Everything from the 7 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
All relative masses are measured against 1/12 of the mass of one atom of carbon-12 (the unified atomic mass unit, u). Relative isotopic mass is the mass of one atom of an isotope on that scale. Relative atomic mass Ar is the weighted mean mass of the atoms of an element. Relative molecular mass Mr adds the Ar values in a molecule; for giant ionic compounds the same sum is called relative formula mass. None of them has units.
A molecule such as Cl2 gives three molecular ion peaks, at m/z 70, 72 and 74, because each atom can be either isotope. With 75% 35Cl and 25% 37Cl the chances are 0.75 × 0.75, 2 × 0.75 × 0.25 and 0.25 × 0.25, a ratio of 9 : 6 : 1. Do not forget the factor of 2 for the mixed molecule, which can form two ways.
One mole is the amount of substance that contains 6.02 × 1023 particles, the Avogadro constant L. That number is chosen so that one mole of carbon-12 atoms has a mass of exactly 12 g, which makes the molar mass M of any substance in g mol−1 numerically equal to its relative mass. So n = m ÷ M and N = n × L. Always say which particle you are counting: atoms, molecules, ions or formula units.
When a question asks for the number of atoms or ions, multiply the number of formula units by how many of that particle each one contains. 0.100 mol of MgCl2 contains 0.200 mol of chloride ions, not 0.100 mol.
In an ionic compound the total positive charge equals the total negative charge, so Al3+ and SO42− combine 2 : 3 as Al2(SO4)3. A balanced equation has the same number of each atom on both sides; balance only with the large numbers in front, never by changing a formula. An ionic equation shows only the species that change, and its charges must balance as well as its atoms.
Never balance an equation by changing a subscript. Writing H2O2 to make H2 + O2 → H2O balance turns water into hydrogen peroxide, a different substance. And an ionic equation that balances atoms but not charge is still wrong: check the total charge on each side.
The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound; the molecular formula is the actual number of atoms of each element in one molecule. To find the empirical formula, divide each mass or percentage by Ar and then divide by the smallest answer. The molecular formula is n × the empirical formula, where n = Mr ÷ empirical formula mass.
Do not round a ratio like 1 : 1.5 to 1 : 2 or 1 : 1.33 to 1 : 1. Those are real ratios of 2 : 3 and 3 : 4, and rounding them early gives the wrong formula (Fe2O3 is 1 : 1.5). In the hydrate experiment, heat to constant mass: if some water remains, the water lost is too small and x comes out too low.
Mass of A → moles of A (divide by M) → moles of B (use the equation's ratio) → mass of B (multiply by M). The limiting reagent is the reactant that is completely used up; it sets the maximum amount of product, and the other reactant is in excess. Percentage yield = actual yield ÷ theoretical yield × 100, with the theoretical yield calculated from the limiting reagent.
Use the ratio from the equation, not a ratio of masses, and always work the theoretical yield from the limiting reagent. If you use the reagent in excess, the answer will be too big and the percentage yield too small.
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So one mole of any gas occupies the same volume: 24.0 dm3 at room conditions, the Data Booklet value. Amount of gas n = V ÷ 24.0 with V in dm3 (or V ÷ 24 000 with V in cm3). For reactions between gases, the volume ratio equals the mole ratio in the equation.
Watch the units: 1 dm3 = 1000 cm3, so use 24.0 with dm3 and 24 000 with cm3. Water formed in a combustion reaction is a liquid at room temperature, so it does not count in the final gas volume. And the molar volume applies only to gases, never to solids, liquids or solutions.
Concentration c = n ÷ V, in mol dm−3 with V in dm3. So n = c × V, remembering that V in cm3 must be divided by 1000. Concentration in g dm−3 = c × M. In a titration: find the moles of the solution of known concentration, use the mole ratio from the equation, then divide by the volume of the other solution. Use the mean of concordant titres, those within 0.10 cm3 of each other.
Two slips cost the most marks: using a volume in cm3 as if it were in dm3 (answers 1000 times too big), and ignoring the mole ratio when the acid is diprotic, as with H2SO4. Also leave out the rough titre and any titre not within 0.10 cm3 of the others when you take the mean.