The specific rotation of the pure (+)-enantiomer of a chiral compound is $+ 50. 0^{\circ}$ (path length in dm, concentration in g/mL). A scalemic sample of 4.00 g is dissolved in the solvent to make 100 mL of solution, and the observed rotation in a 2.00 dm tube is $+ 2.6 0^{\circ}$ . What mass of the (-)-enantiomer is present in the 4.00 g sample?

Answer: A

Concentration

c = 4.00/100 = 0.0400

g/mL, so the specific rotation of the sample is

2.60/ (2.00 \times 0.0400) = + 32. 5^{\circ}

. Enantiomeric excess

= 32.5/50.0 = 0.650

, so the fraction of (+) is

(1 + 0.65) /2 = 0.825

and of (-) is

0.175

. Mass of (-)

= 0.175 \times 4.00 = 0.70

g. (3.30 g is the (+) mass; 1.40 g forgets that ee is a difference.)