A pure sample of the R enantiomer of a compound has a specific rotation, [ α], of + 20 °. A solution containing 0.2 g/mL of a mixture of enantiomers rotates plane polarized light by - 2 ° in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?

The specific rotation of the sample is - 2 degrees/0.2 g/mL of mixture

This equals - 10 degrees/g/mL of sample.

let the proportion of the R (+) enantiomer be x. The proportion of the S (-) enantiomer in the mixture will be given by (1-x).

specific rotation of the mixture = proportion of R enantiomer * its specific rotation + proportion of S enantiome * its specific rotation

i. e.

-10 = x * (+20) + (1-x) * (-20)

-10 = 20x-20 + 20x

-10+20 = 40x

+10 = 40 x

x=10/40 = 25%

Since the proportion of the other enantiomer is 1-x, it is 0.75 or 75%

So the mixture contains 25% R, 75% S, giving you an excess of 50%.

10%

Explanation:

Enantiomeric excess is a way of describing how optically pure a mixture is by calculating the purity of the major enantiomer. It can range from 0%-100%. Enantiomeric excess (ee) can also be defined as the absolute difference between the mole fractions of two enantiomers.

Enantiomeric excess is also called optical purity. This is because chiral molecules cause the rotation of plane-polarized light and are said to be optically active. An enantiomerically pure sample has an enantiomeric excess of 100 percent

Enantiomeric excess = observed specific rotation/specific rotation of the pure enantiomer x 100

From the data given in the question;

observed specific rotation = - 2°

specific rotation of the pure enantiomer = + 20°

Therefore;

ee = 2/20 * 100

ee = 10%