Stoichiometry and Chemical Calculations
Stoichiometry marks are lost on units and on deciding which reagent runs out, not on arithmetic. The paper expects moles from mass, concentration or gas volume, correct limiting-reagent logic, titration and dilution calculations, and sensible significant figures.
The core toolkit
- n = m/M; n = cV with V in dm3; at 273 K and 1 bar, 1 mol of gas occupies 22.7 dm3 (22.4 dm3 at 1 atm).
- Balance the equation first: coefficients give the mole ratio, never the mass ratio.
- Limiting reagent: divide each reagent's moles by its coefficient; the smallest quotient limits.
- Dilution: c1V1 = c2V2. On mixing, use total moles divided by total volume.
- Percentage yield = actual/theoretical × 100. Atom economy = M_r(desired product)/sum of M_r(reactants) × 100.
- Empirical formula from percentage composition: divide by A_r, then by the smallest quotient; molecular formula = empirical formula × (M_r/empirical mass).
Worked example
10.0 g of Al reacts with 10.0 g of Cl2: 2Al + 3Cl2 → 2AlCl3. n(Al) = 10.0/26.98 = 0.371 mol; n(Cl2) = 10.0/70.90 = 0.141 mol. Divide by coefficients: Al gives 0.371/2 = 0.185; Cl2 gives 0.141/3 = 0.0470. Chlorine is limiting. n(AlCl3) = (2/3) × 0.141 = 0.0940 mol, so m = 0.0940 × 133.34 = 12.5 g. Aluminium left over = 0.371 - 0.0940 = 0.277 mol, about 7.5 g.
Hydrates and back-titrations
When a hydrate is heated, the mass lost is water; moles of water per mole of anhydrous salt gives the formula. In a back-titration, the excess titrant gives the unreacted acid or oxidant, and subtracting it from the initial amount gives the sample content.
Traps
- Comparing reagent masses instead of mole quotients, the classic error.
- Forgetting that a diatomic gas such as Cl2 or H2 has M_r = 2A_r.
- Using cm3 in cV without dividing by 1000.
- Ignoring that total volume changes on mixing two solutions.
- Reporting 12.54321 g: match the significant figures of the data and carry guard digits only inside the working.