The balanced chemical equation for the decomposition of magnesium carbonate is: $\text{MgCO}_3(s) \xrightarrow{\Delta} \text{MgO}(s) + \text{CO}_2(g)$ From the stoichiometry of the reaction, $1\text{ mole}$ of $\text{MgCO}_3$ yields $1\text{ mole}$ of $\text{MgO}$. Molar mass of $\text{MgCO}_3 = 24 + 12 + (3 \times 16) = 84\text{ g/mol}$. Molar mass of $\text{MgO} = 24 + 16 = 40\text{ g/mol}$. According to the equation, $40\text{ g}$ of $\text{MgO}$ is obtained from $84\text{ g}$ of pure $\text{MgCO}_3$. Therefore, $8.0\text{ g}$ of $\text{MgO}$ will be obtained from $= \left(\frac{84}{40}\right) \times 8.0 = 16.8\text{ g}$ of pure $\text{MgCO}_3$. The mass of pure $\text{MgCO}_3$ in the sample is $16.8\text{ g}$. The total mass of the impure sample is given as $20.0\text{ g}$. Percentage purity $= \left(\frac{\text{Mass of pure MgCO}_3}{\text{Total mass of the sample}}\right) \times 100 = \left(\frac{16.8}{20.0}\right) \times 100 = 84\%$.