Question

a 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen.calculate the percentage composition of the compound by weight.

Answer 1

Percentage composition=M(element)/M(compound)*100

Boron=0.096/0.24 *100=40%

Oxygen=0.144/0.24 *100=60%

Answer 2

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Given,

First case :

• mass of boron = 0.096g
• mass of the compound = 0.24g

Second case :

• mass of oxygen = 0.144g
• mass of the compound = 0.24g

We know ,

percentage of boron

=>  $\frac{mass of the boron }{mass of the compound } \times 100$

=> $\frac{0.096g}{0.24g} \times 100$

=> 40% Answer

percentage of oxygen

=> $\frac{mass of oxygen  }{mass of the compound }\times 100$

=> $\frac{0.144g}{0.24g}\times 100$

=> 60 % Answer