If you start with 0.173 g of anisole and end up with 0.105 g of o-bromoanisole, what is the percent yield of o-bromoanisole?
mass of anisole = 0.173 g
moles of anisole = 0.173 / 107.13 = 1.615 x 10^-3 mol
1 mol anisole gives --------------> 1 mol o-bromoanisole
1.615 x 10^-3 mol ----------------> 1.615 x 10^-3 mol o-bromoanisole
mass of o-bromoanisole = 1.615 x 10^-3 x 187.036
= 0.302 g
percent yield = (actual / theoretical ) x 100
= (0.105 / 0.302 ) x 100
= 34.77 %
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