Question

50 g of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and 22.4 g calcium oxide. The percentage purity of calcium carbonate in the sample is

Answer 1

Answer: 80% of the calcium carbonate is pure.

Explanation: Decomposition of calcium carbonate is written as:

$CaCO_3\overset{\Delta }{\rightarrow}CaO+CO_2$

Molar mass of $CaCO_3$ = 100g/mol

Molar mass of CaO = 56g/mol

Let the pure sample of $CaCO_3$ = y grams.

Mass of CaO produced after decomposition = 22.4 grams

If 100g of pure sample of $CaCO_3$ produces 56g of CaO, then "y" gram of pure sample of $CaCO_3$ produces 22.4 grams of CaO.

Calculating for y, we get

$y=\frac{100\times 22.4}{56}$

y = 40 grams

% purity of $CaCO_3$ can be calculated by:

$\% purity=\frac{\text{mass of pure sample}}{\text{given impure mass}}\times 100$

Putting the values, we get

$\%purity=\frac{40}{50}\times 100$

% purity of $CaCO_3$ in the sample = 80%