Answer:
(+)-cholesterol = 71.88%
(-)-cholesterol = 28.12%
Explanation:
Asuming 1 gram of sample is dissolved in 1mL of water and the sample cell was 1dm long.
Enantiomeric excess is defined as the amount of pure enantiomer in a sample. The formula is:
ee = [α]mixture / [α]pure enantiomer.
Replacing:
ee = 14° / 32°×100 = 43.75%
As the sample is 14°, There is an excess of (+)-cholesterol and 56.25% is a 1:1 mixture of enantiomers.
That means percent composition of enantiomers is:
(+)-cholesterol = 43.75% + 56.25%/2 = 71.88%
(-)-cholesterol = 56.25%/2 = 28.12%
