# 5.5C: Separation Theory

• • Contributed by Lisa Nichols
• Professor (Chemistry) at Butte College

Although steam distillation appears almost identical to "regular" distillation, the principles behind the separation of components are quite different. In a "regular" distillation, separation is attempted on a mixture of components that dissolve in one another. The vapor produced from these mixtures can be described by a combination of Raoult's law and Dalton's law, as shown in Equation \ref{14} for a two-component mixture.

$\text{Miscible components:} \: \: \: \: \: P_\text{solution} =P_A + P_B = P_A^o \chi_A + P_B^o \chi_B \label{14}$

When the components in the distilling flask do not dissolve in one another, such as when water and nonpolar organic compounds are present, the vapor produced from these mixtures is different. The components act independently from one another (which makes sense considering they do not mix), and the partial pressure from each component is no longer determined by its mole fraction. The partial pressure of each component is simply its vapor pressure, and the vapor composition for a two-component mixture is described by Equation \ref{15}. Although the components do not mix in the liquid phase, they do in the gas phase, which allows for co-distillation of "incompatible components".

$\text{Immiscible components:} \: \: \: \: \: P_\text{solution} = P_A^o + P_B^o \label{15}$

The implications of Equation \ref{15} are several. First, since mole fraction is not a factor, it is possible that a minor component in the distilling flask can be a major component in the distillate if it has an appreciable vapor pressure. In the steam distillation of volatile plant materials, this means that the distillate composition is independent of the quantity of water or steam used in the distilling flask.

Secondly, since the vapor pressures of each component add, the mixture will always boil at a lower temperature than the boiling point of the lowest boiling component. For example, at $$100^\text{o} \text{C}$$ water has a vapor pressure of $$760 \: \text{torr}$$ since it is at its normal boiling point. If another volatile, non-water-soluble component was present with the water in a distilling flask at $$100^\text{o} \text{C}$$ (for example toluene, which has a boiling point of $$111^\text{o} \text{C}$$), it too would produce vapors that contributed to the total pressure. As boiling occurs when the combined pressure matches the atmospheric pressure (let's say $$760 \: \text{torr}$$ for the sake of this calculation), boiling would occur below $$100^\text{o} \text{C}$$. For example, a mixture of toluene and water boils at $$85^\text{o} \text{C}$$, as shown in Equation (16).

\text{Water/toluene mix at } 85^\text{o} \text{C}: \: \: \: \: \: \begin{align} P_\text{solution} &= P_\text{water}^o + P_\text{toluene}^o \\ &= \left( 434 \: \text{torr} \right) + \left( 326 \: \text{torr} \right) = 760 \: \text{torr} \end{align} \label{16}

The distilling temperature in steam distillation is always below $$100^\text{o} \text{C}$$ (the boiling point of water), although in many cases the distilling temperature is very near or just under $$100^\text{o} \text{C}$$. This feature allows for plant essential oils (complex mixtures that often include components with very high boiling points, $$> 250^\text{o} \text{C}$$), to be extracted at lower temperatures than their normal boiling points, and thus without decomposition.