I have always been led to understand that the mass of an element on the periodic table is the weighted average atomic mass over all naturally occurring isotopes. This seems to make sense with all the elements I have looked at except for uranium. Wikipedia's featured article on the element assures me that uranium's relative atomic mass is 238.03. The commonest isotope is U-238, mixed with 0.72% of U-235 and 0.005% of U-234, plus traces of a couple of isotopes with smaller mass numbers.

How then can uranium's relative atomic mass exceed 238? Either I have completely misunderstood something, or there is something else at work - relativistic maybe?

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    $\begingroup$ (From my deleted answer) There also are isotopes of uranium with greater masses than 238 - though their contribution would be tiny. $\endgroup$
    – user15489
    May 29, 2015 at 0:10
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    $\begingroup$ @Santiago I'm not sure they would contribute. They all have half-lives less than an hour and are not part of natural abundance uranium. $\endgroup$
    – ron
    May 29, 2015 at 0:33
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    $\begingroup$ Any uranium sample that contains U-238 (including natural minerals) inevitably contains a very tiny amount of U-239 due to (n,γ) reactions. However, that has no measurable impact on the average atomic mass of uranium. $\endgroup$
    – user7951
    May 29, 2015 at 14:16
  • $\begingroup$ Yes, I am getting that point now $\endgroup$
    – user15489
    May 29, 2015 at 19:43

1 Answer 1


Approximately 99.3% of uranium on Earth is the $\mathrm{^{238}U}$ isotope, and this specific isotope has an atomic mass of $\mathrm{238.05\ u}$, where $\mathrm{u}$ is the atomic mass unit, equivalent to 1/12 the mass of a $\mathrm{^{12}C}$ atom. Including the other isotopes to obtain the average atomic mass drags the value down a little, but it still ends up being slightly larger than $\mathrm{238\ u}$.

It is interesting to point out why the atomic mass is slightly different from the mass number. The masses of a free proton or neutron are both slightly larger than $\mathrm{1\ u}$ ($\mathrm{1.0073\ u}$ and $\mathrm{1.0087\ u}$, respectively). Some of this mass is converted to energy and lost when neutrons and protons (collectively called nucleons) bind to form nuclei, which means bound nuclei will always be lighter than the sum of the masses of their component nucleons. Different nuclei release different amounts of energy per nucleon depending on how many protons and neutrons are inside the nucleus, and on their relative proportions. Very interestingly, the curve for nuclear binding energy per nucleon versus mass number starts off at zero for $\mathrm{^{1}H}$ (there can be no attractive nuclear forces between nucleons if there's only one nucleon!), then rises until it shows a maximum around nickel (specifically for the isotope $\mathrm{^{62}Ni}$) and from there on slowly decreases.

Since the atomic mass unit is standardized with respect to the $\mathrm{^{12}C}$ isotope of carbon, then in approximate terms, the nuclei with a lower binding energy per nucleon than $\mathrm{^{12}C}$ will have less of the initial free proton/neutron mass lost to energy (and thus an isotope will have an atomic mass slightly higher than its mass number), while for nuclei with a higher binding energy per nucleon than $\mathrm{^{12}C}$, more mass will be lost to energy and the resulting isotope will have an atomic mass slightly lower than its mass number.

Taking a horizontal line through the $\mathrm{^{12}C}$ isotope in this graph and looking at its intersections with the nuclear binding energy curve will show the regions where this flip happens; isotopes for atoms below carbon will tend to have higher atomic mass than mass number, isotopes for atoms between carbon and thorium tend to have a lower atomic mass than mass number, and then isotopes for atoms above thorium (such as $\mathrm{^{238}U}$) will again tend to have higher atomic masses than mass numbers. This is only an approximate analysis, as protons and neutrons don't have the same mass, so isotopes very close to the transition region don't behave perfectly smoothly.

Edit: The flipping effect I mention can be seen very nicely in the list of isotopes shown in this site. Look at the (atomic) masses for the isotopes, and see how the fractional part of the mass starts out as circa $.01$ (above the isotope's mass number), then decreases to $.001$, reaches exactly zero (by definition) for $\mathrm{^{12}C}$, then soon after flips under to $.999$ and keeps going lower all the way to $.90$ around zirconium and tin isotopes before slowly rising up to $.98$ for bismuth $\mathrm{^{209}Bi}$ and then flipping back over to $.04$ for thorium and $.05$ for uranium.

Now here's a question: I said the maximum of nuclear binding energy per nucleon happened at $\mathrm{^{62}Ni}$, so why does the fractional part of the masses in the table keep decreasing all the way to tin $\mathrm{^{116}Sn}$? It's because these masses are not normalized with respect to how many protons and neutrons there are in the nucleus. If you were to take all the isotopic masses and divide them by their respective mass numbers, you would see the minimum around $\mathrm{^{62}Ni}$, as expected (actually, the minimum ratio of isotope mass to mass number happens at $\mathrm{^{56}Fe}$ because it has a slightly higher proportion of protons to neutrons. As I said, the small difference between proton and neutron masses creates slight irregularities which I did not take into account for simplicity).

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    $\begingroup$ Aha - the atomic mass of an isotope is not quite the same thing as the mass number. I see (light bulb lights up above head!) $\endgroup$
    – marktwo
    May 28, 2015 at 23:35

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