Conservation of mass and balanced equations, relative formula mass, the mole and Avogadro's number, calculating amounts and masses in reactions, limiting reactants, and concentration of solutions.

What this topic is asking

AQA wants you to apply conservation of mass to balanced equations, calculate relative formula mass, use the mole and Avogadro's number, calculate reacting masses, identify limiting reactants, and calculate the concentration of solutions.

Conservation of mass and relative formula mass

The relative formula mass ($M_r$) of a compound is the sum of the relative atomic masses ($A_r$) of all the atoms in its formula. For example, water $\text{H}_2\text{O}$ has $M_r = (2 \times 1) + 16 = 18$, and calcium carbonate $\text{CaCO}_3$ has $M_r = 40 + 12 + (3 \times 16) = 100$. In a balanced equation the total $M_r$ of the reactants equals the total $M_r$ of the products, because mass is conserved. In a non-enclosed reaction the mass may appear to change: it increases if a gas from the air is taken in (for example a metal gaining mass as it oxidises) and decreases if a gas is given off (for example a carbonate decomposing), but in a sealed container the total mass is always constant.

The mole and reacting masses

The key relationship is number of moles $= \dfrac{\text{mass}}{M_r}$, which rearranges to mass $= \text{moles} \times M_r$. A balanced equation gives the ratio in which substances react in moles (the stoichiometry), so to find the mass of one substance from another you convert to moles, apply the equation ratio, then convert back to mass. The three steps (mass to moles, ratio, moles to mass) are the backbone of every reacting-mass question.

The limiting reactant is the reactant that is completely used up first; it determines the maximum amount of product that can form. The other reactant is then in excess (some is left over). The amount of product is always calculated from the moles of the limiting reactant, never from the excess one.

Concentration of solutions

A cubic decimetre (dm$^3$) is the same volume as a litre, equal to 1000 cm$^3$, so to convert a volume in cm$^3$ to dm$^3$ you divide by 1000. Forgetting this conversion is the single most common reason candidates lose marks on concentration questions. Concentration can also be measured in mol/dm$^3$ by using the number of moles of solute instead of its mass; the two are linked because moles $= \dfrac{\text{mass}}{M_r}$.