Mass of the given sample compound = 0.24g
Mass of boron in the given sample compound = 0.096g
Mass of oxygen in the given sample compound = 0.144g
% composition of compound = % of boron and % of oxygen
Therefore % of boron = (mass of boron)/(mass of the sample compound) x 100
= 40%
Therefore % of oxygen = (mass of oxygen) / (mass of the sample compound) x 100
= 60%