Question

Steam distillation lab?

One can prepare homemade lavender perfume by steam distillation,one of the main components in the lavender oil is linalyl acetare (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 all that was given??

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 added up so:

Pt = 760 = vapor pressure of linalyl acetate + vapor pressue of water

Vapor pressure of water= 760 - 20 = 740 mmHg

this is also the saturation pressure 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 moles

We do know that for Raoult law: P° = XaPt so, applying this to the water:

Xb = P° / Pt = 740/760 = 0.974

And Xa+Xb = 1 so:

Xa = 1-0.974 = 0.026

This means that 1 mole of mixture contains 0.026 moles of linaly acetate (And corresponds to 2.55 moles)

Then 1 mole corresponds 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 g

Mass of lianyl acetate= 500 g

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

The rest is lynale acetate. Hope this helps