Amines Part 3

Concept:Acetylation of aniline with acetic anhydride gives acetanilide in a 1:1 mole ratio.The percent yield is the ratio of actual mass to theoretical mass, multiplied by 100.Explanation:The reaction is: $\mathrm{C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH}$Molar mass of aniline ($\mathrm{C_6H_7N}$) = $93\ \mathrm{g\ mol^{-1}}$Moles of aniline taken = $\frac{9.3}{93} = 0.10\ \mathrm{mol}$Molar mass of acetanilide ($\mathrm{C_8H_9NO}$) = $135\ \mathrm{g\ mol^{-1}}$Theoretical mass of acetanilide = $0.10 \times 135 = 13.5\ \mathrm{g}$Actual mass obtained = $11\ \mathrm{g}$Percent yield = $\frac{11}{13.5} \times 100 = 81.5\%$Answer:Option D: $81.5\%$