Question

SOC (CH, CH2), NH -(CH₃ CH₂)₂NH₂ cr CN(CH₂ CH₂), - HD, -50, mame molecular formula Quantity density maar gwel thinyl chloride sou, 118.97 g/mol 1.5 mL (1.631 g/mL m-toluic acid Coty | 136.15 g/ml 1.35 g (1.054 g/ml) DMF (CH.), NCH | 73.09g/mol I drop (0.948 glane diethylamine (CH₃ CH2)₂ NH | 73,14 g lomol 5 mL 0.707g/ml diethyl (Cats) 1 74.12g/mol 20mL 0.706 glu
The top half of the image depicts the reaction of the synthesis of N,N-diethyl-m-toluamide (DEET). The bottom half of the image attached shows the table of chemicals used in the reaction. How would I calculate the theoretical yield and furthermore, the percent yield? This experiment had a lot of chemicals used, and I don't know where to start.

Answer 1

I can help you in calculating the theoretical yield as the percent yield is obtained from the observed weight of the product and taking the ratio of the actual yield and the theoretical yield. Here the major reactants are m-toluic acid, thionyl chloride and diethyl amine. 1 mole m-toluic acid reacts with 1 mole of thionyl chloride to form 1 mole of m-toluic chloride which reacts with 1 mole of diethylamine to form 1 mole of N, N-diethyl-m-toluamide. Moles of m-toluic acid= 1.35/136.15 moles= 0.0099 moles Moles of thionyl chloride= (1.5x 1.631) g/118.97 = 0.02 moles So, m-toluic acid is limiting reagent and the theoretical yield of m-toluic chloride is 0.0099 moles. Moles of diethylamine = (5x0.707) g / 73.14 = 0.0488 moles So, m-toluic chloride is limiting reagent and the theoretical yield of N, N- diethyl-m-toluamide is 0.0099 moles. The molecular weight is 191.27 g/mol. So theoretical yield is = 0.0099x191.27 = 1.89 g.