Question

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:-

• Weight of Compound = 0.24g

• Weight of Boron = 0.096g

• Weight of Oxygen = 0.144g

To Find:-

• The Percentage Composition of Compound by weight.

Formulae used:-

• $\sf{Percentage\:Composition = \dfrac{Weight\:of\: elements\times{100}}{weight\:of\:Compound}}$

Now,

→ weight of Compound = 0.24g

→ weight of Boron element = 0.096g

→ $\sf{Percentage\:Composition = \dfrac{Weight\:of\:Boron\times{100}}{weight\:of\:Compound}}$

→ $\sf{ Percentage\: Composition = \dfrac{0.096\times{100}}{0.24}}$

→ $\sf{\dfrac{9.6}{0.24}}$

→$\sf{ 40}$

Hence, The Percentage Composition of Boron in Compound is 40%.

→ weight of Compound = 0.24g

→ weight of oxygen element = 0.144g

→ $\sf{Percentage\:Composition = \dfrac{Weight\:of\:oxygen\times{100}}{weight\:of\:Compound}}$

→ $\sf{ Percentage\: Composition = \dfrac{0.144\times{100}}{0.24}}$

→ $\sf{\dfrac{14.4}{0.24}}$

→$\sf{60}$

Hence, The Percentage Composition of Oxygen in Compound is 60%.

Answer 2

GIVEN :-

• Mass of Compound = 0.24 g

• Mass of Oxygen = 0.144 g

• Mass of Boron = 0.096 g

TO FIND :-

• The percentage composition of the compound by weight .

SOLUTION :-

Fᴏʀ Bᴏʀᴏɴ :-

$\red\checkmark\:\bf{\purple{\%\:of\:Boron\:=\:\dfrac{Mass\:of\:Boron}{Mass\:of\:Compound}\times{100}\:}}$

• $\bf\red{Mass\:of\:Boron}$ = 0.096 g .

• $\bf\red{Mass\:of\:Compound}$ = 0.24 g .

$\rm{:\implies\:\%\:of\:Boron\:=\:\dfrac{0.096}{0.24}\times{100}\:}$

$\rm{:\implies\:\%\:of\:Boron\:=\:0.4\times{100}\:}$

$\bf\green{:\implies\:\%\:of\:Boron\:=\:40\%\:}$

Fᴏʀ Oxʏɢᴇɴ :-

$\red\checkmark\:\bf{\purple{\%\:of\:Oxygen\:=\:\dfrac{Mass\:of\:Oxygen}{Mass\:of\:Compound}\times{100}\:}}$

• $\bf\red{Mass\:of\:Oxygen}$ = 0.144 g .

• $\bf\red{Mass\:of\:Compound}$ = 0.24 g .

$\rm{:\implies\:\%\:of\:Oxygen\:=\:\dfrac{0.144}{0.24}\times{100}\:}$

$\rm{:\implies\:\%\:of\:Oxygen\:=\:0.6\times{100}\:}$

$\bf\green{:\implies\:\%\:of\:Oxygen\:=\:60\%\:}$

$\huge\red\therefore$ Tʜᴇ ᴘᴇʀᴄᴇɴᴛᴀɢᴇ ᴏғ Oxʏɢᴇɴ ɪɴ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪs "60%" .

$\huge\red\therefore$ Tʜᴇ ᴘᴇʀᴄᴇɴᴛᴀɢᴇ ᴏғ Bᴏʀᴏɴ ɪɴ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪs "40%" .