$16.8 \mathrm{~g}$ of an impure magnesium carbonate ($\mathrm{MgCO}_3$) sample completely decomposes

Question

$16.8 \mathrm{~g}$ of an impure magnesium carbonate ($\mathrm{MgCO}_3$) sample completely decomposes on heating to give carbon dioxide and exactly $4.0 \mathrm{~g}$ of solid magnesium oxide. What is the percentage purity of the magnesium carbonate in the sample? (Atomic mass of $\mathrm{Mg}=24, \mathrm{C}=12, \mathrm{O}=16$)

Professor PC Thomas Classes · Answer

(b) : The balanced decomposition reaction is $\mathrm{MgCO}_3 \xrightarrow{\Delta} \mathrm{MgO} + \mathrm{CO}_2$. The molar mass of $\mathrm{MgCO}_3$ is $84 \mathrm{~g/mol}$ and $\mathrm{MgO}$ is $40 \mathrm{~g/mol}$. According to stoichiometry, $40 \mathrm{~g}$ of $\mathrm{MgO}$ is produced by the decomposition of exactly $84 \mathrm{~g}$ of pure $\mathrm{MgCO}_3$. Therefore, $4.0 \mathrm{~g}$ of $\mathrm{MgO}$ is produced from $\frac{84}{40} 	imes 4.0 = 8.4 \mathrm{~g}$ of pure $\mathrm{MgCO}_3$. The percentage purity of the sample is $\frac{	ext{Mass of pure substance}}{	ext{Total mass of sample}} 	imes 100 = \frac{8.4}{16.8} 	imes 100 = 50\%$.