Question

One can prepare homemade lavender perfume by steam distillation,one of the main components in the lavender oil is linalyl acetate (MM 196 g/mol), which has a vapor pressure of 20 mm Hg at the boiling point of the mixture with water. If one collect 500 g of lavender from a garden and performs a codistillation with water, what would be the compositions of the distillate?

This is what I have so far:

Wt of water = (18g/mol x 20 mmHg)/(196 g/mol x (760 mmHg - 20 mmHg)) = 0.0025 g water

I don't understand how to get the percent composition when only given the amount of lavender used since I don't know how much linalyl acetate I could get from 500 g of lavender

Answer 1

In case of water insoluble liquids, the mixture boils at a temperature at which sum of the vapor pressures of individual components get addded up

total pressure, 760= vapor pressure of linalyl acetate + vapor pressue of water

Vapor pressure of water= 760-20 =740 mm Hg

this is also the saturation pressure of water

moles of water/ moles of water/ Mole of linalyl acetate= partial pressure of water/ Partial pressure of linalyl acetat

moles of lianyl acetate= 500/196=2.55

basis : 1 gm of lianyl acetate= 1/196 g/mol

Mole fraction = 740/760 for water= 0.973

mole fraction = 1-0.973-= 0.227 for lianyl acetate

1mole of mixture contains 0.227 mole of linaly acetate

mole of linayl acetate= 500/196=2.55

0.227 mole correspond to 2.55 mole

1 mole correspond to 2.55/0.227=11.23

Moles of water= 11.23- 2.55=8.68 moles, mass of water= 8.68*18=156.24 gm

Mass of lianyl acetate= 500 gm

mass fraction water= 156.24/(500+156.24)=0.238 water