Stoichiometry

What the exam asks

The two items you will meet most are a limiting reagent scenario and a formula determination, often a combustion. Both are procedural, and both are lost to the same two mistakes. The first is picking the limiting reagent by mass, or by moles without dividing by the coefficient: 1 mol of nitrogen and 1.5 mol of hydrogen looks like nitrogen is scarce, but hydrogen is consumed three times as fast and runs out first. The second is combustion analysis, where candidates count the oxygen in the CO2 and H2O as if it came from the sample. It came from the O2. Oxygen is always the one you get by subtraction, never by weighing an oxide. Expect the numbers to be clean, and treat an untidy intermediate as a sign you have misread rather than a reason to approximate.

Stoichiometry is bookkeeping in moles. Grams, litres of solution and litres of gas are all just packaging: convert to moles, use the balanced equation as an exchange rate, convert back to whatever the question asked for. Almost every error in this topic happens because someone reasoned in the packaging rather than in moles.

The exam poses it as a scenario. Two reagents are mixed and you are asked how much product forms, or a compound is burnt and you are asked what it was. Neither rewards memorising a formula. What they reward is a fixed procedure: moles first, divide by the coefficient, work from whichever reagent runs out, and only convert back at the end.

There is no calculator, so the arithmetic is part of the question rather than an afterthought. GAMSAT numbers are chosen to cancel. If you are reaching for long division, you have usually missed a factor: round the molar mass, keep quantities as fractions, and cancel before you multiply. A quantity that looks ugly halfway through is a signal to look again, not to push on.

What to hold

• Moles are the currency: n = m/M for a solid, n = cV for a solution, and n = V/Vm for a gas, and the balanced equation only ever speaks in moles.
• The limiting reagent is found by dividing each reagent's moles by its coefficient in the balanced equation, and taking the smallest result.
• It cannot be found by mass, and it cannot be found from moles alone, because the coefficients say how fast each reagent is being consumed.
• Everything the reaction produces scales off the limiting reagent, and the excess reagent left over is what you started with minus what the limiting reagent consumed.
• Percentage yield is actual divided by theoretical, times 100, where theoretical yield is what the limiting reagent would give if nothing went wrong.
• A yield above 100% is not a triumph, it is a wet or impure product, because the mass includes something that is not the product.
• An empirical formula is the simplest whole number ratio of atoms; a molecular formula is a whole number multiple of it, and that multiple is the molecular mass divided by the empirical formula mass.
• To get an empirical formula from percentages, divide each percentage by the element's relative atomic mass, then divide all the results by the smallest.
• If that ratio lands near x.5, multiply everything by 2 rather than rounding, because 1:1.5 is 2:3 and rounding destroys the compound.
• In combustion analysis, all the carbon in the CO2 and all the hydrogen in the H2O came from the sample, but the oxygen did not.
• Oxygen in the sample is found by difference: sample mass minus the mass of carbon minus the mass of hydrogen, because the rest of the oxygen came from the O2 that was burnt.
• The arithmetic is designed to cancel, so round molar masses to whole numbers, keep values as fractions until the end, and cancel factors before multiplying.