Distillation material balance ($F=D+B,Fz=D{x}_D+B{x}_B$)

Consider a nonreacting two-component mixture processed into just two products, with no losses or unaccounted inventory. Let $F,D,B$ be feed, distillate, and bottoms amount of substance values in moles, and let $z,x_D,x_B$ be their component-A mole fractions. These are whole-stream compositions; $x_D$ is not automatically the instantaneous equilibrium vapor composition $y$ above the pot. Total and component balances giveThe component-B balance follows by subtracting the second equation from the first. Eliminating $B$ gives, for $x_D\ne x_B$,Nonnegative product amounts require $z$ between $x_B$ and $x_D$. For an original stipulated feed $F=1\mathrm{mol}$, $z=0.35$, $x_D=0.80$, and $x_B=0.10$,Check the component inventories:The distillate contains a fraction $(D{x}_D)/(Fz)=40/49\approx 0.816327$ of the original component A, despite its higher mole fraction $x_D=0.80$. This distinction between composition and recovery is useful when deciding how much enriched product to collect. These balances determine neither temperatures, heat input, nor the number of equilibrium stages; they do not prove that the specified products are attainable in a particular fractionating column.