Exam Style Question - 4

QuestionA student made 1-bromobutane from 7.40 g of butan-1-ol. After purification, 9.60 g of 1-bromobutane was obtained.C₄H₉OH + HBr → C₄H₉Br + H₂OM_r: butan-1-ol = 74.0; 1-bromobutane = 136.9(a) Calculate the percentage yield.(b) Suggest two reasons why the yield is less than 100%.Answer(a) Step 1: moles of butan-1-oln = m ÷ M_r = 7.40 ÷ 74.0 = 0.100 molStep 2: the reaction ratio is 1:1, so the theoretical amount of 1-bromobutane is 0.100 mol.Step 3: theoretical massm = n × M_r = 0.100 × 136.9 = 13.7 gStep 4: percentage yieldpercentage yield = (actual mass ÷ theoretical mass) × 100= (9.60 ÷ 13.69) × 100 = 70.1%(b) Any two of:• the reaction did not go to completion• side reactions formed other products (for example, elimination to form but-1-ene)• product was lost when transferring between apparatus, in the separating funnel or on the drying agent• some product was lost as vapour during distillationExaminer tip: always calculate the theoretical yield from the limiting reagent and keep unrounded values until the final answer, then give it to three significant figures.