6
$\begingroup$

Yes, atomic masses are non-integral due to isotopes and binding energy. These are the two reasons I found on google and other answers on stackexchange.

But there is something else - isn't the most obvious reason for it the way atomic masses are calculated? We defined 1 amu as one twelfth the mass of a C-12 atom. Now C-12 has 6 protons and 6 neutrons (thus 1 amu isn't mass of 1 proton or 1 neutron). And mass of proton and neutron is different. And each element has a different ratio of protons to neutrons. So of course when we convert their atomic masses in terms of amu it is going to be non-integral.

But this reason isn't mentioned anywhere so now I am starting to doubt it. Is there a flaw in this logic?

$\endgroup$
5
  • $\begingroup$ Yes it will be non integral for example atomic mass of hydrogen is 1.0008 amu but for calculations we round them off to nearest integer $\endgroup$
    – Lllt
    Jun 2 at 4:13
  • 3
    $\begingroup$ 1 amu = 1/12 ($\ce{6 x m_e^- + 6 x m_p^+ + 6 x m_n^o - (nuclear binding energy of C-12 + atomic binding energy of C-12))}$ When you apply this to determine the masses of other isotopes, you are essentially doing a similar calculation for them (effectively--in practice, these would typically obtained by measurement), and then dividing by the above. So clearly you wouldn't expect atomic masses to be integral. For an explanation of nuclear and atomic binding energies, see en.wikipedia.org/wiki/Binding_energy $\endgroup$
    – theorist
    Jun 2 at 7:06
  • 1
    $\begingroup$ Please note that the unit symbol "amu" is obsolete (in chemistry since 1961), and 1 amu was not "one twelfth the mass of a C-12 atom". $\endgroup$
    – Loong
    Jun 2 at 17:39
  • 4
    $\begingroup$ Does this answer your question? Why atomic masses aren't integers? $\endgroup$
    – Mithoron
    Jun 3 at 0:26
  • $\begingroup$ @Mithoron- No, actually that was the question I read before posting mine. But as I said, it doesn't include anything about differences in masses of proton and neutron. So my question is a little different $\endgroup$ Jun 3 at 2:57
10
$\begingroup$

Yes this is a valid reason, albeit a small one, almost negligible. After all, 1amu and the masses of 1 proton and 1 neutron are pretty close to each other.

Then again, the binding energy would also be negligible, if not for oxygen's $15.999$ which catches the student's eye right away and begs for an explanation. And of course binding energy is the only possible explanation, as oxygen's other isotopes work in the wrong direction, and the ratio of protons and neutrons in $^{12}\rm C$ and $^{16}\rm O$ is the same.

The lack of a similar clear example in your case is probably the reason why your reason is never mentioned anywhere.

So it goes.

$\endgroup$
0
3
$\begingroup$

The mass difference between a bare proton and a bare neutron is 1293 keV, while the mass of an electron is 511 keV. That means that the effect you're talking about is often going to be on the same order of magnitude as the difference in mass between an atom and one of its ions. In something like liquid-phase chemistry, as opposed to mass spectroscopy, I doubt that an ion ever really even maintains its ionization state for long enough to make the electron-mass correction even meaningful, and if that's not meaningful then the correction you're talking about, which is on the same order of magnitude, is also not very meaningful.

When you want to start talking about corrections at this level, you probably need to define more carefully what purpose you're going to use the resulting numbers for. For example, if you're going to look at the infrared spectrum of the NaCl molecule in a gas, which comes from its end-over-end rotation, then it's got some moment of inertia that is mainly determined by the nuclei, but when the bond forms, one of the electrons shifts its mass around, and that will change the result.

Keep in mind also that, conceptually, the protons and neutrons in a nucleus aren't really the same creatures as the particles that exist in free space -- the "bare" proton and neutron I referred to above. A free neutron is a little ball of quarks and gluons, with pairs of particles appearing and disappearing spontaneously in a kind of quantum soup. The mass that we measure for the neutron is some kind of average or composite value that represents the masses and kinetic energies of all those sub-particles. When a proton or neutron is incorporated in a nucleus, it loses its individual identity, and it's no longer really a proton or neutron. It promiscuously shares all its quarks and gluons with its neighbors. It's only a kind of shorthand terminology when we refer to the nucleus as being made of protons and neutrons. So it's not even really totally obvious whether it makes sense to try to add the masses of all "the protons and neutrons" inside a nucleus. It does sort of work as an approximation (it's what's done in standard versions of the nuclear liquid drop model), but it's only an approximation.

Another thing to realize is that most chemistry is done with light elements, for which the proton number and neutron number are about the same. Since the atomic mass standard is based on 12C, which has equal neutron and proton numbers, you will only see any effect in a.m.u. masses when you go to heavy nuclei.

$\endgroup$
4
  • $\begingroup$ Wow that was a great explanation! $\endgroup$ Jun 2 at 16:41
  • $\begingroup$ I understood most of it but can you explain the NaCl rotation part in more detail? $\endgroup$ Jun 2 at 16:46
  • 1
    $\begingroup$ @DhruviKataria: The rotational band forms a series of energy levels with energies proportional to $J(J+1)/I$, where $I$ is the moment of inertia and $J$ is the angular momentum in units of $\hbar$. What you observe in spectroscopy is the differences among energy levels in this band, so those are $\propto 1/I$. $\endgroup$
    – user6999
    Jun 2 at 16:48
  • 1
    $\begingroup$ The NaCl infrared part is indeed not clear. What is meant by "when the bond forms, one of the electron shifts its mass around" in "For example, if you're going to look at the infrared spectrum of the NaCl molecule in a gas, which comes from its end-over-end rotation, then it's got some moment of inertia that is mainly determined by the nuclei, but when the bond forms, one of the electrons shifts its mass around, and that will change the result"? $\endgroup$
    – M. Farooq
    Jun 2 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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