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

(i) Mass of boron in compound = 0.096 g

And, Mass of compound = 0.24 g

So, Percentage of boron = Mass of boron in compound/Mass of compound × 100

= 0.096/0.24 × 100

= 40   ……………….(1)

(ii)  Mass of oxygen in compound = 0.144 g

And, Mass of compound = 0.24 g

So, Percentage of oxygen = Mass of oxygen in compound/Mass of compound × 100

= 0.144/0.24 × 100

= 60   …………..(2)

Thus, the percentage composition of the compound is: Boron = 40%; Oxygen = 60%