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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.
Answers
Answer :
Given :-
Mass of the compound = 0.24g
Mass of boron = 0.096g
Mass of oxygen = 0.144g
Now ,
Percentage of Boron = Mass of boron in compound/ Mass of compound*100
= 0.096 g / 0.024 g * 100
= 0.4 *100
= 40
Again ,
= Mass of oxygen in compound/Mass of compound *100
= 0.144 g/0.24 g *100
= 0.6 *100
= 60
Answer:
Total mass of compound = 0.24 g
Mass of boron in it = 0.096 g
Therefore % composition by weight of boron = (Mass of boron/ Total mass of compound)*100
= (0.096/0.24)*100
= 40 %
Mass of oxygen in the compound = 0.144 g
Therefore % composition by weight of oxygen = (Mass of oxygen/ Total mass of compound)*100
= (0.144/0.24)*100
= 60 %.
The compound is made up of only boron and oxygen, therefore the % composition of the componenets sum to 100% (60% + 40%)