Calculations in chemistry: relative formula mass, percentage by mass, empirical formulae from masses, reacting masses from balanced equations, concentration in grams per cubic decimetre, and the mole.

What this dot point is asking

Edexcel wants you to calculate relative formula mass and percentage by mass, find empirical formulae from reacting masses or percentage composition (including the magnesium-oxide core practical), calculate reacting masses from balanced equations, calculate concentration in g/dm$^3$, and recall the definition of the mole. These calculations carry many marks and recur across the course; the mole-based work continues in separate chemistry.

Relative formula mass and percentage by mass

The relative formula mass ($M_r$) is found by adding the relative atomic masses of all the atoms in the formula. For $H_2SO_4$: $(2 \times 1) + 32 + (4 \times 16) = 98$.

The percentage by mass of an element in a compound is:

For example, the percentage of nitrogen in ammonium nitrate $NH_4NO_3$ ($M_r = 80$) is $\dfrac{2 \times 14}{80} \times 100 = 35\%$, because there are two nitrogen atoms.

Empirical formulae

The method works whether you are given masses or percentages:

1. Write the mass (or percentage) of each element.
2. Divide by the relative atomic mass to get moles of each.
3. Divide all the answers by the smallest to get the ratio.
4. If needed, scale up to whole numbers.

Reacting masses

A reacting-mass calculation relates the mass of one substance to the mass of another using the balanced equation. Use the three-step method, which works for any equation:

1. Convert the known mass to moles: $\text{moles} = \dfrac{\text{mass}}{M_r}$.
2. Use the mole ratio from the balanced equation.
3. Convert back to mass: $\text{mass} = \text{moles} \times M_r$.

The same steps work in grams, kilograms or tonnes, because the mole ratio is a pure number.

Concentration in g/dm3

Concentration measures how much solute is dissolved in a given volume:

Remember $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so divide a volume in cm$^3$ by $1000$ first. For example, $5$ g of salt in $250$ cm$^3$ is $5 / 0.25 = 20$ g/dm$^3$.

The mole

So one mole of carbon dioxide ($M_r = 44$) has a mass of $44$ g, and $\text{moles} = \dfrac{\text{mass}}{M_r}$.

Try this

Q1. Calculate the relative formula mass of $Ca(OH)_2$ ($A_r$: Ca = 40, O = 16, H = 1). [2 marks]

• Cue. $40 + 2 \times (16 + 1) = 40 + 34 = 74$.

Q2. A compound contains $2.4$ g of carbon and $0.8$ g of hydrogen. Find its empirical formula ($A_r$: C = 12, H = 1). [3 marks]

• Cue. Moles C $= 2.4 / 12 = 0.2$; moles H $= 0.8 / 1 = 0.8$; ratio $0.2 : 0.8 = 1 : 4$, so $CH_4$.

Q3. Calculate the concentration in g/dm$^3$ of $12$ g of solute in $400$ cm$^3$ of solution. [2 marks]

• Cue. $400$ cm$^3 = 0.4$ dm$^3$; concentration $= 12 / 0.4 = 30$ g/dm$^3$.