Using the Gas Equation to Find Molecular Mass

The gas equation (n = PV/RT) is your best mate for finding molecular masses of gases. It connects pressure, volume, temperature, and the number of moles in a simple relationship.

Here's how it works in practice: imagine you've got a lighter fuel canister and want to find the molecular mass of the gas inside. You weigh the canister, dispense exactly 1000 cm³ of gas (making sure the pressure equals atmospheric pressure), then weigh it again.

The calculation is straightforward. If you lost 2.28g of mass and collected 1000 cm³ at 14°C and 100,000 Pa, you can work out the number of moles: n = (100,000 × 1000 × 10⁻⁶) ÷ (8.31 × 287) = 0.042 mol. Since 0.042 mol weighs 2.28g, one mole weighs 54.5g - that's your molecular mass!

Quick Tip: Always convert temperature to Kelvin (add 273) and volume to m³ before using the gas equation.

Finding Empirical Formulae

Empirical formulae show the simplest ratio of atoms in a compound - think of them as the most basic recipe. Finding them is like following a three-step recipe that never fails.

First, find the mass of each element (usually given in exam questions). Then convert these masses to moles by dividing by each element's atomic mass. Finally, turn these mole amounts into the simplest whole number ratio.

Let's say you've got 10.01g of calcium carbonate containing 4.01g calcium, 1.20g carbon, and 4.80g oxygen. Convert to moles: Ca = 4.01÷40.1 = 0.10 mol, C = 1.20÷12.0 = 0.10 mol, O = 4.80÷16.0 = 0.30 mol. The ratio is 0.10:0.10:0.30, which simplifies to 1:1:3, giving CaCO₃.

For molecular formulae, just divide the actual molecular mass by the empirical formula mass. If ethene has a molecular mass of 28.0 but its empirical formula CH₂ has a mass of 14.0, then 28.0÷14.0 = 2, so the molecular formula is C₂H₄.

Remember: Empirical formula = simplest ratio, molecular formula = actual number of atoms in the molecule.

Combustion Analysis

Combustion analysis is brilliant for finding formulae of compounds containing carbon, hydrogen, and oxygen. When you burn these compounds completely, all the carbon becomes CO₂ and all the hydrogen becomes H₂O.

Here's the trick: work backwards from the products to find what was in the original compound. If burning 0.53g of compound X produces 1.32g CO₂ and 0.54g H₂O, you can calculate the original amounts.

The carbon calculation: 1.32g CO₂ ÷ 44.0 = 0.03 mol CO₂, which means 0.03 mol carbon atoms originally. The hydrogen calculation: 0.54g H₂O ÷ 18.0 = 0.03 mol H₂O, but each water molecule has 2 hydrogens, so that's 0.06 mol hydrogen atoms originally.

For oxygen, it's trickier - you calculate by difference. The carbon and hydrogen masses add up to 0.42g, so the remaining 0.11g must be oxygen (0.01 mol). The ratio becomes C:H:O = 3:6:1, giving the empirical formula C₃H₆O.

Top Tip: Always check your mass calculations add up - if they don't, you've made an error somewhere!