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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.
$O_2=60\%\;;\;B=40\%$
$O_2=40\%\;;\;B=60\%$
$O_2=70\%\;;\;B=30\%$
$O_2=35\%\;;\;B=65\%$
Solution
Mass of the compound $=0.24 \,g$
Mass of boron $=0.096\, g$
Mass of oxygen $=0.144\, g$
Percentage of boron $=$ $\frac {\rm {Mass\, of \,boron}}{\rm {Mass \,of \,compound}}$ $\times 100$
$ = \frac{{0.096\,g}}{{0.240\,g}} \times 100 = 40\%$
Percentage of oxygen $=$ $\frac {\rm {Mass\, of \,oxygen}}{\rm {Mass \,of \,compound}}$ $\times 100$
$ = \frac{{0.144\,g}}{{0.240\,g}} \times 100 = 60\%$
Alternative method
Percentage of oxygen $=$ $100$ percentage of boron
$=100 – 40 = 60\%$
