Question

1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br (l) If 18.54 mL of 1-butanol produced 15.65 g of 1-bromobutane, the percentage yield of the product equals: (Assume the density of 1-butanol is 0.81 g/mL, the molar mass of 1-butanol is 74 g/mol, and the molar mass of 1-bromobutane is 137 g/mol.)

Answer 1

Answer:

The answer to your question is: yield = 56.27%

Explanation:

Data

             CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br

                18.54 ml 1-butanol            15.65 g of 1-bromobutane

% yield = ?

density = 0.81 g/ml

MM = 74 g  1- butanol

MM = 137 g 1-bromobutane

Process

Calculate mass of 1- butanol

              density = mass/volume

              mass = density x volume

              mass = 0.81 x 18.54

              mass = 15.02 g of 1-butanol

Theoretical yield

              74 g of 1- butanol -----------------  137 g of 1-bromobutane

              15.02 g of 1- butanol -------------   x

              x = (15.02 x 137) / 74

              x = 27.81 g of 1-bromobutane

% yield = experimental yield / theoretical yield x 100

% yield = 15.65 / 27.81 x 100

% yield = 56.28