Question

A O.24g SAMPLE OF COMPOUND OF OXYGEN AND BORON WAS FOUND BY ANALYSIS TO CONTAIN 0.144g OF OXYGEN. THEN CALCULATE THE PERCENTAGE COMPOSITION OF THE COMPOUND BY WEIGHT.

Answer 1

Total amount of sample = 0.24 g
As the amount of oxygen present in the given sample = 0.144 g
Therefore, amount of boron present in the same sample = (0.24 - 0.144)
= 0.096
Percent composition by weigh or mass composition = (Mass of the element / Total mass of the sample ) $\times$ 100 %
Mass percent of oxygen = (0.144/0.24 ) ​$\times$ 100 %
= 60 %
Mass percent of boron = (100-60) %
= 40 %
Alternatively,
Mass percent of boron = (0.096/0.24) ​​$\times$ 100 %
= 40 %