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:
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%)
Explanation:
Please mark me the brainliest answer
Answer:
Mass of compound = 0.24 g and, mass of boron = 0.096 g percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% mass of oxygen = 0.144 g again, mass of compound = 0.24 g percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60% thus ...