Reacting masses and volumes: core method

· Stoichiometry = using a balanced equation to calculate amounts of reactants/products.
· Always start with a balanced symbol equation; the coefficients give the mole ratio.
· Main conversion route: given quantity → moles → mole ratio → required quantity.
· Key formula: moles = mass ÷ Mr for solids/pure substances.
· Key formula: mass = moles × Mr.
· Key formula: moles = concentration × volume for solutions, with volume in dm³.
· For gases at the same temperature and pressure, volume ratio = mole ratio.
· In calculations, keep extra digits during working, then round the final answer to the correct significant figures.

This diagram shows why moles are the central bridge in reacting-mass and volume calculations. Use it to decide which conversion is needed before applying the mole ratio from the balanced equation. Source

Reacting masses

· Use moles = mass ÷ Mr to convert the known mass into moles.
· Use the balanced equation mole ratio to find moles of the required substance.
· Convert back using mass = moles × Mr if the answer is needed in grams.
· For combustion or decomposition questions, include all reactants/products before balancing.
· Percentage yield = actual yield ÷ theoretical yield × 100.
· Theoretical yield = maximum calculated amount from stoichiometry.
· Actual yield = amount actually obtained in the experiment.
· Percentage yield is usually below 100% due to incomplete reaction, side reactions, loss during transfer/filtration, or impurities.

Volumes and concentrations of solutions

· Concentration usually means mol dm⁻³.
· Convert volume carefully: 1000 cm³ = 1 dm³, so 25.0 cm³ = 0.0250 dm³.
· Formula: moles = concentration × volume in dm³.
· Rearrangements: concentration = moles ÷ volume and volume = moles ÷ concentration.
· In titration-style questions, use the titre to calculate moles of the known solution.
· Then apply the mole ratio to find moles of the unknown solution.
· Finally calculate unknown concentration using concentration = moles ÷ volume in dm³.
· For acids/alkalis, do not assume a 1:1 ratio unless the balanced equation shows it.

Titration calculations are solution stoichiometry problems. The key idea is that the equivalence point occurs when reacting substances are present in the exact mole ratio from the balanced equation. Source

Volumes of gases

· For gases at the same temperature and pressure, use the balanced equation coefficients as volume ratios.
· Example pattern: if 1 mol CH₄ reacts with 2 mol O₂, then 1 volume CH₄ reacts with 2 volumes O₂ under the same conditions.
· For gas calculations, check whether the question gives molar gas volume or asks for volume ratio.
· If a molar gas volume is given, use: moles = gas volume ÷ molar gas volume.
· If gases are compared under identical conditions, you can often skip moles and use direct volume ratios.
· In hydrocarbon combustion, balance in this order: C → CO₂, H → H₂O, then O₂.
· Remember: water may be shown as H₂O(l) or H₂O(g) depending on conditions; only gases count in gas-volume ratios.

This diagram shows why gas volumes follow the same ratio as moles when temperature and pressure are constant. It is especially useful for exam questions involving combustion gases or reacting gas mixtures. Source

Limiting reagent and excess reagent

· A limiting reagent is the reactant that is used up first and controls the maximum amount of product.
· An excess reagent is present in more than the amount needed; some remains unreacted.
· Method 1: calculate product from each reactant; the reactant giving less product is the limiting reagent.
· Method 2: divide each reactant’s moles by its equation coefficient; the smaller value identifies the limiting reagent.
· Once the limiting reagent is found, use only its moles to calculate theoretical yield.
· Do not use the excess reagent to calculate product unless the question confirms it is the limiting reactant.
· In exam answers, clearly state: limiting reagent = … because it produces fewer moles of product / is used up first.

Deducing stoichiometric relationships

· Stoichiometric relationship = the simplest whole-number mole ratio between reacting substances.
· Use calculated moles to find ratios, then divide by the smallest mole value.
· Convert near-whole numbers carefully: e.g. 1 : 1.5 becomes 2 : 3.
· Ratios can be deduced from masses, solution volumes/concentrations, or gas volumes.
· For experimental data, repeat titres or measurements may be used to improve reliability before calculating the ratio.
· A deduced ratio should match a balanced equation or help construct one.

Significant figures and exam traps

· Final answers should match the significant figures given or requested in the question.
· Do not round too early; premature rounding can change the final answer.
· Always include units: g, mol, dm³, cm³, mol dm⁻³.
· Use dm³, not cm³, in moles = concentration × volume.
· Check whether the question asks for mass, moles, volume, concentration, percentage yield, or limiting reagent.
· If the answer is a ratio, it normally has no units.
· In gas-volume questions, check whether all gases are at the same temperature and pressure before using direct volume ratios.