CCEA A-Level Chemistry

Writing formulae of ionic compounds

The charge on a simple ion can be predicted from the position of its element in the Periodic Table. Metals in Groups I, II and III (13) lose one, two and three electrons to form $\ce{Na+}$, $\ce{Mg^2+}$ and $\ce{Al^3+}$, and non-metals in Groups V, VI and VII gain three, two and one electrons to form $\ce{N^3-}$, $\ce{O^2-}$ and $\ce{Cl-}$. Each of these ions has the electron arrangement of a noble gas. An ionic compound is electrically neutral, so the formula is the smallest whole-number ratio of ions in which the charges cancel: magnesium and chlorine give $\ce{MgCl2}$, and aluminium and oxygen give $\ce{Al2O3}$, because two $\ce{Al^3+}$ (+6) balance three $\ce{O^2-}$ (−6).

Where an ion contains more than one atom, its formula goes in brackets when more than one is needed, as in $\ce{Ca(NO3)2}$ and $\ce{(NH4)2SO4}$. The molecular ions to know, with their charges, are in the table.

Balanced equations

A balanced equation has the same number of atoms of each element on both sides, and in an ionic equation the total charge is also the same on both sides. State symbols show the physical state: (s), (l), (g), and (aq) for a substance dissolved in water.

A full equation shows every formula, and an ionic equation leaves out spectator ions, which are present unchanged on both sides. The reaction of aqueous barium chloride with aqueous sodium sulfate is written in full as

and as an ionic equation as

Sodium and chloride ions take no part, so they do not appear. Equations for reactions not met before can be worked out when the reactants and products are given, by writing the formulae first and only then balancing with coefficients, never by changing a subscript in a formula. For example, the reaction of aluminium with dilute sulfuric acid gives aluminium sulfate and hydrogen. The formulae are $\ce{Al}$, $\ce{H2SO4}$, $\ce{Al2(SO4)3}$ and $\ce{H2}$, and balancing gives

Avogadro's constant, the mole and molar mass

The mole is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant, $L = 6.02 \times 10^{23}\ \pu{mol-1}$. The particles may be atoms, molecules, ions or electrons, so the kind of particle must always be stated: one mole of $\ce{H2}$ molecules contains $\pu{6.02e23}$ molecules and $\pu{1.20e24}$ atoms.

The molar mass, $M$, is the mass of one mole of a substance, in $\pu{g mol-1}$. Numerically it equals the relative atomic mass, relative molecular mass or relative formula mass.

where $n$ is the amount in moles and $m$ the mass in grams. The number of particles is $n \times L$.

For $\pu{3.65 g}$ of hydrogen chloride ($M = 36.5$):

The same quantity is obtained for solutions and gases by other routes. For a solution, $n = c \times V$, with $c$ in $\pu{mol dm-3}$ and $V$ in $\pu{dm3}$, and a volume in $\pu{cm3}$ is divided by 1000 to convert it, because $\pu{1 dm3} = \pu{1000 cm3}$.

Reacting masses

The coefficients in a balanced equation give the ratio of the amounts in moles that react. A reacting-mass calculation has the same steps every time.

1. Write the balanced equation.
2. Convert the known mass to moles.
3. Use the ratio in the equation to find the moles of the substance wanted.
4. Convert back to a mass or volume.

Where the quantities of two reactants are given, one is usually in excess, and the other, the limiting reagent, decides how much product forms. The limiting reagent is the one that would be used up first.

Worked example with an excess

$\pu{2.43 g}$ of magnesium is added to $\pu{100 cm3}$ of $\pu{1.00 mol dm-3}$ hydrochloric acid.

All the magnesium would need $2 \times 0.100 = 0.200$ mol of acid, and only 0.100 mol is present, so the acid is limiting and magnesium is in excess. The acid decides the yield:

The magnesium used is $0.0500 \times 24.3 = \pu{1.215 g}$, so $\pu{1.215 g}$ is left over.

Hydrated salts and water of crystallisation

An anhydrous substance contains no water. A hydrated crystal has water molecules built into its lattice, called water of crystallisation, in a fixed ratio to the ions, and it is written with a dot, as in $\ce{CuSO4.5H2O}$. Heating drives the water off, and heating to constant mass (heating, cooling in a desiccator, weighing, and repeating until two weighings agree) shows that no water remains.

Finding the number of molecules of water

$\pu{2.86 g}$ of hydrated sodium carbonate, $\ce{Na2CO3.xH2O}$, leaves $\pu{1.06 g}$ of anhydrous solid after heating.

The formula is $\ce{Na2CO3.10H2O}$. The same method works from percentage composition: treat the percentages as masses in a 100 g sample, divide each by its molar mass and find the simplest ratio. A hydrate whose mass loss is large compared with its anhydrous mass has a large $x$.

Percentage of water can also be calculated from the formula: for $\ce{CuSO4.5H2O}$ ($M = 249.6$), five waters weigh $5 \times 18.0 = 90.0$, and $\frac{90.0}{249.6} \times 100 = 36.1\%$.