HookThe chemist who weighed the world — and lost his head
In the 1770s and 1780s, in Paris, Antoine Lavoisier did something chemists before him had barely thought to do: he weighed everything, precisely, in sealed vessels. When he heated mercury in a closed flask of air, the total mass never changed, even as a red powder formed and part of the air was used up. Reaction after reaction told the same story — the mass of the products always equalled the mass of the reactants. That overturned centuries of vague "phlogiston" thinking and founded quantitative chemistry: chemistry you can count. Lavoisier was guillotined in 1794 during the Revolution; a judge is said to have remarked that "the Republic has no need of scientists."
Conservation of mass is the hinge for the whole of C3. Because atoms are only ever rearranged, a balanced equation must have equal numbers of atoms on each side, and the mole lets you count those atoms simply by weighing. From that one idea flow all the tools in this section — balancing, relative formula mass, moles, reacting masses, limiting reactants, concentration, percentage yield, atom economy and gas volumes. It is also the section that rewards discipline: show every line of working, because the method carries most of the marks.
ModelConservation of mass and balancing equations
In a chemical reaction, atoms are not created or destroyed — they are simply rearranged — so the total mass of the reactants equals the total mass of the products, and every element must have the same number of atoms on both sides of the equation. To balance an equation you place large numbers (multipliers) in front of the formulae. You must never change a formula's subscripts to make it balance, because that changes the substance itself.
Half equations do the same bookkeeping for electrons at electrodes, showing electrons lost or gained — for example, a sodium atom losing one electron, \(\text{Na} \rightarrow \text{Na}^{+} + \text{e}^{-}\). Whether you are balancing atoms or electrons, the principle is identical: what goes in must come out.
Balance the combustion of methane, which starts unbalanced as \(\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). Balance hydrogen first: there are 4 H atoms on the left, so you need \(2\text{H}_2\text{O}\) on the right. That right-hand side now holds \(2\) oxygen atoms in the carbon dioxide and \(2\) in the water, \(4\) in total, so you need \(2\text{O}_2\) on the left. The balanced equation is \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] Check every element: 1 carbon each side, 4 hydrogen each side, 4 oxygen each side. Note that 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 a different chemical entirely.
ModelRelative formula mass
The relative formula mass \(M_r\) of a compound is the sum of the relative atomic masses \(A_r\) of every atom in its formula. The only real trap is brackets: a subscript outside a bracket multiplies everything inside it. Because mass is conserved, the total \(M_r\) of the reactants in a balanced equation equals the total \(M_r\) of the products — a quick check that your equation and your arithmetic agree.
Find the relative formula mass of calcium hydroxide, \(\text{Ca(OH)}_2\), using \(A_r\) values Ca \(= 40\), O \(= 16\), H \(= 1\). The subscript \(2\) applies to the whole bracket, so you have one calcium plus two lots of (one oxygen and one hydrogen): \[ M_r = 40 + 2\times(16 + 1) = 40 + 34 = 74 \] For comparison, calcium carbonate \(\text{CaCO}_3\) is \(40 + 12 + (3\times16) = 100\). The classic error is to ignore the bracket and write \(40 + 16 + 1 = 57\); multiplying the whole bracket by \(2\) is what makes the difference.
MechanismWhen mass seems to change — reactions involving gases
If a reaction is not sealed, its mass can appear to change, which trips up students who think mass has been created or destroyed. There are two cases. Mass seems to increase when a substance combines with a gas from the air: a piece of magnesium or steel wool gains mass as it reacts with oxygen, because that oxygen was not sitting on the balance to begin with.
Mass seems to decrease when a reaction gives off a gas that escapes: a metal carbonate breaking down, or a candle burning, leaves less behind because a gaseous product has drifted away. In both cases mass is really conserved — if you trapped every gas involved, the total would not change at all. The apparent change is just the mass of a gas that crossed the boundary of your apparatus.
Heat \(100\) g of calcium carbonate until it fully decomposes: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\). The solid left in the crucible (calcium oxide) weighs only \(56\) g, so it looks as though \(44\) g has vanished. It has not: \(44\) g of carbon dioxide has escaped as a gas, and \(56 + 44 = 100\) g — exactly the mass you started with. Account for the gas and conservation of mass holds perfectly; ignore it and you "lose" \(44\) g that were never lost.
