Question

At the start of Lab 7, a student was given 5.7 mL of 1-butanol. He converted the 1-butanol into the alkyl halide using HBr. At the end of the experiment the student obtained 5.5 mL of product. Assuming 1-butanol is the limiting reagent, calculate the theoretical (using grams) and percent yields. Remember to include references and show all calculations and units.

Answer 1

The balanced chemical equation is -

CH3CH2CH2CH2OH + HBr à C4H9Br + H2O

We need to know the following –

Density of butanol (0.81 g/ml) and butyl bromide (1.27 g/ml)

Molar mass of butanol (74.12 g/mol) and butyl bromide (137 g/mol).

According to the balanced equation 1 mol of butyl bromide is produced from 1 mol of butanol.

1-Butanol = 5.7 ml = 5.7 ml *0.81 g/ml (density of 1-butanol = 0.81 g/ml) = 4.62 g = 4.62 g/74.12 g/mol = 0.062 mols.

As 1-butanol is the limiting reagent, the maximum alkyl halide that can form = 0.062 mols

So, 100% yield or theoretical yield = 137 g/mol*0.062mol = 8.49 g

Actual product obtained = 5.5 ml of Butylbromide = 5.5 ml *1.27 g/ml = 6.98 g

So, the percentage yield = 100*6.98/8.49 % = 82.21%