= Aerobic respiration
<Aerobic respiration> uses <molecular oxygen> as the final <electron acceptor>. For complete <oxidation> of <glucose>, the net material balance is
$$
\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6+6\mathrm{O}_2\longrightarrow6\mathrm{CO}_2+6\mathrm{H}_2\mathrm{O}.
$$
This records the overall conversion of <glucose>, <molecular oxygen>, <carbon dioxide> and <water>; it is not one <enzyme>-catalyzed step, and it omits the separate <ATP> accounting. For an original <stoichiometry> check, each side contains six <carbon>, twelve <hydrogen> and eighteen <oxygen> <chemical atoms>. If exactly $1\,\mathrm{mmol}$ of <glucose> undergoes this complete conversion, the equation requires $6\,\mathrm{mmol}$ of <molecular oxygen> and yields $6\,\mathrm{mmol}$ each of <carbon dioxide> and net <water>. This is an equation-based calculation, not an observed seed measurement.
The <molecular oxygen> is reduced to <water>, rather than being the source of the <carbon dioxide>'s <carbon>. A formal <redox electron balance>, using acidic half-reaction notation, makes the roles explicit:
$$
\begin{aligned}
\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6+6\mathrm{H}_2\mathrm{O}&\longrightarrow6\mathrm{CO}_2+24\mathrm{H}^{+}+24e^{-},\\
6\mathrm{O}_2+24\mathrm{H}^{+}+24e^{-}&\longrightarrow12\mathrm{H}_2\mathrm{O}.
\end{aligned}
$$
Adding and cancelling gives the net equation above. Both half-reactions conserve <chemical atoms> and <electric charge>. The $24$ <Electrons> balance a net transfer; they are carried through cellular intermediates, rather than released as a pool of free <Electrons>. The final respiratory <reduction> can be written $\mathrm{O}_2+4\mathrm{H}^{+}+4e^{-}\longrightarrow2\mathrm{H}_2\mathrm{O}$. These equations are bookkeeping, not a tracing experiment that assigns every product <oxygen> <chemical atom> to its original <molecule>.<Biology 2e by Mary Ann Clark and others>[Biology 2e, §7.4, Electron Transport Chain, Figure 7.12: oxygen reduction to water]{ref}
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