1-methoxybutane (butyl methyl ether) was synthesized by the following reaction C
ID: 588334 • Letter: 1
Question
1-methoxybutane (butyl methyl ether) was synthesized by the following reaction CH OH Use the data from the Aldrich Catalog below to calculate the following Sodium methoxide FW 54.02,d 0.945 1-Bromobutane, FW 137.03, d 1.276 Butyl methyl ether, FW 88.15, d 0.744 a) (4pts)A student used 20 ml of 1-bromobutane in this experiment. How many grams and how many moles of bromobutane did she use? b) (4pts)She used 45 ml of a 25% (wt/wt% solution) of sodium methoxide in methanol. Use the density of this solution to calculate the mass of sodium methoxide she used. Use the mass to calculate the number of moles of sodium methoxide she used. (4pts)Compare the number of moles of reactants and decide which is the limiting reagent. In other words, how many moles of product are theoretically possible? c) (4pts)How many grams of product are theoretically possible? This is commonly called the theoretical yield. d) e) (4pts)If a student obtained 12 g of product, what is the % yield?Explanation / Answer
from equation : 1 mol 1-bromobutane = 1 mol NaOCH3
a) mass of 1-bromobutane used = 20*1.276 = 25.52 mol
no of mol of 1-bromobutane used = 25.52/137.03 = 0.186 mol
b) mass of NaOCH3 solution she used = 45*0.945 = 42.525 g
mass of NaOCH3 = 25*42.525/100 = 10.63 g
no of mol of NaOCH3 = 10.63/54 = 0.197 mol
c) limiting reactant = 1-bromobutane
d) no of mol of product possible theoretically = 0.186 mol
mass of product possible theoretically = 0.186*88.15 = 16.4 g
e) % yield = P.Y/T.Y*100
= 12/16.4*100
= 73.17 %
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