= Linear step-growth molecule count
{title2=$N=N_0-J$}
Use an ideal feed in which every starting <molecule> has exactly two suitable <functional groups>: either one $A$ and one $B$, or two $A$ groups mixed with two $B$ groups. Only an $A$–$B$ pairing can form a joining <chemical bond>. Assume no loss of feed-derived material, no splitting of chains, no competing consumption of the counted <functional groups>, no branching, and no <cyclization in step-growth polymerization>. Linear here describes connectivity; a chain can bend in space. Small byproducts of <polycondensation> do not enter the chain-bearing <molecule> count.
Let $N_0$ count all initial feed <molecules>, $J$ the number of intermolecular joins, and $N$ all remaining chain-bearing <molecules>, including unreacted <monomers>. Each join turns two <molecules> into one, so $N=N_0-J$ exactly. If chain $i$ contains $j_i$ feed-derived <monomeric units>, conservation gives $\sum_{i=1}^{N}j_i=N_0$. Consequently the number-average <degree of polymerization> is
$$
X_n=\frac{\sum_{i=1}^{N}j_i}{N}=\frac{N_0}{N_0-J}.
$$
This is a count of both feed species' <monomeric units>, rather than a count of interior <constitutional repeating units>. For $AA+BB$ chemistry an interior repeat often contains two feed-derived units, as explained in <monomer>. Equal <functional group> reactivities are not needed for this bookkeeping at a specified $J$; predicting a time course or a chain-length <probability distribution> requires additional assumptions.
\Image[/step-growth-molecular-joining-count.png]
{title=Exact <step-growth polymerization> count: joining a two-unit chain to a three-unit chain consumes one pair of <functional groups>}
{id=image-step-growth-molecular-joining-count}
{description=Original stipulated eight-$AB$-molecule example. Circles are feed-derived <monomeric units>, lines joining <chemical bonds>, and letters the surviving <functional groups>. Internal $A$ and $B$ groups have already reacted and are not drawn. The depicted layouts are schematic connectivity, not molecular geometry or sampled experimental chains.}
For the figure's eight $AB$ feed <molecules>, three joins can give lengths $(2,3,1,1,1)$, so $N=5$ and $X_n=8/5$. Joining the first two chains gives $(5,1,1,1)$: the incorporated total stays eight, $N$ becomes four, and $X_n=2$. A <graph> makes the bookkeeping visible: each starting <monomer molecule> is a <vertex of a graph>, and each intermolecular joining <chemical bond> connects two previously distinct <connected components of a graph>. The connected pieces are the counted <molecules>.
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