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?

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Mergus Mergus · Answer 1 · 2019-10-10T08:45:02+02:00

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 = [tex]\frac {Excess}{Total\ Concentration}\times 100[/tex] = [tex]\frac {11.5-1.5}{11.5+1.5}\times 100[/tex] = 76.9231 %

Optical purity = 76.9231 %

Also,

Optical purity = [tex]\frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100[/tex]

Optical rotation of pure enantiomer = −127 °

[tex]76.9231=\frac {optical\ rotation\ of\ mixture}{-127^0}\times 100[/tex]

Specific rotation of mixture = - 97.6923 °