As you know, there are not many choices for the test designers to make a problem in ochem that needs calculations. Here is one of them that I think is important, and everyone should know: enantiomerics excess or optical purity! [Two different names for the same concept] In order to find the enantiomeric excess, we divide the specific rotation of the mixture by the specific rotation of one of the pure enantiomers, and multiply it by 100 to get the % [Remember: the specific rotation of the two enantiomers are the same numbers with different signs]. Let's say we get 30% for optical purity [for example for the (+) enantiomer].This means 30% of the mixture is (+), and the other 70% is racemic [meaning you have another 35% (+) and finally 35% (-) ]. Add the 30 and 35 to get the percentage of (+). You will get 65% (+) and 35% (-). Another Example: The specific rotation of 2-butanol is +13.52 for the (d) enantiomer, and -13.52 for the (l) enantiomer. Suppose we have a sample of 2-butanol which gives [alpha]D = -5.41 . Find the percentage of each of the two enantiomers in the mixture.