$20.0 ext{ g}$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 ext{ g}

Question

$20.0	ext{ g}$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0	ext{ g}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of Mg = 24)

Accepted Answer

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

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