Question

1. 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.

Please help me ​

Answer 1

Explanation:

Mass of compound = 0.24 g and, mass of boron = 0.096 g percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% mass of oxygen = 0.144 g again, mass of compound = 0.24 g percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60% thus

the percentage of oxygen and boron in the compound is 60% and 40% respectively.

Answer 2

Explanation:

(i)Mass of boron in compound is 0.096g

and, Mass of compound is 0.24g

So, Percentage of boron = (Mass of boron in compound/Mass of compound)*100

=0.096/0.24. *100

= 40

(ii) Mass of oxygen in compound is 0.144g

Mass of compound is 0.24g

Percentage of oxygen= Mass of oxygen in compound/Mass of compound. *100

=0.144/0.24. *100

60

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