DataChemical measurements — repeats, means and uncertainty
Real measurements scatter, so good practice is to repeat a reading and take the mean of the concordant results — the ones that agree closely — after discarding any anomaly that clearly does not fit the pattern. The resolution of an instrument is the smallest change it can detect. The uncertainty in a mean can be estimated as plus or minus half the range of the repeats, and answers should be quoted to a sensible number of significant figures — no more precise than your least precise measurement allows.
Three repeats of a temperature rise give \(8.5\), \(8.7\) and \(8.9\)°C. The mean is \[ \frac{8.5 + 8.7 + 8.9}{3} = \frac{26.1}{3} = 8.7\ \text{°C} \] The range is \(8.9 - 8.5 = 0.4\), so the uncertainty is \(\pm 0.2\)°C, and you report the result as \(8.7 \pm 0.2\)°C. If a fourth reading of \(9.9\)°C had appeared, you would treat it as an anomaly and leave it out of the mean rather than let one stray value distort the average.
ModelThe mole — counting atoms by weighing
Atoms are far too small and too numerous to count one by one, 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 equation in the section: \(\text{mass} = M_r \times \text{moles}\), or rearranged, \(\text{moles} = \dfrac{\text{mass}}{M_r}\). It is the bridge between the grams you can weigh on a balance and the vast number of atoms you cannot see.
How many moles are there in \(80\) g of sodium hydroxide, and how many formula units is that? First the relative formula mass of \(\text{NaOH}\) is \(23 + 16 + 1 = 40\). Then \[ \text{moles} = \frac{\text{mass}}{M_r} = \frac{80}{40} = 2\ \text{mol} \] To turn moles into a particle count, multiply by the Avogadro constant: \(2 \times 6.02\times10^{23} = 1.20\times10^{24}\) formula units. Weighing out \(80\) g has let you count more than a trillion trillion particles — that is the power of the mole.
MechanismReacting masses, balancing from moles and limiting reactants
The big numbers in a balanced equation are a mole ratio: they tell you how many moles of each substance react. So you can predict the mass of a product from the mass of a reactant by going through moles — convert mass to moles, apply the ratio, convert moles back to mass. The same idea lets you deduce balancing numbers: find the moles of each substance that actually reacted and simplify the ratio. When two reactants are mixed, one usually runs out first — the limiting reactant — and it controls how much product forms, while the other is left in excess.
For example, if \(4.8\) g of magnesium (\(0.2\) mol) reacts with \(3.2\) g of oxygen (\(0.1\) mol of \(\text{O}_2\)), the ratio \(0.2 : 0.1\) simplifies to \(2 : 1\), which gives the balanced equation \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\). The mole ratio is not just for balancing — it drives every reacting-mass calculation.
Burn \(4.8\) g of magnesium fully in oxygen: \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\), with \(A_r\) Mg \(= 24\), O \(= 16\). Moles of Mg \(= \frac{4.8}{24} = 0.2\) mol; the \(2 : 2\) ratio means \(0.2\) mol of MgO forms, so its mass is \(0.2 \times 40 = 8.0\) g. Now suppose only \(0.05\) mol of oxygen is available. The equation needs \(2\) mol of Mg for every \(1\) mol of \(\text{O}_2\), so \(0.05\) mol of \(\text{O}_2\) can react with just \(0.10\) mol of Mg — the oxygen is the limiting reactant, \(0.10\) mol of Mg is left over, and the product is \(2 \times 0.05 = 0.10\) mol of MgO, which is \(4.0\) g. Always identify the limiting reactant first, then base every amount on it.
ModelConcentration of solutions
Concentration measures how much solute is dissolved in a given volume of solution. In grams per cubic decimetre it is the mass of solute divided by the volume in \(\text{dm}^3\), remembering that \(1\ \text{dm}^3 = 1000\ \text{cm}^3\). In moles per cubic decimetre (Higher tier) it is the moles of solute divided by the volume in \(\text{dm}^3\), so \(c = \dfrac{n}{V}\); you convert between the two using the relative formula mass. Rearranged as \(n = c \times V\), this is the equation that unlocks reacting volumes in a titration.
