Capstone: The full calculation journey
Every part of this topic is really one chain of steps: a real industrial problem starts with a word equation and ends with a percentage yield. Here you will take a single problem all the way through — balance → relative formula mass → moles → reacting mass → percentage yield — one stage at a time. Each stage unlocks the next.
A factory makes iron by reducing iron(III) oxide, Fe₂O₃, with excess carbon monoxide, CO. It heats 320 g of Fe₂O₃ and collects 196 g of iron. Work out the percentage yield of iron. (Ar: O = 16, Fe = 56.)
Work through it one stage at a time — each stage unlocks when the one before it is correct.
Step 1 of 4 — Balance the equation
Type the balancing number in front of each substance (use 1 if none is needed), then press Check.
Step 2 of 4 — Mr, moles & reacting mass
Build the table one box at a time to find the theoretical mass of iron. Because CO is in excess, Fe₂O₃ is the limiting reactant, so the iron is based on it.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
| Fe₂O₃known | Fefind this | |
|---|---|---|
| Ratio | ||
| Moles | ||
| Mass | 320 ggiven | g |
| Mr |
🔒 Locked — balance the equation in Step 1 first.
Step 3 of 4 — Line up the two masses
For a percentage yield you compare the actual mass made with the theoretical mass. From Step 2 the theoretical mass of iron is 224 g; the factory actually collected 196 g. The actual yield always goes on top.
🔒 Locked — finish the table in Step 2 first.
Step 4 of 4 — Calculate the percentage yield
percentage yield = (actual ÷ theoretical) × 100 = (196 ÷ 224) × 100. Type your answer, then press Check.
🔒 Locked — finish the table in Step 2 first.
That is exactly the chain every multi-step quantitative question follows.
- Conservation of mass: no atoms are created or destroyed in a reaction, so the total mass of products equals the total mass of reactants. This is why equations must be balanced.
- Balanced equation: same number of each type of atom on both sides. Balance by adding numbers in front of formulae — never change a formula.
- Mass change with a gas: mass appears to increase when a gas is absorbed from the air (e.g. a metal forming an oxide) and to decrease when a gas is released to the air (e.g. a carbonate giving off CO₂). In a sealed system the mass is unchanged.
- Relative formula mass (Mr) = sum of the Ar values of all the atoms in the formula. In a balanced equation the total Mr of the reactants equals the total Mr of the products.
- Uncertainty: report measurements to a sensible number of significant figures (round once, at the end, matching the least precise data); the uncertainty of a mean can be estimated from the range of repeat results. Tare the balance before weighing to avoid a zero error.
- The mole H: one mole contains the Avogadro number of particles, 6.02 × 10²³. Mass = Mr × moles, so moles = mass ÷ Mr.
- Balancing using masses H: convert masses to moles, divide by the smallest to get the whole-number ratio — this gives the balancing numbers.
- Reacting masses H: use the balanced equation to find the mole ratio, then convert between mass of one substance and mass of another.
- Limiting reactant H: the reactant fully used up; it limits the amount of product. The reactant in excess is left over. Amount of product is proportional to the moles of the limiting reactant.
- Concentration: in g/dm³, concentration = mass ÷ volume (dm³). The greater the mass dissolved in a given volume, the higher the concentration. (1 dm³ = 1000 cm³.)
- Titration T H: concentration in mol/dm³ = moles ÷ volume (dm³). Use the equation's mole ratio with one known concentration to find the other.
- Percentage yield T = (actual yield ÷ theoretical yield) × 100. Never 100% — losses occur, reactions may be incomplete or reversible, or there are side reactions.
- Atom economy T = (Mr of desired product ÷ sum of the Mr of all reactants) × 100. High atom economy means less waste and is more sustainable and economical.
- Gas volumes T H: equal volumes of any gases at the same temperature and pressure contain equal numbers of moles. One mole of any gas occupies 24 dm³ at RTP (20 °C, 1 atm), so volume (dm³) = moles × 24.
That completes C3 — the maths that underpins every reaction in the course. It rests on the bonding and formulae of C2 (you need a correct chemical formula before you can find an Mr), and it leads straight into C4 — Chemical Changes, where these moles, masses and concentrations are put to work on real reactions such as titrations of acids.