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Co-distillation is the vaporization and collection of multiple components together during distillation. A component can enter the collected vapor alongside another volatile substance instead of being isolated as a pure vapor. Steam distillation is the case in which water supplies a volatile carrier; co-distillation is not restricted to water.

The result depends on vapor-liquid equilibrium. In an ideal, fully mixed binary liquid at a specified temperature, Raoult's law gives $p_A=x_A{p}_A^{\ast }$ and $p_B=x_B{p}_B^{\ast }$. The vapor fraction is $y_A=p_A/(p_A+p_B)$. The McMaster distillation manual uses liquid/vapor composition diagrams to explain why the vapor can be enriched in one component while still containing both.^[ref]

For original hypothetical data, let $x_A=0.25$, $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. Then $p_A=20$ and $p_B=15\mathrm{kPa}$, giving $y_A=20/35\approx 0.571$. Both components co-distil, but the vapor is richer in $A$ than the starting liquid. These are equilibrium pressures at the stated temperature; boiling additionally requires their sum to match the external pressure.

With two persistent immiscible liquid phases, a different limiting model uses the pure-component vapor pressures directly, rather than multiplying each by its fraction of a homogeneous mixture. An azeotrope is another special equilibrium situation. Specify the phase behavior before assigning a distillate composition, and use a distillation material balance to distinguish enrichment from total recovery. Collecting a component together with a carrier does not itself establish complete separation from every impurity.

L. J. Zatman coauthored a 1964 investigation of bacterial methanol dehydrogenase with Chris Anthony. The original byline gives the initials retained here. Their work investigated how an enzyme transfers reducing equivalents during methanol oxidation.^[ref]

A particularly informative part of the study concerned product identity. The researchers separated material by steam distillation before testing for formaldehyde, because an assay reagent interfered with direct measurement. They explicitly distinguished detecting formaldehyde after analysis from proving that free formaldehyde was the immediate enzymic product.^[ref]

The underlying oxidation can be represented by the balanced half-reactionBoth sides contain one carbon, one oxygen and four hydrogen atoms, and the total charge remains zero. This bookkeeping establishes the two-electron change for that net conversion. It does not identify the electron acceptor, the cofactor or every intermediate.

The same distinction applies to an analytical treatment that releases a detectable molecule from a bound precursor. Recovering the molecule establishes what the whole treatment yields; an independent observation is needed to establish its state before treatment. This makes the paper a useful example of how careful product analysis can support a reaction assignment while leaving a mechanistic question open.

Steam distillation is co-distillation with water, in which water vapor carries volatile material into a laboratory condenser and receiver. It can recover volatile organic compounds from less volatile material without requiring the isolated compound to reach its own pure-component boiling point. The University of York's teaching apparatus combines material and water in a boiling flask connected to a distillation head, condenser and collection flask.^[ref]

A useful ideal case is liquid immiscibility: two separate liquid phases remain present, and each supplies approximately its pure-component vapor pressure. Their sum reaches the external pressure at boiling:Because both terms contribute, the required temperature can be below either pure liquid's boiling point at that same external pressure. This limit assumes both liquid phases persist and neglects mutual solubility and vapor nonideality.

For invented equilibrium data, let the water and organic partial pressures be $95$ and $5\mathrm{kPa}$ at the boiling temperature. The vapor mole ratio is $5/95$. If the organic compound has molar mass $150\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and water is approximated by $18\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the condensed mass ratio isThis is an equilibrium model, not an extraction yield.

A 1964 methanol dehydrogenase study used steam distillation before a formaldehyde assay. Its authors explicitly left unresolved whether the product had been free or bound before separation.^[ref] Recovering a volatile product and establishing its earlier molecular state are different experimental claims.