Question.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:
Mass of boron $=0.096 \mathrm{~g}$ (Given)
Mass of oxygen $=0.144 \mathrm{~g}$ (Given)
Mass of sample $=0.24 \mathrm{~g}$ (Given)
Percentage of boron $=\frac{\text { Mass of boron }}{\text { Mass of sample }} \times 100$
$=\frac{0.096}{0.24} \times 100=40 \%$
Percentage of oxygen $=\frac{\text { Mass of oxygen }}{\text { Massof sample }} \times 100$
$=\frac{0.144}{0.24} \times 100=60 \%$
Thus, percentage of boron by weight in the compound = 40%
And, percentage of oxygen by weight in the compound = 60%
Mass of boron $=0.096 \mathrm{~g}$ (Given)
Mass of oxygen $=0.144 \mathrm{~g}$ (Given)
Mass of sample $=0.24 \mathrm{~g}$ (Given)
Percentage of boron $=\frac{\text { Mass of boron }}{\text { Mass of sample }} \times 100$
$=\frac{0.096}{0.24} \times 100=40 \%$
Percentage of oxygen $=\frac{\text { Mass of oxygen }}{\text { Massof sample }} \times 100$
$=\frac{0.144}{0.24} \times 100=60 \%$
Thus, percentage of boron by weight in the compound = 40%
And, percentage of oxygen by weight in the compound = 60%
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