Stoichiometry

Formulae & balanced equations

Chemistry has its own shorthand. A formula tells you what atoms a substance contains; a balanced equation tells you exactly what reacts with what, and in what ratio.

Chemical formulae

A chemical formula uses element symbols with subscripts to show how many atoms of each are in one unit of the substance. Water is H₂O: two hydrogen atoms, one oxygen. Sulfuric acid is H₂SO₄: two hydrogens, one sulfur, four oxygens.

Seven elements always travel in pairs — the diatomic elements: hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine. Their formulae are H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂. Writing them without the subscript is a quick mark lost.

Balanced symbol equations

A symbol equation describes a reaction. The substances that react sit on the left, the products on the right, with an arrow between. To balance an equation, place coefficients — numbers in front of each formula — so that each kind of atom appears the same number of times on both sides. Subscripts inside a formula must never be changed; that would change the substance itself.

Worked example: balancing an equation

Iron reacts with chlorine to form iron(III) chloride — balance Fe + Cl₂ → FeCl₃. Count atoms: the left has 1 Fe and 2 Cl, the right has 1 Fe and 3 Cl, so chlorine does not balance. The lowest common multiple of 2 and 3 is 6, so use 3 Cl₂ on the left and 2 FeCl₃ on the right. Rechecking iron, the right now has 2 Fe, so put a 2 in front of Fe. The balanced equation is 2 Fe + 3 Cl₂ → 2 FeCl₃ — and the coefficients also tell us 2 mol Fe reacts with 3 mol Cl₂.

State symbols

After each formula, a small italicised state symbol shows the physical state: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.

Ar, Mr & the mole

Atoms are far too small to count one by one, but they can be weighed in vast numbers. The mole is the conversion: a bridge between the world of grams and the world of single atoms.

Relative atomic mass, Ar

An atom’s mass in kilograms is tiny and inconvenient. Instead, chemists use the relative atomic mass Ar: a comparison against one-twelfth of a carbon-12 atom, taken as exactly 12. Hydrogen’s Ar is 1, oxygen’s is 16, sodium’s is 23. The values are pure numbers — no units.

Relative molecular mass, Mr

For a compound, add up the Ar values of every atom in one formula unit. This sum is the relative molecular mass Mr, or for ionic compounds the relative formula mass. For carbon dioxide, CO₂: Mr = (1 × 12) + (2 × 16) = 44.

ExtendedThe mole

The mole (symbol mol) is just a count — like a dozen, but huge. One mole of any substance contains 6.02 × 10²³ particles: atoms, molecules or formula units. This number, the Avogadro constant, is set so that the mass of one mole of a substance in grams equals its Ar or Mr. One mole of carbon atoms therefore weighs 12 g; one mole of CO₂ weighs 44 g.

From this comes the central equation of stoichiometry: n = m / Mr, linking moles, mass in grams and relative molecular mass.

ExtendedWorked example: mass to moles

How many moles of CO₂ are in 88 g of carbon dioxide? First find Mr of CO₂: (1 × 12) + (2 × 16) = 44. Then apply n = m / Mr: n = 88 / 44 = 2 mol. That 88 g contains exactly twice the number of molecules as 44 g, which holds one Avogadro number.

Reacting masses

Once you can convert mass to moles, you can predict exactly how much product a reaction will give. Every problem follows the same recipe.

ExtendedThe recipe

A balanced equation does more than show what reacts. Its coefficients give the mole ratio — the exact proportions in which the substances react. For 2 Mg + O₂ → 2 MgO, every 2 moles of magnesium reacts with 1 mole of oxygen and produces 2 moles of magnesium oxide.

Every reacting-mass calculation follows the same four steps: (1) balance the equation; (2) convert the known mass to moles using n = m / Mr; (3) use the mole ratio from the equation to find moles of what you want; (4) convert those moles back to the required quantity — mass, volume, and so on.

ExtendedWorked example: reacting masses

Limestone (calcium carbonate, CaCO₃) decomposes on heating to give quicklime (CaO) and carbon dioxide. What mass of quicklime comes from 50 g of pure calcium carbonate? (Ar: Ca = 40, C = 12, O = 16.) The equation CaCO₃ → CaO + CO₂ is already balanced, mole ratio 1 : 1 : 1. Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100, so n = 50 / 100 = 0.50 mol. From the 1 : 1 ratio, moles of CaO = 0.50 mol. Mr of CaO = 40 + 16 = 56, so mass = n × Mr = 0.50 × 56 = 28 g. The 22 g difference is the carbon dioxide that has escaped as gas.

