Chapter 3 - Equations, the Mole, and Chemical Formulas - Questions and Exercises - Exercises - Page 134: 3.79

62.04% carbon, 10.41% hydrogen and 27.55% oxygen.

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%