0.24g of a sample of compound having boron and oxygen was found on analysis to contain 0.096g of boron and oxygen calculate the percentage comparison of the compound
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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, the percentage of oxygen and boron in the compound is 60% and 40% respectively.
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