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.​

Answer 1

GIVEN :-

• Mass of the compound = 0.24 g
• Mass of boron in the compound = 0.096 g
• Mass of oxygen in the compound = 0.144 g

TO FIND :-

• Percentage composition of the compound

SOLUTION :-

Percentage composition of an element in the compound is given by ,

$\large {\underline {\boxed {\bigstar { \red {\sf{ \: p = \dfrac{mass \: of \: the \: particular \: element \: in \: compound}{mass \: of \: the \: compound} \times 100}}}}}}$

Percentage composition of oxygen in the Compound ,

$\implies \large \sf \: P_1 = \dfrac{0.144}{0.24} \times 100 \\ \\ \implies \sf \: P_1 = \frac{14.4}{0.24} \\ \\ \implies {\underline {\boxed {\blue {\sf {\: P_1 = 60\%}}}}}$

Percentage composition of boron in the compound is,

$\implies \large \sf \:P_2 = \frac{0.096}{0.24} \times 100 \\ \\ \implies \sf \: P_2 = \frac{9.6}{0.24} \\ \\ \implies {\underline {\boxed {\blue {\sf{P_2 = 40\%}}}}}$

∴ The compound contains 60% of oxygen and 40% of boron

Answer 2

Answer:

GIVEN :-

Mass of the compound = 0.24 g

Mass of boron in the compound = 0.096 g

Mass of oxygen in the compound = 0.144 g

TO FIND :-

Percentage composition of the compound

SOLUTION :-

Percentage composition of an element in the compound is given by ,

\large {\underline {\boxed {\bigstar { \red {\sf{ \: p = \dfrac{mass \: of \: the \: particular \: element \: in \: compound}{mass \: of \: the \: compound} \times 100}}}}}}

★p=

massofthecompound

massoftheparticularelementincompound

×100

Percentage composition of oxygen in the Compound ,

\begin{gathered} \implies \large \sf \: P_1 = \dfrac{0.144}{0.24} \times 100 \\ \\ \implies \sf \: P_1 = \frac{14.4}{0.24} \\ \\ \implies {\underline {\boxed {\blue {\sf {\: P_1 = 60\%}}}}}\end{gathered}

⟹P

1

=

0.24

0.144

×100

⟹P

1

=

0.24

14.4

⟹

P

1

=60%

Percentage composition of boron in the compound is,

\begin{gathered} \implies \large \sf \:P_2 = \frac{0.096}{0.24} \times 100 \\ \\ \implies \sf \: P_2 = \frac{9.6}{0.24} \\ \\ \implies {\underline {\boxed {\blue {\sf{P_2 = 40\%}}}}}\end{gathered}

⟹P

2

=

0.24

0.096

×100

⟹P

2

=

0.24

9.6

⟹

P

2

=40%

∴ The compound contains 60% of oxygen and 40% of boron