Question

An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a specific rotation of −127°. If this solution were mixed with 500 mL of a solution containing 3 g of a racemic mixture of the compound, what would the specific rotation of the resulting mixture of the compound? What would be its optical purity?

Answer 1

Answer:

Optical purity = 76.9231 %

Specific rotation of mixture = - 97.6923 °

Explanation:

The mass of the racemic mixture = 3 g

It means it contains R enantiomer = 1.5 g

S enantiomer = 1.5 g

Amount of Pure R = 10 g

Total R = 11.5 g

Total volume = 500 mL + 500 mL = 1000 mL = 1 L

[R] = 11.5 g/L

[S] = 1.5 g/L

Enantiomeric excess = $\frac {Excess}{Total\ Concentration}\times 100$ = $\frac {11.5-1.5}{11.5+1.5}\times 100$ = 76.9231 %

Optical purity = 76.9231 %

Also,

Optical purity = $\frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100$

Optical rotation of pure enantiomer = −127 °

$76.9231=\frac {optical\ rotation\ of\ mixture}{-127^0}\times 100$

Specific rotation of mixture = - 97.6923 °