Gas volumes & concentration

Two more conversions complete the toolkit: one for gases (volume to moles) and one for solutions (concentration to moles). Both reduce to a single multiplication.

ExtendedGas volumes at r.t.p.

Avogadro’s law states that, at the same temperature and pressure, equal volumes of any gases contain equal numbers of particles. The species do not matter — only the count does. At room temperature and pressure, one mole of any gas occupies 24 dm³ (24 000 cm³): V = n × 24 dm³.

Worked example: what volume of carbon dioxide at r.t.p. is produced when 25 g of CaCO₃ decomposes completely? The mole ratio CaCO₃ : CO₂ is 1 : 1. Moles of CaCO₃ = 25 / 100 = 0.25 mol, so moles of CO₂ = 0.25 mol. Volume = 0.25 × 24 = 6 dm³ — six litres of gas from a small pile of solid.

ExtendedSolution concentration

For a substance dissolved in water, concentration is moles of solute per unit volume of solution, in mol / dm³: c = n / V, with V in dm³. If a question gives volume in cm³, divide by 1000 first. So 0.50 mol of NaCl dissolved in 250 cm³ of solution gives c = 0.50 / 0.250 = 2.0 mol / dm³.

Empirical & molecular formulae

If you know the masses (or percentages) of each element in a compound, you can work out its formula from scratch. The principle: turn masses into moles, then read off the ratio.

ExtendedEmpirical formula from composition

The empirical formula gives the simplest ratio of atoms. For sodium chloride this is NaCl — one Na to one Cl. For ethanoic acid CH₃COOH the empirical formula is CH₂O, because the actual 2 : 4 : 2 ratio simplifies. To deduce an empirical formula from experimental data, convert each element’s mass (or percentage) to moles, then simplify the mole ratio to whole numbers: (1) divide each mass or % by Ar; (2) divide all answers by the smallest; (3) multiply up if needed to get whole numbers.

Worked example: a compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass (Ar: C = 12, H = 1, O = 16). Assume a 100 g sample so percentages become grams, then divide by Ar: C 40.0 / 12 = 3.33 mol, H 6.7 / 1 = 6.7 mol, O 53.3 / 16 = 3.33 mol. Divide by the smallest (3.33): C 1.00, H 2.01, O 1.00 — a ratio of 1 : 2 : 1, so the empirical formula is CH₂O. This is the empirical formula of every simple sugar; glucose, C₆H₁₂O₆, is six times the empirical unit.

ExtendedMolecular formula from empirical

The molecular formula gives the real number of atoms in one molecule — always a whole-number multiple of the empirical formula. If you know the empirical formula and the relative molecular mass Mr, divide Mr by the empirical formula mass to get the multiplier. CH₂O has empirical mass 12 + 2 + 16 = 30; if the compound’s Mr is 180, the multiplier is 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆.

Limiting reactants, yield & purity

Real reactions rarely use perfectly matched reactant amounts, and never produce exactly the predicted mass. Three calculations capture the gap between theory and the bench.

ExtendedThe limiting reactant

When two reactants are mixed in amounts that do not exactly match the mole ratio, one runs out before the other. The reactant that runs out first is the limiting reactant: it determines how much product can form, and the other is left over in excess. To find which is limiting, convert each reactant’s mass to moles, then divide by its coefficient in the balanced equation. The reactant with the smaller value is limiting; use its moles to calculate everything else.

ExtendedPercentage yield

The theoretical yield is the mass of product predicted by perfect stoichiometry; the actual yield is what you obtain in the lab, and the two are rarely equal. Percentage yield = (actual / theoretical) × 100. Worked example: decomposing 50 g of pure CaCO₃ should give 28 g of CaO, but a student obtains 22.4 g, so % yield = (22.4 / 28) × 100 = 80%. Common causes of less than 100%: incomplete reaction, side reactions, and product lost on transfer between vessels.

ExtendedPercentage purity

Raw materials are often impure. Percentage purity tells you what fraction of a sample is the substance you actually want: % purity = (mass of pure substance / mass of sample) × 100. If a 25 g sample of limestone contains 20 g of CaCO₃, the purity is (20 / 25) × 100 = 80%.