Question

A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.

Answer 1

Given that:

The Mass of the sample = 0.24g

Mass of boron in the given sample = 0.096 g

Mass of oxygen in the given sample = 0.144 g

% composition of compound = [% of boron] + [% of oxygen]

Therefore,

$\%\mathrm{composition}\mathrm{of}\mathrm{boron}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{boron}}{\mathrm{Mass}\mathrm{of}\mathrm{sample}\mathrm{compound}}\times 100$

$\%\mathrm{composition}\mathrm{of}\mathrm{boron}=\frac{0.096}{0.24}\times 100=40\%$

Therefore,

$\%\mathrm{composition}\mathrm{of}\mathrm{oxygen}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{oxygen}}{\mathrm{Mass}\mathrm{of}\mathrm{sample}\mathrm{compound}}\times 100$

$\%\mathrm{composition}\mathrm{of}\mathrm{oxygen}=\frac{0.144}{0.24}\times 100=60\%$

The percentage composition of the boron and oxygen in the compound by weight is 40% and 60% respectively.