Part 3: Reacting quantities, yield and atom economy

Use the limiting reactant to establish the theoretical yield, then judge both the experimental recovery and the waste designed into a process.

Follow the balanced equation

Convert the known mass, gas volume or solution quantity to moles. Use the equation’s coefficient ratio to find the required substance’s moles. Finally convert to the requested quantity. Label the substance at each step.

Worked example: 4.20 g NaHCO₃ (M = 84.0) decomposes by 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O. n(NaHCO₃) = 0.0500 mol, so n(CO₂) = 0.0250 mol and mass CO₂ = 0.0250 × 44.0 = 1.10 g.

A reactant can remain when the reaction stops

Compare available moles divided by the coefficient for each reactant. The smallest value identifies the limiting reactant. Do not compare masses directly.

For 2Mg + O₂ → 2MgO, 0.100 mol Mg and 0.0400 mol O₂ have n/coefficient values 0.0500 and 0.0400. O₂ limits: it uses 0.0800 mol Mg and makes 0.0800 mol MgO. Magnesium left = 0.0200 mol. If M(MgO) = 40.3, theoretical product mass = 3.224 g.

Worked limiting reagent from two measured masses

For 2Al + 3Cl₂ → 2AlCl₃, take 2.70 g Al and 7.10 g Cl₂, with M values 27.0, 71.0 and 133.5 g mol⁻¹. Available amounts are 0.100 mol Al and 0.100 mol Cl₂. Equal amounts do not mean exact reacting proportions: the equation needs more chlorine than aluminium.

Divide each amount by its coefficient: Al gives 0.100/2 = 0.0500; Cl₂ gives 0.100/3 = 0.033333…. Chlorine limits. Product amount = 2 × 0.033333… = 0.066666… mol, giving theoretical product mass 8.90 g. Aluminium consumed = 0.066666… mol, so 0.033333… mol or 0.900 g remains.

An alternative is to calculate chlorine needed for all aluminium: 0.100 × 3/2 = 0.150 mol, but only 0.100 mol is present. Either method is valid if the ratio is shown. Comparing reactant masses or unadjusted mole counts can select the wrong limiting reagent.

Yield measures actual recovery

Percentage yield = actual yield/theoretical yield × 100, using the same units. If the previous reaction gives 2.58 g dry MgO, yield = 2.58/3.224 × 100 = 80.0%. Losses can arise during transfer or purification, through side reactions or incomplete conversion. A result over 100% suggests contamination, retained solvent or an incorrect calculation.

Atom economy instead measures the fraction of reactant mass incorporated into a chosen useful product, based on the balanced equation. Include coefficients; do not include catalysts or solvents that are not stoichiometric reactants. For 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O, atom economy for Na₂CO₃ = 106/(2 × 84.0) × 100 = 63.1%.

Work backwards from an actual yield

If a process must deliver 5.34 g AlCl₃ at 75.0% isolated yield, the theoretical target is 5.34/0.750 = 7.12 g. This is larger than the required actual mass, as it must be. The theoretical amount is 7.12/133.5 = 0.053333… mol, requiring the same amount of Al, or 1.44 g, when chlorine is in excess.

For the same equation, AlCl₃ is the only stoichiometric product, so atom economy is 100%. A 75% yield can coexist with 100% atom economy: the former concerns actual product recovery, while the latter concerns where atoms go in the ideal balanced equation. Actual solvent waste and purification losses remain relevant.

Compare whole processes

A reaction can have 100% atom economy and a poor practical yield. Increasing atom economy reduces stoichiometric waste, but a fair process comparison also considers energy, feedstock renewability, solvent use, toxicity, separation, recycling and whether by-products have a use. A catalyst can reduce energy demand without changing the atom economy of the same balanced equation.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources