= Two-population molar mass example
Suppose a synthetic <polymer> population contains three chains of <molar mass> $10000\,\mathrm{g\,mol^{-1}}$ for every one chain of <molar mass> $40000\,\mathrm{g\,mol^{-1}}$. Multiplying both chain counts by any common factor leaves the averages unchanged. Substitution into the existing <number-average molar mass> and <weight-average molar mass> definitions gives
$$
M_n=\frac{3(10000)+40000}{4}\,\mathrm{g\,mol^{-1}}=17500\,\mathrm{g\,mol^{-1}},\qquad M_w=\frac{3(10000)^2+(40000)^2}{3(10000)+40000}\,\mathrm{g\,mol^{-1}}\approx27143\,\mathrm{g\,mol^{-1}}.
$$
The larger chains are only one quarter of the chain count but supply $40000/70000=4/7$ of the <mass>. Thus a <weighted mean> that emphasizes <mass> gives a larger result than one that weights each <molecule> equally. Total <mass> divided by total chain <amount of substance> gives $M_n$, while $M_w/N_{\mathrm A}$ is not the mean <mass> of a randomly chosen chain. Inferring chain counts from <size-exclusion chromatography> still requires the existing <SEC detector response> and <SEC molar mass calibration> assumptions.
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