Question

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

Answer 1

the main components in the lavender oil is linalyl acetate : MM 196g/mol

linalyl acetate a vapor presure ~ 20mm Hg ( at the boiling point of the mixture with water )

thus, P^o(linalyl acetate) ~ 20mm Hg

and P^o(water) ~ 760 mm Hg   ( at the boiling point of the mixture with water )

we have,

composition of the distillate ?

mass of oil / mass of water ~ P^o(linalyl acetate)*MM(linalyl acetate) / P^o(water)*MM(water)

mass of oil / mass of water = 20 mmHg*196 g/mol / 760 mmHg*18 g/mol

mass of oil / mass of water =0.2865

mass of oil = 0.2865 * mass of water

mass fraction of linalyl acetate = 0.2865 * mass of water / (0.2865 * mass of water+ mass of water) = 0.2227

thus, composition of the distillate ~ 22.27 wt % ( linalyl acetate) + 77.73 wt % (water)