Question

A0.24 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​

Answer 1

Answer:

Percentage Composition(w/w) of Boron and Oxygen in the compound are 40 % and 60 % respectively.

Explanation:

GIVEN

Total mass of sample

M = 0.24 gram

Mass of Boron,

m(B) = 0.096 gram

Mass of Oxygen,

m(O) = 0.144 gram

CALCULATION

1. Percentage composition(w/w) of Boron in the compound

$= \frac{m(B)}{M} \times100$

$= \frac{0.096}{0.24} \times100$

= 40 %

2. Percentage composition(w/w) of Oxygen in the compound

$= \frac{m(O)}{M} \times100$

$= \frac{0.144}{0.24} \times100$

= 60 %

Answer 2

Answer:

Percentage Composition(w/w) of Boron and Oxygen in the compound are 40 % and 60 % respectively.

Explanation:

GIVEN

Total mass of sample

M = 0.24 gram

Mass of Boron,

m(B) = 0.096 gram

Mass of Oxygen,

m(O) = 0.144 gram

CALCULATION

1. Percentage composition(w/w) of Boron in the compound

= \frac{m(B)}{M} \times100=

M

m(B)

×100

= \frac{0.096}{0.24} \times100=

0.24

0.096

×100

= 40 %

2. Percentage composition(w/w) of Oxygen in the compound

= \frac{m(O)}{M} \times100=

M

m(O)

×100

= \frac{0.144}{0.24} \times100=

0.24

0.144

×100

= 60 %