An organic compound undergoes elemental analysis and yields C: $40.0\%$, H: $6.67\%$, and the remain

Question

An organic compound undergoes elemental analysis and yields C: $40.0\%$, H: $6.67\%$, and the remainder is oxygen. The correct empirical formula of the compound is

Professor PC Thomas Classes · Answer

(c) : The percentage of oxygen = $100 - (40.0 + 6.67) = 53.33\%$. To find the empirical formula, divide each percentage by its atomic mass to get relative moles: Moles of C = $\frac{40.0}{12} = 3.33$; Moles of H = $\frac{6.67}{1} = 6.67$; Moles of O = $\frac{53.33}{16} = 3.33$. The simplest whole-number ratio is obtained by dividing by the smallest value ($3.33$): C = 1, H = 2, O = 1. Thus, the empirical formula is $\mathrm{CH}_2\mathrm{O}$.