Calculate the mass percent of $O$ present in sodium sulphate $({N a}_{2} {S O}_{4})$ .

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Solution

The molar mass of sodium sulphate is $2 (23) + 32 + 4 (16) = 46 + 32 + 64 = 142 g / m o l$

Mass percentage $= \frac{Mass of oxygen}{Mass of N a_{2} S O_{4}} \times 100$

The mass percentage of oxygen in sodium sulphate $= \frac{64}{142} \times 100 = 45.07 %$ .

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