Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of (+) isomer is +30°. The optical purity is __________%.

Solution

Chemistry ma’am · 3 steps

Optical purity just asks what fraction of the full rotation your mixture actually shows.

1. What it means. A pure (+) isomer would rotate light by +30°. In the mixture, the (-) isomer cancels part of that rotation, so only +12.6° is seen.

   Optical purity = (observed rotation)/(rotation of pure enantiomer) × 100

2. Put in numbers. 12.6/30 = 0.42, and 0.42 × 100 = 42%.
3. So. The optical purity of the mixture is 42%.

The common slip: 71. Reporting the percentage of the (+) isomer, (100 + 42)/2, gives 71, not 42. That is the composition of the mixture, not its optical purity.

The idea to keep. Optical purity is observed rotation over the pure enantiomer's rotation, times 100. It equals the enantiomeric excess, not the percentage of the major isomer.

Answer from the NTA final answer key. Solution by PrepWiser.