Why do all atomic masses have decimals if there are some elements that don't have isotopes?

Question

I understand that the way to calculate the atomic mass is to obtain a weighted average of the masses of all the isotopes of an element. I also understand that there are about 26 elements that are monoisotopic. If the reason the atomic masses have decimals is because of this weighted average, then it should follow that the monoisotopic elements should have a whole-number mass. Therefore, why do all the atomic masses have decimals?

Answer 1

This is a simple question which has a complicated answer.

In simplest terms, there is one isotope, $\ce{^{12}C}$, which does have an integer atomic mass by definition.

unified atomic mass unit (u) - Non-SI unit of mass (equal to the atomic mass constant), defined as one twelfth of the mass of a carbon-12 atom in its ground state and used to express masses of atomic particles.

So for $\ce{^{12}C}$, there are exactly 12.000... grams of carbon per mole of the $\ce{^{12}C}$ atoms.

Now using the $\mathrm u$ as a stake in the ground, the weight of the various atomic particles are:

• neutron = 1.008 664 915 88(49) u
• proton = 1.007 276 466 879(91) u
• electron = 0.000 548 579 909070(16) u = $5.48579909070(16)\times10^{-4}$ u

Now a carbon atom has 6 neutrons, 6 protons and 6 electrons. A simple addition of all the individual particle masses yields about 12.09893977602 u. This is not exactly 12.000...

You can also see that the mass of a neutron isn't equal to the mass of an electron and a proton (1.007276466879 u + 0.000548579909 u = 1.007825046788 u).

You can also see that an electron and a proton have a different mass than the $\ce{^1H}$ isotope (1.00782503224 u)

The problem here is the binding energies of the nucleons and to a lesser extent the electrons. From Einstein's famous formula, $E=mc^2$, some of the mass of the particles is converted to the energies necessary to hold the particles together.

We could use a different scale other than $\ce{^{12}C}$, but the neutron and proton couldn't both have an integer mass. When you also consider the binding energies of the nucleus and the electrons, only a maximum of one isotope will be able to have an integer mass. (Why $\ce{^{12}C}$ was picked as the standard is another complicated answer.)

Answer 2

The answer to your question has to do with Einstein's most famous formula $$ E=mc^2 $$ which relates energy to mass, but what does it mean? Let's consider a simple example first. The simplest atom is of course the hydrogen atom which consists of one electron and of one proton. If you add the mass of the proton $1.672621898\times 10^{-27}$kg, to the mass of the electron $9.10938356\times 10^{-31}$ kg, you find a value of $1.67353284\times 10^{-27}$ kg. If you compare this to the mass of the $^1$H atom of $1.6735326915926\times 10^{-27}$ kg, you see that the hydrogen atom is lighter than the combined mass of the proton and the electron by $2.44017148\times 10^{-35}$ kg. What happened to the mass when the electron and proton got bounded together? Well, according to Einstein the difference in mass is converted to the binding energy of the electron! If you use Einsteins formula you find a binding energy of $-2.19311675\times 10^{-18}$ J or $-13.7$ eV which is very close to the Rydberg constant!

Let's now get back to other atoms. In the same way that the formation of a hydrogen atom from an electron and a proton lowers the mass of the atom by the corresponding binding energy, the binding energy between the neutrons, protons and electrons that make up heavier atoms also results in a reduction of the mass of the atom. Because the binding of the neutrons and protons is much stronger than the binding energy of the electrons to the core, the effect of the mass is also much larger. See for instance here for further details.