Quantitative chemistry

Moles, the mole calculations, empirical formulae, reacting masses, gas volumes, concentrations and yield.

11 min read

The relative molecular/formula mass $M_{r}$ is the sum of the $A_{r}$ values. One mole is the $A_{r}$ or $M_{r}$ in grams and contains Avogadro's number of particles. So $moles = \frac{mass}{M _{r}}$ .

Worked example

How many moles are in $36 g$ of water ( $M_{r} = 18$ )?

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

Percentage composition by mass of an element $= \frac{n \times A _{r}}{M _{r}} \times 100$ . The empirical formula is the simplest whole-number ratio of atoms; find it by dividing each element's mass (or %) by its $A_{r}$ and simplifying. The molecular formula is a whole-number multiple of it (use $M_{r}$ ).

Worked example

A compound is $40%$ C, $6.7%$ H, $53.3%$ O. Find the empirical formula.

Divide by $A_{r}$ : C $\frac{40}{12} = 3.3$ , H $\frac{6.7}{1} = 6.7$ , O $\frac{53.3}{16} = 3.3$ . Ratio $1 : 2 : 1$ → $CH_{2} O .$

Use the balanced equation's mole ratio to find masses of reactants and products. The limiting reactant is the one that runs out first, and it sets how much product forms.

Worked example

What mass of MgO forms from $48 g$ of Mg? ( $2 Mg + O_{2} \to 2 MgO$ ; $A_{r}$ : Mg 24, O 16.)

Moles Mg $= \frac{48}{24} = 2$ . Ratio Mg:MgO is $1 : 1$ , so $2$ mol MgO; mass $= 2 \times 40 = 80 g .$

At room temperature and pressure, one mole of any gas occupies $24 dm^{3}$ , so $moles = \frac{volume}{24}$ . Concentration is measured in $mol dm^{- 3}$ or $g dm^{- 3}$ : $concentration = \frac{moles}{volume in dm ^{3}}$ . A saturated solution holds the maximum dissolved solute.

In a titration, use the concentration and volume of one solution plus the equation's mole ratio to find the other. Percentage yield $= \frac{actual yield}{predicted yield} \times 100$ .

Worked example

$25 cm^{3}$ of NaOH is neutralised by $20 cm^{3}$ of $0.1 mol dm^{- 3}$ HCl ( $1 : 1$ ). Find the NaOH concentration.

Moles HCl $= 0.1 \times \frac{20}{1000} = 0.002$ . Same moles NaOH in $0.025 dm^{3}$ : $\frac{0.002}{0.025} = 0.08 mol dm^{- 3} .$

Exam tip

Always convert volumes to $dm^{3}$ (divide cm³ by 1000) before using concentration. Mixing up cm³ and dm³ is the classic 1000× error.

Using cm³ instead of dm³ in concentration.
Ignoring the mole ratio from the balanced equation.
Forgetting to check which reactant is limiting.

Quantitative chemistry

Relative mass & the mole

Composition & empirical formula

Reacting masses & limiting reactant

Gas volumes & solutions

Titrations & percentage yield

Common mistakes