Chemistry: general, organic and analytical
Chemistry on this examination tests quantitative general chemistry, the structure and reactions of organic molecules, and the logic of analytical measurements. Expect mole calculations, pH and buffers, titration curves, functional-group identification and isomer counting.
Core rules
- Stoichiometry: balance equations first; the limiting reagent sets the maximum product. Moles n = m/M; at STP, 1 mol of gas occupies 22.4 L.
- Solutions: molarity M = mol solute / L solution. Dilution follows M1V1 = M2V2. pH = −log[H3O+]; pOH = 14 − pH at 25 °C.
- Equilibrium and buffers: Le Chatelier's principle predicts shifts; Henderson–Hasselbalch gives pH = pKa + log([A−]/[HA]). A buffer works best near its pKa.
- Organic chemistry: functional groups determine reactivity; IUPAC naming; constitutional, geometric and optical isomerism; nucleophilic substitution (SN1/SN2) and elimination.
- Analytical chemistry: primary standards, equivalence point versus endpoint, indicator choice and significant figures.
Worked example
10.0 g of CaCO3 (M = 100 g/mol) reacts with 100 mL of 1.00 M HCl. Which reagent is limiting, and what volume of CO2 forms at STP? CaCO3 + 2HCl → CaCl2 + CO2 + H2O. n(CaCO3) = 10.0/100 = 0.100 mol; n(HCl) = 0.100 × 1.00 = 0.100 mol. The equation requires 2 mol HCl per mol CaCO3, so 0.100 mol HCl consumes only 0.050 mol CaCO3. HCl is limiting; CO2 = 0.050 mol; volume = 0.050 × 22.4 = 1.12 L. Reasoning: compare available moles with the stoichiometric ratio, not raw moles.
Traps that cost marks
- Using molarity as moles without multiplying by volume.
- Confusing the equivalence point with the indicator endpoint.
- Ignoring coefficients when comparing limiting reagents.
- Applying pH + pOH = 14 outside 25 °C.