Relative masses of atoms and molecules — practice questions

To find the weighted average mass, we multiply the mass of each isotope by its fractional abundance and sum the results.

Step 1: Convert percentages to fractional abundances. Abundance of $^{10}$B = 19.9 / 100 = 0.199 Abundance of $^{11}$B = 80.1 / 100 = 0.801

Step 2: Calculate the weighted average. $A_r = (\text{abundance of } ^{10}\text{B} \times \text{mass of } ^{10}\text{B}) + (\text{abundance of } ^{11}\text{B} \times \text{mass of } ^{11}\text{B})$ $A_r = (0.199 \times 10.01) + (0.801 \times 11.01)$ $A_r = 1.99199 + 8.81901$ $A_r = 10.811$

Step 3: Round to the required number of decimal places. $A_r = 10.8$ (to 1 d.p.)

The formula shows one unit of copper(II) sulfate is associated with five molecules of water. The dot (·) means we add the masses together.

Step 1: Calculate the $M_r$ of the anhydrous part (CuSO₄). $M_r(\text{CuSO}_4) = A_r(\text{Cu}) + A_r(\text{S}) + 4 \times A_r(\text{O})$ $M_r(\text{CuSO}_4) = 63.5 + 32.1 + (4 \times 16.0)$ $M_r(\text{CuSO}_4) = 63.5 + 32.1 + 64.0 = 159.6$

Step 2: Calculate the mass of the water of crystallisation (5H₂O). $M_r(\text{H}_2\text{O}) = (2 \times A_r(\text{H})) + A_r(\text{O}) = (2 \times 1.0) + 16.0 = 18.0$ Mass of 5H₂O = $5 \times 18.0 = 90.0$

Step 3: Add the two parts together. $M_r(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}) = 159.6 + 90.0 = 249.6$

The relative formula mass is 249.6.