States of matter and stoichiometry
States of matter and stoichiometry connects the mole, gas behaviour and quantitative reactions. The exam tests multi-step calculations: formula determination, limiting reagents, solution concentrations and gas volumes, with units and significant figures.
Core rules
- The ideal gas equation is PV = nRT, with P in Pa, V in m3, n in mol, T in K and R = 8.314 J mol−1 K−1. Dalton's law: total pressure is the sum of partial pressures.
- Real gases deviate most at high pressure and low temperature, where molecular volume and attractions matter.
- The mole is the amount containing 6.02 × 1023 particles; molar mass is mass per mole. Percentage composition gives empirical formula; molecular formula is a whole-number multiple of it.
- Concentration c = n/V, usually mol dm−3; 1 dm3 = 1000 cm3. Molar gas volume is about 22.7 dm3 mol−1 at 273 K and 100 kPa (22.4 dm3 mol−1 at 101.3 kPa).
- For reactions, convert to moles, divide by stoichiometric coefficients to find the limiting reagent, then use the ratio to find product moles. Percentage yield = actual/theoretical × 100.
Worked example
A 0.500 g sample of a gas occupies 200 cm3 at 100 kPa and 298 K. Find its molar mass. Convert: V = 200 cm3 = 2.00 × 10−4 m3; P = 1.00 × 105 Pa. Then n = PV/RT = (1.00 × 105)(2.00 × 10−4)/(8.314 × 298) = 20.0/2478 = 8.07 × 10−3 mol. M = m/n = 0.500/8.07 × 10−3 = 62.0 g mol−1. The gas could be, for example, NO2 or a small organic molecule; the calculation alone does not identify it.
Traps that cost marks
- Using °C instead of K in the gas equation.
- Forgetting to convert cm3 to dm3 or m3; 1 cm3 = 10−3 dm3.
- Choosing the limiting reagent by smallest moles rather than smallest moles divided by coefficient.
- Mixing up concentration units or forgetting that dilution changes moles per volume, not total moles.
- Quoting too many significant figures, or ignoring state symbols when calculating gas volumes.