Talk:Avogadro constant

The following statement is incorrect:

Such an atom consists of 6 protons, 6 neutrons and 6 electrons, and N_A is therefore equal to 12 grams divided by the sum of the masses of a proton, neutron and electron.

While an atom of Carbon-12 does include the various subatomic particles mentioned above, it also has some mass due to the nuclear binding energy. Therefore, the sum of the masses of 6 free protons, neutrons and electrons will be different from that of an atom of Carbon-12. -- Matt Stoker

Ah yes, that makes sense. I originally added the above sentence, and I was a little worried that the numbers didn't come out quite right... :-)

But now you've got me thinking. The example on mole unit adds molecular masses like this: atomic mass of carbon is 12, atomic mass of hydrogen is 1, so molecular mass of C₂H₆ is 2*12+6 = 30. This is incorrect, isn't it? It neglects the mass in the binding energy. AxelBoldt

Strictly speaking, you are correct that the chemical binding energy contributes mass. However, this mass is very small compared to the mass due to the nuclear binding energy and so can usually be neglected. For example, the chemical binding energy in a hydrogen molecule contributes a mass of ~2.4E-9 g/mol. By contrast the nuclear binding energy in Deuterium contributes a mass of ~0.0029 g/mol (over six orders of magnitude greater than the chemical binding energy). For the purpose of defining Avogadros Number to arbitrary precision, both nuclear binding energy and chemical binding energy must be accounted for and I would assume the definition refers to 0.012 kg of free Carbon-12 atoms (ie. no chemical bonding), since graphite or diamond would have a slightly different mass. However, for most other applications the chemical binding energy contribution to mass can be neglected, so the example under mole unit should be fine. --Matt Stoker

I see, thanks. This is interesting stuff, too interesting to be hidden away in Talk:. It would be nice if we had this information either in mole unit or Avogadro's number.

Also, am I correct in assuming that free protons have a higher mass than nuclear ones? AxelBoldt

This last question is not as simple to answer as it sounds, since there is no way to independently determine the mass of a proton bound into a nucleus. However, it is true that the atomic mass of an atom is always less than the sum for a corresponding number of free electrons, protons and neutrons. This mass deficiency corresponds to the nuclear binding energy (E=mc^2) of the atom and represents the amount of energy that would have to be added to break the atom into its component subatomic particles. If you plot the mass per nucleon (neutrons and protons) vs. number of nucleons for all of the isotopes, you will notice that it initially decreases, reaches a minimum at 56 nucleons, then gradually increases. (This is why fusion is possible for light elements and fission for heavy elements). --Matt Stoker