1The idea in one paragraph

A balanced equation is a recipe written in particles. The big numbers in front of each formula, the coefficients, say how many particles of each substance react with how many of the others, and because a mole is a fixed number of particles, they give the ratio in moles too. Every calculation then makes the same three moves: turn what you measured into moles, use the ratio, and turn the answer back. On top of that sit three questions about any reaction: which reactant runs out first, how much of the possible product was collected, and how much of the reactants' mass could ever have become product.

2Writing the equation

A chemical equation shows the reactants and products with their formulas, the arrow, and the ratio in which they react. Deduce it in three steps.

Write the correct formula of every substance first. Formulas are fixed by bonding, not by balancing: aluminium chloride is AlCl₃ because Al³⁺ needs three Cl⁻ ions. Once a formula is right you never touch it again.

Balance by changing only the coefficients. Count each element on both sides and adjust the numbers in front. Leave elements that appear in only one substance on each side until the end, and deal with oxygen and hydrogen last in combustion. Figure 1 counts the atoms in the combustion of methane: one carbon, four hydrogen and four oxygen atoms on each side.

Add state symbols. (s) solid, (l) liquid, (g) gas, (aq) dissolved in water. A missing one costs the mark when the question asks for them. Water formed at room temperature is (l), not (aq); an acid or salt in solution is (aq), not (l).

Work one from words. Aluminium reacts with hydrochloric acid to give aluminium chloride solution and hydrogen gas.

Use the lowest whole-number ratio unless the question asks for something else. For some reactions only the particles that change are shown. When barium chloride solution meets sodium sulfate solution, the sodium and chloride ions stay in solution unchanged, so the ionic equation is Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). The same idea written for electrons is the half-equation you meet in Reactivity 3.2.

3What the coefficients mean, and the road map

In 2Al + 6HCl → 2AlCl₃ + 3H₂, the coefficients say that 2 aluminium atoms react with 6 HCl to give 2 AlCl₃ and 3 H₂. Multiply every term by the Avogadro constant and the statement becomes: 2 mol of Al react with 6 mol of HCl to give 2 mol of AlCl₃ and 3 mol of H₂. That is the mole ratio, 2 : 6 : 2 : 3, and it is the only thing the equation tells you directly.

It does not give a ratio of masses: two grams of aluminium do not react with six grams of acid. So every calculation runs along the road in Figure 2.

Given quantity → moles of A → × (coefficient of B ÷ coefficient of A) → moles of B → quantity wanted

The ratio is always wanted over given: for 2Al + 6HCl, moles of HCl needed = moles of Al × 6/2. Writing it upside down is the commonest error here, and easy to catch: should the answer be more moles or fewer?

4Reacting masses

In a blast furnace, carbon monoxide reduces iron(III) oxide to iron:

Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)

Worked example 1. Calculate the mass of iron that can be made from 80.0 g of iron(III) oxide.

Every line names its substance, intermediate values keep an extra figure, and only the final answer is rounded to the precision of the data. An examiner can award method marks for any line they can follow, even if the last line is wrong.

5Gas volumes

Avogadro's law says equal volumes of gases at the same temperature and pressure contain equal numbers of particles. So, for gases measured under the same conditions, the ratio of volumes is the same as the ratio of moles, and the equation can be read directly in volumes. Figure 3 does this for propane burning.

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

2.00 dm³ of propane needs 5 × 2.00 = 10.0 dm³ of oxygen and gives 3 × 2.00 = 6.00 dm³ of carbon dioxide, all at the same temperature and pressure. No moles needed. The water is a liquid at room temperature, with almost no volume compared with a gas, so it drops out of the volume sum: another reason state symbols matter.

When you need to move between a gas volume and a mass, use the molar volume of an ideal gas. At STP (273 K and 100 kPa), a typical value is 22.7 dm³ mol⁻¹; use the figure printed in your data booklet.

Worked example 2. Calculate the volume of carbon dioxide, at STP, produced when 11.0 g of propane burns completely.

The molar volume changes with temperature and pressure, which is Structure 1.5. At any conditions other than STP use pV = nRT, with p in Pa, V in m³ and T in K.

Avogadro's law describes without explaining. It was proposed from measured combining volumes, long before anyone knew why it held. The explanation came later from the kinetic model: gas particles are so far apart that the space a gas fills depends on how many there are, not on their size.

6Solutions and titrations

For a reaction in solution, the amount comes from concentration and volume:

n = c × V, with c in mol dm⁻³ and V in dm³ (divide cm³ by 1000)

A titration measures the volume of one solution that exactly reacts with a known volume of another. One is a standard solution, of accurately known concentration, and the result gives the concentration of the other.

Worked example 3. 25.00 cm³ of sodium hydroxide solution is exactly neutralised by 18.40 cm³ of 0.100 mol dm⁻³ sulfuric acid. Calculate the concentration of the sodium hydroxide.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Start from the solution with both a concentration and a volume. The trap is the ratio: sulfuric acid releases two H⁺ per formula unit, so it needs two NaOH, and assuming 1 : 1 halves the answer.

Concentration is sometimes given in g dm⁻³. Divide by the molar mass to get mol dm⁻³ before you use n = c × V.

