Question

(OChem)

In a Limonene isolation lab (from orange peels), we performed the steam distillation and measured the temperature corresponding to the distillate volume. (solvent: water, solute: limonene and they are immiscible)

If the limonene distillate was 35ml and the temperature at this point was 95°C, how do I calculate the mole fraction of limonene? (PV=nRT)

Is it possible to get the mole fraction of limonene by using these data and some reference data?

The P equation for the Steam Distillation is P=P°_H2O + P°_Limonene and I could not figure out how to do it.

Answer 1

By using the ideal gas equation PV=nRT,

P = Total pressure of limonene distillate

V = total volume of distillate

n = total number of moles

From the above equation,(T=273+98 Kelvin), V=35*10^(-3) liter

P = n*8.314*368/35*10^(-3)

But the formula to calculate mole fraction is: X(mole fraction)=n(moles of limonene)/n(total no of moles)

so from this formula, there is a need of limonene moles and total pressure to calculate moles fraction.

Another formula for moles fraction is : X=partial pressure of limonene( P°Limonene)/total pressure(P)

There is no data available of partial pressure.