1g ~of an impure sample of magnesium carbonate (containing no...

Consider,

${MgCO}_{3} \to MgO + {CO}_{2} ↑$

${MgCO}_{3}$:~\left(1 \times 24\right) + \left(1 \times 12\right) + \left(3 \times 16\right) = 84g$</p><p>${CO}_{2}$:~</p><p>\left(1 \times 12\right) + \left(2 \times 16\right) = 44g$

$100$~pure $84g$~${MgCO}_{3}$ on heating gives $44g$~${CO}_{2}$

Given that~$1g$ ~of ${MgCO}_{3}$~on heating gives $0 . 44g$~${CO}_{2}$

Therefore, $84g$~${MgCO}_{3}$~sample on heating gives $36 . 96g$~${CO}_{2}$

Percentage of purity of the sample =~$\dfrac{36 . 96gof{CO}_{2}}{44gof{CO}_{2}} \times 100$

Percentage of impurity =~$16$