Question

laboratory work. We only are doing molecular modeling work. The laboratory work involves refluxing menthone in a mixture of glacial acetic acid and 1.0M hydrochloric acid. After the equilbirium is established where menthone and isomenthone are both present, these molecules may be extracted and the mixture isolated so the optical rotation of the mixture can be measured. In reported results, the experimental optical rotation of the menthone and isomenthone mixture was measured and found to be A=-3.0 degrees. The specific rotation of the isomenthone is Ai= +92 degrees and menthone is Am = -30 degrees. The measured angle of rotation, A, for this mixture is given by A = (Am)(Xm ) + (Ai )(Xi) or combining with Eq. (4), the above can be written as A = (Am) Xm + (Ai ) (1- Xm) (6) Use the information above to calculate the experimental mole fraction of menthone, Xm, and isomenthone, Xi. Next calculate the experimental equilibrium constant K. See the Special Answer sheet to fill in your results. These will be compared to your modelling results.

Answer 1

The measured angle of rotation, A , of the mixture is given by

A = (Am) (Xm) + (Ai) (Xi)

But, sum of the mole fractions of the substances in a mixture is equal to 1 i.e Xm + Xi = 1

     Xi =1-Xm

Now, A = (Am) (Xm) + (Ai) (1-Xm)

The measured angle of rotation, A , of the mixture is -3.0 degrees

The specific rotation of the menthone is Am= -30 degrees

The specific rotation of the isomenthone is Ai = +92 degrees

-3.0 = (Xm)(-30) + 92(1-Xm)

mole fraction of menthone, Xm = 0.779

mole fraction of isomenthone, Xi = 1-0.779 = 0.221

The equilibrium constant is the ratio of mole fraction of isomenthone to the mole fraction of menthone

The equilibrium constant, K = 0.221/0.779 = 0.283