Chemical Formulae, Equations & the Mole

Writing Chemical Formulae

A chemical formula tells you which elements are present in a substance and how many atoms of each are joined together. For elements you simply use their symbol (e.g. Fe for iron), but molecules of non-metal elements are often written to show how their atoms pair up, such as $O_2$, $H_2$ and $Cl_2$.

For ionic compounds, the formula shows the simplest whole-number ratio of ions that gives an overall neutral charge. You balance the positive and negative charges so they cancel out.

Key terms

Ion – a charged particle formed when an atom loses or gains electrons.

Compound – a substance containing two or more elements chemically bonded together.

To build an ionic formula, swap the charges and write them as subscripts:

Exam tip

When a compound contains more than one of a molecular ion (like $OH^-$, $NO_3^-$ or $SO_4^{2-}$), put it in brackets before the subscript, e.g. $Ca(OH)_2$, not $CaOH_2$.

Balancing Symbol Equations

In a chemical reaction, atoms are never created or destroyed — they are only rearranged. This means a balanced equation must have the same number of each type of atom on both sides.

Worked example

Balance: $\;CH_4 + O_2 \rightarrow CO_2 + H_2O$

Carbon: 1 each side — fine. Hydrogen: 4 on the left, so we need $2H_2O$.

Now oxygen: right side has $2 + 2 = 4$ oxygens, so we need $2O_2$ on the left.

Final balanced equation: $\;CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

You can only change the big numbers (coefficients) in front of formulae — never the small subscripts inside a formula, as that would change the substance itself.

State Symbols

State symbols are added in brackets after each formula to show its physical state:

$(s)$ solid
$(l)$ liquid
$(g)$ gas
$(aq)$ aqueous — dissolved in water

For example: $\;Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$

The Mole and Avogadro's Constant

Atoms are far too small and too numerous to count individually, so chemists group them into a unit called the mole.

Key terms

Mole (mol) – the amount of a substance that contains the same number of particles as there are atoms in 12 g of carbon-12.

Avogadro's constant – the number of particles in one mole, equal to $6.02 \times 10^{23}$.

So one mole of any substance always contains $6.02 \times 10^{23}$ particles, whether those particles are atoms, molecules or ions.

Relative Formula Mass ($M_r$)

The relative formula mass ($M_r$) of a compound is found by adding up the relative atomic masses ($A_r$) of all the atoms in its formula.

Worked example

Find the $M_r$ of calcium carbonate, $CaCO_3$. ($A_r$: Ca = 40, C = 12, O = 16)

$M_r = 40 + 12 + (3 \times 16)$

$M_r = 40 + 12 + 48 = 100$

Moles, Mass and $M_r$

These three quantities are linked by a single key equation:

$$moles = \frac{mass}{M_r}$$

The mole triangle below helps you rearrange it: cover the quantity you want to find and the triangle shows the calculation.

Worked example

How many moles are in 36 g of water, $H_2O$? ($M_r = 18$)

$moles = \frac{36}{18} = 2 \text{ mol}$

And the reverse: what is the mass of 0.25 mol of $CaCO_3$ ($M_r = 100$)?

$mass = moles \times M_r = 0.25 \times 100 = 25 \text{ g}$

Calculating Reacting Masses

A balanced equation gives the mole ratio in which substances react. This lets you predict the mass of product formed, or the mass of reactant needed.

Worked example

What mass of magnesium oxide forms when 6 g of magnesium burns completely?

$2Mg + O_2 \rightarrow 2MgO$ ($A_r$: Mg = 24, O = 16)

Moles of Mg $= \frac{6}{24} = 0.25 \text{ mol}$

The ratio of Mg : MgO is $2:2$, so moles of MgO $= 0.25 \text{ mol}$.

$M_r$ of MgO $= 24 + 16 = 40$

Mass of MgO $= 0.25 \times 40 = 10 \text{ g}$

Percentage Yield

In practice you rarely obtain the full amount of product an equation predicts — some is lost in handling, reactions may not finish, or side reactions occur. Percentage yield compares what you actually got with the theoretical maximum.

$$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Worked example

A reaction should produce 8 g of product, but only 6 g is collected. Find the percentage yield.

$\% \text{ yield} = \frac{6}{8} \times 100 = 75\%$

Concentration of Solutions

The concentration tells you how much solute is dissolved in a given volume of solution. It can be measured in $mol/dm^3$ or in $g/dm^3$.

$$concentration = \frac{moles}{volume \; (dm^3)}$$

Watch out

Volumes are often given in $cm^3$. To convert to $dm^3$, divide by 1000 ($1 \; dm^3 = 1000 \; cm^3$).

Worked example

0.5 mol of $NaCl$ is dissolved to make $250 \; cm^3$ of solution. Find the concentration in $mol/dm^3$.

Volume $= \frac{250}{1000} = 0.25 \; dm^3$

$concentration = \frac{0.5}{0.25} = 2 \; mol/dm^3$

To convert between the two units, multiply by $M_r$:

$$\text{concentration } (g/dm^3) = \text{concentration } (mol/dm^3) \times M_r$$

Empirical Formula

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. You can work it out from masses or from percentage composition.

Worked example

A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula. ($A_r$: C = 12, H = 1)

Moles of C $= \frac{2.4}{12} = 0.2$

Moles of H $= \frac{0.8}{1} = 0.8$

Divide by the smallest (0.2): C $= 1$, H $= 4$.

Empirical formula $= CH_4$

If you are given percentages, treat them as if they were masses out of 100 g and follow exactly the same steps.

Moles of Gas and Molar Volume

At room temperature and pressure (rtp), one mole of any gas occupies the same volume: $24 \; dm^3$ (or $24\,000 \; cm^3$).

$$moles \; of \; gas = \frac{volume \; (dm^3)}{24}$$

Worked example

What volume does 0.5 mol of carbon dioxide occupy at rtp?

$volume = moles \times 24 = 0.5 \times 24 = 12 \; dm^3$

Real world

Mole calculations let manufacturers scale recipes precisely — from baking soda in food to the tonnes of ammonia made for fertilisers — so no expensive reactant is wasted.