Dissolve \(20\) g of sodium chloride to make \(250\ \text{cm}^3\) of solution. The volume is \(\frac{250}{1000} = 0.25\ \text{dm}^3\), so \[ \text{concentration} = \frac{20}{0.25} = 80\ \text{g/dm}^3 \] Higher-tier extension: a titration shows that \(25.0\ \text{cm}^3\) of \(0.100\ \text{mol/dm}^3\) sodium hydroxide is exactly neutralised by \(20.0\ \text{cm}^3\) of hydrochloric acid. Moles of NaOH \(= 0.100 \times 0.0250 = 2.5\times10^{-3}\) mol, and the \(1 : 1\) reaction means the moles of HCl are the same, so \[ c(\text{HCl}) = \frac{2.5\times10^{-3}}{0.0200} = 0.125\ \text{mol/dm}^3 \] The key habit is converting \(\text{cm}^3\) to \(\text{dm}^3\) before you divide.
ModelPercentage yield and atom economy
Percentage yield compares how much product you actually made with the theoretical maximum from the balanced equation: \(\text{percentage yield} = \dfrac{\text{actual mass}}{\text{theoretical mass}}\times100\). It can never exceed \(100\%\); yield is lost to reactions that do not go to completion, reversible reactions, side reactions, and product left behind when transferring or filtering.
Atom economy measures something different — how much of the reactant mass ends up as the useful product: \(\text{atom economy} = \dfrac{M_r\text{ of desired product}}{\text{total }M_r\text{ of reactants}}\times100\). A high atom economy means less waste, which is both more sustainable and usually more profitable. A reaction can have a high yield but a low atom economy if much of the starting mass leaves as by-products.
Iron is extracted in a blast furnace by \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). The desired product is iron, of mass \(2\times56 = 112\). The total \(M_r\) of the reactants is \(\text{Fe}_2\text{O}_3 = (2\times56)+(3\times16) = 160\) plus \(3\text{CO} = 3\times28 = 84\), giving \(244\). So \[ \text{atom economy} = \frac{112}{244}\times100 = 45.9\% \] Over half the reactant mass leaves as carbon dioxide — a real driver behind the search for cleaner extraction. Separately, if a process should give \(80\) g of product but yields only \(60\) g, its percentage yield is \(\frac{60}{80}\times100 = 75\%\). The two numbers answer different questions: yield is how much you got, atom economy is how little you wasted.
ModelGases by the litre — molar gas volume
Equal volumes of different gases at the same temperature and pressure contain equal numbers of moles. At room temperature and pressure (rtp), one mole of any gas occupies \(24\ \text{dm}^3\) (that is \(24{,}000\ \text{cm}^3\)). So the volume of a gas is its moles multiplied by \(24\), and its moles are its volume divided by \(24\). This lets you predict gas volumes straight from a balanced equation.
What volume of carbon dioxide, at rtp, is produced when \(50\) g of calcium carbonate fully decomposes by \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)? The relative formula mass of \(\text{CaCO}_3\) is \(100\), so the moles are \(\frac{50}{100} = 0.5\) mol, and the \(1 : 1\) ratio gives \(0.5\) mol of carbon dioxide. Then \[ V = 0.5 \times 24 = 12\ \text{dm}^3 \] Half a mole of gas fills \(12\ \text{dm}^3\) — twelve litres from fifty grams of powder, calculated before you run the reaction.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This is the section where working earns the marks. For every calculation, write the relationship first (\(\text{mass} = M_r \times \text{moles}\), \(c = \dfrac{n}{V}\)), substitute the numbers with their units, then give the answer with a unit — a bare number scores little even when it is correct. Convert units before you calculate: \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by \(1000\), and always check whether the question wants \(\text{g/dm}^3\) or \(\text{mol/dm}^3\).
On reacting-mass and limiting-reactant questions, go via moles every single time — mass to moles, apply the ratio, moles back to mass — and identify the limiting reactant before the final step so you base the answer only on it. Quote answers to a sensible number of significant figures and sanity-check the size of your result: a yield above \(100\%\), or a product mass greater than the reactants, is a signal you have slipped. The Higher-tier calculations — moles, concentrations in \(\text{mol/dm}^3\), and gas volumes — carry the heaviest marks in this topic, so practise them until each is a single clean line of substitution.