## Chemistry: Principles and Practice (3rd Edition)

Mass percentage of an element = (mass of that element in the compound × 100)/(molar mass of the compound) Molar mass of acetone ($C_{3}H_{6}O$) = $(3\times12.011g)+(6\times1.008g)+(1\times15.999g)$ = 58.08 g Mass % of C = $\frac{(3\times12.011g)\times100}{58.08g}$ = 62.04% Mass % of H = $\frac{(6\times1.008g)\times100}{58.08g}$ = 10.41% Mass % of O = $\frac{15.999g\times100}{58.08g}$ = 27.55%