Chemical formulae, equations, calculations — Edexcel IGCSE Chemistry Modular - UNIT 1 Revision Notes & Topical Questions

The relative atomic mass A_r of an element is calculated from the mass numbers and relative abundances of its isotopes, and is found on the Periodic Table (the larger of the two numbers).

Multiply everything inside a bracket by the number outside it.

Amounts of substance are measured in moles (mol).

Balances measure grams, not moles, so convert using: moles = mass ÷ molar mass, and mass = moles × molar mass. The same number of carbon atoms (6.02 × 10²³ in 12 g) is in one mole of carbon as of molecules in one mole of CO₂.

Worked example 1. The mass of 0.250 mol of zinc (A_r = 65.38) = 0.250 × 65.38 = 16.3 g. Worked example 2. Moles in 2.64 g of sucrose C₁₂H₂₂O₁₁ (M_r = 342.3) = 2.64 ÷ 342.3 = 7.71 × 10⁻³ mol. Always show your working: marks can be gained even if the final answer is wrong.

A balanced equation gives the ratio of moles of reactants and products. To find an unknown mass:

Worked example 3. 2Mg + O₂ → 2MgO. Mass of MgO from 6.0 g of Mg (A_r 24, O 16)? Moles Mg = 6.0 ÷ 24 = 0.25. Ratio Mg : MgO = 1 : 1, so 0.25 mol MgO. Mass = 0.25 × 40 = 10 g.

Worked example 4. 2Al₂O₃ → 4Al + 3O₂.

Phrases such as "calculate the mass", "minimum mass" and "maximum mass" signal this method.

If the masses of reactants and products are known, convert each to moles and simplify the ratio.

Worked example. 1.2 g of carbon reacts with 16.2 g of zinc oxide to give 4.4 g of carbon dioxide and 13 g of zinc.

These questions look hard, but they are easy if you organise your working neatly. Remember that the molar ratio of a balanced equation gives the ratio of amounts of each substance.

Yield is the amount of product obtained from a reaction. The theoretical yield is the maximum amount, calculated from the balanced equation and reacting masses, assuming 100% conversion. The actual yield is the mass obtained in practice, and can only be found by experiment.

percentage yield = (actual yield ÷ theoretical yield) × 100

A yield of 100% is never achieved in practice because:

Industry aims for a high percentage yield to reduce costs and waste; companies compare different reaction pathways (sequences of reactions) by their percentage yields. A common mistake is to divide theoretical by actual yield; if the result exceeds 100%, you have inverted it.

Worked example. A student prepares 1.6 g of dry copper(II) sulfate crystals from sulfuric acid and copper(II) oxide. The theoretical yield is 2.0 g. Percentage yield = (1.6 ÷ 2.0) × 100 = 80%.

The formulae of simple compounds can be found by accurate mass measurements before and after a reaction, converting masses to moles and finding the ratio. Experiments are easiest when a gas is lost or gained. Hydrated copper sulfate, CuSO₄·xH₂O, is heated to drive off the water of crystallisation as vapour.

Practical tipavoid overheating, as the salt might decompose and give a larger mass change.

Aimdetermine the empirical formula of magnesium oxide by combustion of magnesium.

Mass of oxygen = mass of oxide − mass of magnesium. Divide each mass by the relative atomic mass (Mg 24, O 16) to find moles, simplify the ratio, and write the formula M_xO_y. For example, 0.24 g Mg and 0.16 g O gives 0.010 : 0.010 = 1 : 1, so MgO.

Aimdetermine the formula of copper(II) oxide by reduction with methane (natural gas).

Mass of copper = final mass − tube; mass of oxygen = starting oxide mass − mass of copper. Calculate moles with A_r (Cu 63.5, O 16), simplify the ratio, and write the formula. For example 1.27 g of copper and 0.32 g of oxygen gives 0.0200 : 0.0200 = 1 : 1, so CuO.

Heat until the mass is constant for any of these experiments, and keep the reduction experiment well ventilated because methane is flammable. Because methane is flammable, burn off excess gas safely.

The molecular formula shows the actual number and type of each atom in a molecule: ethanoic acid is C₂H₄O₂. The empirical formula is the simplest whole-number ratio of atoms of each element in a compound: ethanoic acid is CH₂O. The formula of an ionic compound is always an empirical formula.

Use a table and follow the steps:

Worked example 1. A compound contains 10 g of hydrogen and 80 g of oxygen. Moles: H = 10 ÷ 1 = 10, O = 80 ÷ 16 = 5. Ratio (divide by 5): H 2, O 1. The empirical formula is H₂O.

Worked example 2. Carbohydrate X contains 31.58% C and 5.26% H by mass.

Worked example. The empirical formula of X is C₄H₁₀S (A_r: C 12, H 1, S 32) and its M_r is 180. Empirical formula mass = 48 + 10 + 32 = 90. 180 ÷ 90 = 2. Molecular formula = C₈H₂₀S₂.

Another example. A hydrocarbon has empirical formula CH₂ and M_r = 56. Empirical mass = 14, n = 4, so the molecular formula is C₄H₈.

A hydrated salt contains water molecules in its crystal structure; the formula shows this as, for example, CuSO₄·5H₂O, where the dot indicates water of crystallisation. The formula is found by heating a weighed sample to constant mass to drive off the water, then reweighing the anhydrous salt.

Adapt the empirical formula method, using the salt and water instead of elements: write the mass of salt and of water, their relative formula masses, calculate moles, divide by the smaller number to get the ratio 1 : x, and write the formula.

The formula is CuSO₄·5H₂O. Although the specification is not clear whether you need to deduce hydrated salt formulae, the method is just an application of the empirical formula calculation and is useful to know.

A solute is a substance (often a solid) that dissolves in a liquid solvent to make a solution. The amount of solute can be given as a mass in grams (g) or as an amount in moles (mol). Concentration tells us how much solute is dissolved in a fixed volume of solution, and in chemistry calculations it is usually measured in mol/dm³ (or in g/dm³).

The key relationship is a formula triangle with moles at the top and concentration and volume at the bottom:

The volume must always be in dm³. Volumes in the laboratory are normally measured in cm³, so convert first: to go from cm³ to dm³ divide by 1000, and to go from dm³ to cm³ multiply by 1000. Many wrong answers come from forgetting this one step.

Avogadro’s law states that equal volumes of different gases, measured at the same temperature and pressure, contain the same number of molecules. In other words, equal amounts (moles) of gases occupy equal volumes under the same conditions.

At room temperature and pressure (RTP: 20 °C and 1 atmosphere) one mole of any gas occupies 24 dm³, which is 24 000 cm³.

A second formula triangle links moles of gas, volume and molar volume:

Use dm³ with 24 and cm³ with 24 000; never mix the two. Example: 225.6 dm³ of methane is 225.6 ÷ 24 = 9.4 mol. Going the other way, 0.50 mol of any gas at RTP occupies 0.50 × 24 = 12 dm³.

Because equal moles of gases have equal volumes, the ratio of the coefficients in a balanced equation is also the ratio of gas volumes (all measured at the same temperature and pressure). You do not need to convert to moles at all.

The complete combustion of propaneC₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g).

Questionwhat volume of oxygen is needed to react with 150 cm³ of propane, and what is the total volume of gaseous products?

This shortcut works only for gases. Water formed as a liquid would not be counted, and solids are never included.