7The limiting reactant

Reactants are rarely mixed in exactly the ratio of the equation. The one that is used up first stops the reaction, and it is called the limiting reactant. Anything left over is in excess. Figure 4 shows it with particles. In N₂ + 3H₂ → 2NH₃, six hydrogen molecules can react with only two nitrogen molecules; the third nitrogen molecule has nothing to react with. Hydrogen runs out first, so it limits the amount of ammonia.

To find the limiting reactant from data, convert each reactant to moles and divide each by its own coefficient. The smallest result is the limiting reactant. Comparing moles without dividing by the coefficients is the mistake that loses this mark.

Worked example 4. 5.00 g of aluminium is heated in 10.0 g of chlorine.

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

(a) Deduce the limiting reactant. (b) Calculate the maximum mass of aluminium chloride. (c) Calculate the mass of the excess reactant left over.

Chlorine limits the product even though there are fewer grams of aluminium, because each aluminium atom needs one and a half chlorine molecules. The greater mass is never a guide.

A graph can show the limiting reactant change over. Figure 5 comes from adding barium chloride solution a little at a time to a fixed amount of sodium sulfate solution and weighing the dried precipitate. At first every drop of barium chloride reacts, so barium chloride is limiting and the mass rises in a straight line. At 10.0 cm³ the sulfate is used up; from then on sulfate is limiting and extra barium chloride makes nothing. The turning point is where the reactants were in exactly the equation's ratio, which is how an experiment like this can find an unknown concentration.

Check it: 25.0 cm³ of 0.200 mol dm⁻³ sulfate is 5.00 × 10⁻³ mol, which needs 10.0 cm³ of 0.500 mol dm⁻³ Ba²⁺ and gives 5.00 × 10⁻³ × 233.40 = 1.17 g of BaSO₄.

The same idea explains a danger in combustion (Reactivity 1.3). With too little air, oxygen is the limiting reactant, so not all the carbon can become CO₂: carbon monoxide and soot form instead. Carbon monoxide binds to haemoglobin more strongly than oxygen does, which is why a heater in a poorly ventilated room can kill.

8Theoretical, experimental and percentage yield

The theoretical yield is the mass of product calculated from the limiting reactant, assuming the reaction goes to completion and nothing is lost. The experimental yield is the mass you actually collect. The percentage yield compares them.

percentage yield = (experimental yield ÷ theoretical yield) × 100

Worked example 5. Ethyl ethanoate, a solvent, is made by heating ethanoic acid with excess ethanol and an acid catalyst.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

6.00 g of ethanoic acid gives 5.28 g of pure ester. Calculate the percentage yield.

Figure 6 shows where the missing 3.52 g could have gone. The split is invented for illustration, but each kind of loss is real.

Why the experimental yield is usually lower:

• The reaction does not go to completion. This one is reversible (the ⇌ sign), so it stops at an equilibrium mixture that still contains acid and alcohol. That is Reactivity 2.3.
• Side reactions turn some of the reactant into a different product.
• Mechanical losses: product left on glassware, filter paper and the stirring rod, or lost when a liquid is poured.
• Purification losses: some product stays dissolved when crystals are filtered, or is removed with the impurities in distillation.
• Reactants were impure, so less of the limiting reactant was present than the mass suggests.

Why it can come out higher than 100%, which should never happen with a pure, dry product:

• The product is still wet, with water or solvent, because it was not dried to constant mass.
• The product contains impurities: unreacted excess reactant, or a by-product that came out of solution with it.

Both are systematic errors: repeating the experiment repeats them. The fix is in the method: dry to constant mass, wash the crystals, recrystallize.

9Atom economy

Percentage yield asks how much of the possible product you collected. Atom economy asks a different question: even if every step worked perfectly, what fraction of the mass of the reactants could end up in the product you want?

atom economy = (molar mass of desired product ÷ total molar mass of all reactants) × 100

The equation is in the data booklet. Use the coefficients from the balanced equation: if two moles of product form, count two molar masses on top.

Figure 7 compares two ways of making ethanol.

Worked example 6.

Hydration of ethene: C₂H₄(g) + H₂O(g) → C₂H₅OH(g)

Fermentation of glucose: C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

A reaction with a single product puts every reactant atom into it, which is why every addition reaction, including addition polymerization, has an atom economy of 100%. Fermentation cannot beat 51.1%, whatever the yield, because almost half the mass leaves as carbon dioxide.

Atom economy and waste are inversely related. A low atom economy means a large mass of by-product for every kilogram of product, and in industry every tonne of by-product has to be separated, stored, treated or disposed of. Making iron in the blast furnace, Fe₂O₃ + 3CO → 2Fe + 3CO₂, has an atom economy of only 45.8%, and the rest is carbon dioxide released on an enormous scale. Green chemistry aims for high atom economy because the cheapest waste to deal with is waste that is never made.

Atom economy and yield are not the same measure, and neither is the whole story. A green chemist also asks:

• how much energy the process needs, and where it comes from;
• whether the by-product is useful: fermentation's CO₂ can be sold, and a by-product that is sold is not waste;
• whether the feedstock is renewable: glucose from crops versus ethene from crude oil;
• how hazardous the reactants, solvents and products are;
• how fast the reaction is, and whether a catalyst can be reused (Reactivity 2.2).

That list is what a "discuss the efficiency of a process" question is testing. On it, fermentation can still be the greener route to ethanol despite its low atom economy.