If you had a sample containing an elevated concentration of $\ce{H2^{18}O}$, and bubbled $\ce{C^{16}O2}$ through it, would some of the oxygen-18 isotope be transferred from water to carbon dioxide?

I am aware that this occurs with hydrogen isotopes and am curious if it works for larger and heavier atoms such as oxygen.

  • 8
    $\begingroup$ Yes, absolutely. In (at least) aqueous geochemistry, the phenomena is used to determine source rock and age for various parts of the hydrologic system. See this publication for more information. $\endgroup$ Mar 24, 2022 at 2:24
  • 7
    $\begingroup$ It is quite obvious it does, via its equilibrium reactions with water, forming carbonic acid and bicarbonate/carbonate ions. $\endgroup$
    – Poutnik
    Mar 24, 2022 at 5:21
  • $\begingroup$ @Poutnik, I realized that as I read the paper Todd recommended. I wonder if it would still occur if the molecule did not dissociate. Like if the gas were ethanol instead. $\endgroup$
    – ericnutsch
    Mar 24, 2022 at 5:33
  • 1
    $\begingroup$ @ericnutsch It doesn't have to dissociate, but a reaction is obligatory. With alcohol, if pH was high enough, substitution would be possible. $\endgroup$
    – Mithoron
    Mar 24, 2022 at 14:48
  • 1
    $\begingroup$ Ethanol does not exchange with water in weakly acidic conditions; see the article by J. Dunbar, doi.org/10.1524/zpch.1982.130.2.247. $\endgroup$ Mar 24, 2022 at 15:34

1 Answer 1


As Todd Minehardt points out, not only can the oxygen be exchanged, but this exchange is applied in aqueous geochemistry.

The exchange occurs through the formation of carbonic acid, given in blue below:

$\ce{CO2 + H2O <=> \color{blue}{H2CO3}}$

Once the carbonic acid molecule is formed, its oxygen atoms effectively become equivalent because they rapidly exchange hydrogen ions with the water solvent. So when the above (dynamically equilibrated) reaction is reversed, any one of the carbonic acid oxygens might end up as a water molecule, including atoms that were originally in the carbon dioxide.

  • $\begingroup$ Now I have to wonder if, during a chemical reaction, certain atoms are more likely to end up in certain places. E.g. if (hypothetically) H2CO3 did not exchange hydrogen ions, would the "central" O atom be from the CO2 molecule that went into it? (It "looks like" that arrangement should require less bonds to be broken and re-formed during the reaction) $\endgroup$
    – user253751
    Mar 24, 2022 at 16:49
  • $\begingroup$ That depends on if you allow tautomerization from the direct adduct $\ce{H2O - CO2}$ to $\ce{HO - C(O)OH}$. With the tautomerization, the "central" oxygen atom would come from the carbon dioxide, but the hydroxyl oxygens would come from both reactants and thus mixing would occur. But without the tautomerizaton, the oxygen atoms cannot mix because the water and carbon dioxide oxygens never become equivalent. $\endgroup$ Mar 24, 2022 at 22:09
  • $\begingroup$ Hmmmm... just guessing here: $\ce{CO2 ->[H2O] CO2 (aq)}$, $\ce{CO2 (aq) + H2O <=> H+ (aq) + HCO3+ (aq)}$, $\ce{HCO3+ (aq) <=> H+ (aq) + CO2^2+ (aq)}$. The Grotthuss mechanism accounts for the mobility of protons, and carbonic acid will most likely be in the smallest concentration of all involved species. Like actually lower that $\ce{2 H2O (aq) <=> OH- (aq) + H3O+ (aq)}$. $\endgroup$ Mar 24, 2022 at 22:42
  • 1
    $\begingroup$ @user253751 the particular reaction is not a good example, but otherwise, marking atoms by their (stable or radioactive) isotope and looking where they go in a reaction is a powerful tool for studying reaction mechanisms. Stable isotopes (e.g. Hydrogen-2) are traced by NMR and radioactive isotopes (this is the older approach) by their radiation. $\endgroup$
    – fraxinus
    Mar 27, 2022 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.