Question

if a pure R isomer has a specific rotation of -106.0, and a sample contains 70.0% of the R isomer and 30.0% of it's enantiomer, what is the observed specific rotation of the mixture

Answer 1

percentage of S enantiomer = 30 %
percentage of R enantiomer = 70 %
fraction of S enantiomer = 0.3
fraction of R enantiomer = 0.7
specific rotation of S enantiomer = 106.0
specific rotation of R enantiomer = -106.0

net rotation = fraction of S enantiomer * specific rotation of S enantiomer + fraction of R enantiomer * specific rotation of R enantiomer

net rotation = 0.3 * (106.0) + 0.7 * (-106.0)
net rotation = 31.8 - 74.2
net rotation = -42.4
Answer: -42.4