Mass of boron = 0.096g (Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound ={0.096×100\over0.24}%
= 40%
Thus, percentage of oxygen by weight in the compound ={0.144×100\over0.24}%
= 60 %