toddlail - 29-11-2006 at 18:16
Given the following information, calculate the experimental percent yield of 1-bromobutane:
A student refluxed for 45 min a mixture of 7.5 mL of 1-butanol (MW=74.12 g/mol, d=0.810 g/mL), 15 g of sodium bromide (MW=102.9 g/mol) in 15 mL of
water, and 12 mL of 18M sulfuric acid. After the reaction was worked up and the product was purified (using simple distillation) the student obtained
4.430 g of pure 1-bromobutane (MW=137.03 g/mol, d=1.276 g/mL).
Baphomet - 29-11-2006 at 20:01
7.5 mL of 1-butanol = [(7.5 * 0.810) / 74.12] moles = 81.96 mmol
15 g of sodium bromide = 145.77 mmol
The equation proceeds thus:
C4H9OH + NaBr -> C4H9Br + NaOH
(Side reaction of NaOH and H2SO4 serves to speed the first reaction)
Each mole of 1-butanol is the limiting factor and yields one mole of 1-bromobutane theoretically, since NaBr is in excess
So the theoretical yield is 81.96 mmol = (81.96 * 137.03) / 1000 = 11.23 grams
The percentage of theoretical yield that was obtained is therefore: 4.430 / 11.23 = 39.45%