you can weigh and the enormous number of particles you can never see.
How many moles are there in \(36\) g of water, and how many molecules is that? The relative formula mass of \(\text{H}_2\text{O}\) is \(2\times1 + 16 = 18\), so \[ \text{moles} = \frac{\text{mass}}{M_r} = \frac{36}{18} = 2\ \text{mol} \] To turn moles into a particle count, multiply by the Avogadro constant: \(2 \times 6.02\times10^{23} = 1.20\times10^{24}\) molecules. Two tablespoons of water contains more than a trillion trillion molecules — and you counted them simply by reading a balance.
MechanismReacting masses and balancing from moles (Higher tier)
The big numbers in a balanced equation are a mole ratio — they tell you how many moles of each substance take part. So you can predict the mass of a product from the mass of a reactant by routing through moles: convert the given mass to moles, apply the ratio, then convert the answer back to a mass. The same ratio works in reverse, letting you deduce the balancing numbers of an equation from the masses that actually reacted — find the moles of each substance and simplify the ratio to whole numbers.
Sodium burns in oxygen and you measure that \(9.2\) g of sodium reacts with \(3.2\) g of oxygen. Deduce the equation, then the mass of sodium oxide formed. Moles of Na \(= \frac{9.2}{23} = 0.40\) mol; moles of \(\text{O}_2 = \frac{3.2}{32} = 0.10\) mol. The ratio Na \(:\ \text{O}_2\) is \(0.40 : 0.10 = 4 : 1\), which gives \[ 4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \] From the ratio, \(0.40\) mol of sodium makes \(0.20\) mol of \(\text{Na}_2\text{O}\), and its \(M_r\) is \((2\times23) + 16 = 62\), so the mass is \(0.20 \times 62 = 12.4\) g. Conservation of mass confirms it: \(9.2 + 3.2 = 12.4\) g in, \(12.4\) g out.
MechanismLimiting reactants — the one that runs out first (Higher tier)
When two reactants are mixed, one of them is usually used up before the other. That one is the limiting reactant, and it sets how much product can form; whatever is left over of the second reactant is said to be in excess and simply does nothing once its partner has run out. The method never varies: convert both masses to moles, use the mole ratio to see which reactant runs out first, and then base every amount of product on the limiting reactant alone.
Heat \(5.6\) g of iron with \(4.8\) g of sulfur to make iron sulfide, \(\text{Fe} + \text{S} \rightarrow \text{FeS}\), with \(A_r\) Fe \(= 56\), S \(= 32\). Moles of iron \(= \frac{5.6}{56} = 0.10\) mol; moles of sulfur \(= \frac{4.8}{32} = 0.15\) mol. The equation needs them in a \(1 : 1\) ratio, so \(0.10\) mol of iron reacts with only \(0.10\) mol of sulfur — the iron is the limiting reactant and \(0.05\) mol of sulfur (\(1.6\) g) is left in excess. The product is therefore \(0.10\) mol of FeS, and with \(M_r = 56 + 32 = 88\) its mass is \(0.10 \times 88 = 8.8\) g. Identify the limiting reactant before anything else, then never look at the excess again.
ModelConcentration of solutions
Concentration tells you how much solute is dissolved in a given volume of solution. On this course it is measured in grams per cubic decimetre: the mass of solute divided by the volume in \(\text{dm}^3\), where \(1\ \text{dm}^3 = 1000\ \text{cm}^3\). The one habit that prevents most mistakes is converting the volume from \(\text{cm}^3\) to \(\text{dm}^3\) — dividing by \(1000\) — before you divide. A more concentrated solution has more solute packed into the same volume, which is why it looks darker or tastes stronger.
Dissolve \(15\) g of potassium nitrate to make \(500\ \text{cm}^3\) of solution. First convert the volume: \(500\ \text{cm}^3 = \frac{500}{1000} = 0.5\ \text{dm}^3\). Then \[ \text{concentration} = \frac{\text{mass}}{\text{volume}} = \frac{15}{0.5} = 30\ \text{g/dm}^3 \] The equation runs backwards just as easily: how much potassium nitrate is in \(250\ \text{cm}^3\) of that same \(30\ \text{g/dm}^3\) solution? Mass \(= \text{concentration} \times \text{volume} = 30 \times 0.25 = 7.5\) g. Halve the volume and you halve the mass of dissolved solid — the concentration stays the same because it is a ratio.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This is the section where method earns the marks, and the AQA mark scheme rewards it line by line: a typical calculation gives a mark for the relationship or balanced equation, a mark for the correct substitution, and a mark for the final answer with its unit. Write the relationship first (\(\text{moles} = \dfrac{\text{mass}}{M_r}\), \(\text{concentration} = \dfrac{\text{mass}}{\text{volume}}\)), substitute with units, then state the answer — a bare number scores little even when it is right. At least a fifth of the marks in GCSE Chemistry are for maths skills, and most of them are cashed out here.
Across AO1 and AO2, convert units before you calculate — \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by \(1000\) — and, on any reacting-mass or limiting-reactant question, go via moles every single time: mass to moles, apply the ratio, moles back to mass, identifying the limiting reactant before the final step. For AO3 'evaluate the method' questions, name the resolution of the instrument, discard the anomaly explicitly, average only the concordant repeats and quote the uncertainty as half the range. Finally, sanity-check the size of your result: a product heavier than the reactants, or an answer with more significant figures than your data, is a signal you have slipped — and the interpretation is the mark candidates most often drop.