Question

A 0.24g sample of compound of oxygen and Born was found by analysis to contain 0.096 of Born and 0.144 g of oxygen calculate the percentage composition of the compound by the weight​

Answer 1

Given,

• A 0.24 g of sample of compound of Oxygen and Boron has the composition Oxygen = 0.144 g and Boron = 0.096 g.

We have to find the percentage composition,

Which can be calculated as,

⇒ Mass % = Mass of Element in Compound / Mass of Compound × 100

So, Mass % of Oxygen

⇒ Mass % of Oxygen = 0.144 / 0.24 × 100

⇒ Mass % of Oxygen = 14.4 / 0.24

⇒ Mass % of Oxygen = 60%

Similarly, for Boron,

⇒ Mass % of Boron = 0.096 / 0.24 × 100

⇒ Mass % of Boron = 9.6 / 0.24

⇒ Mass % of Boron = 40%

So, The percentage composition of the given compound is,

• 60% Oxygen
• 40% Boron

Some Information :-

• PPM = Mass of solute / Mass of Solution × 10⁶
• PPB = Mass of solute / Mass of Solution × 10⁹

Answer 2

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Question:

A 0.24 g of sample of compound of Oxygen and Boron has the composition Oxygen = 0.144 g and Boron = 0.096 g.

For finding the percentage composition,

First we have to calculate ,

⇒ Mass % = Mass of Element in Compound / Mass of Compound × 100

Hence , Mass % of Oxygen

⇒ Mass % of Oxygen = 0.144 / 0.24 × 100

⇒ Mass % of Oxygen = 14.4 / 0.24

⇒ Mass % of Oxygen = 60%

Similarly, for Boron,

⇒ Mass % of Boron = 0.096 / 0.24 × 100✔

⇒ Mass % of Boron = 9.6 / 0.24✔

⇒ Mass % of Boron = 40%✔

Hence , The percentage composition of the given compound is,

60% Oxygen✔

40% Boron ✔

extra Information :-

• PPM = Mass of solute / Mass of Solution × 10⁶
• PPB = Mass of solute / Mass of Solution × 10⁹