Answer to Question #112756 in Organic Chemistry for Asma Iqbal

Question

Linalyl acetate is a component of essential oil of lavender. It is immiscible in water and exerts a vapor pressure of 12 torr near the boiling point of water. Predict the amount of water (in mL) that must co-distill to isolate 1.0 g of linalyl acetate from lavender using a steam distillation.

Answer 1

One should determine the volume of vapor which consists of organic compound and water steam. For this, one uses the ideal gas law:

p*V = m/M*R*T

where p = 12/760*101325 = 1600 Pa, T = 373 K (boiling point of water), R = 8.31 J/(K*mole),

M = 196.29 g/mole for linalyl acetate.

V = 1/196.29*8.31*373/1600 = 9.869*10^-3 m³ = 9.869 L.

This volume is as well occupied by steam. The conditions of water evaporation are: T = 373 K, p = 1 atm = 101325 Pa. Now using the ideal gas law one calculates the mass of water (M = 18 g/mole) vapor contained in this volume:

m = 101325*9.869*10^-3*18/8.31/373 = 5.807 g

Since density of liquid water is 1 g/mL, one obtains that the amount of water that must be evaporated is 5.807 mL