With the recent redefinition of the kilogram, what is the mass of $N_A$ (Avogadro's constant) of carbon-12 atoms? $N_A$ was defined as exactly 6.02214076×$10^{23}$ atoms. Then how close would the mass of $N_A$ atoms of carbon-12 be to 12 grams?

Is it still true that the mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 gram of carbon-12? Is a mole exactly $N_A$?

  • $\begingroup$ Have you read the relevant sections of physics.stackexchange.com/questions/147433/… ? $\endgroup$ – PM 2Ring Nov 21 '18 at 7:58
  • $\begingroup$ oh, I don't know why I didn't find that! Thanks. But reading over the section on "The Mole" leaves me puzzled as to the answer to my question ... $M_u=m_u N_A$ is the molar mass constant, which ceases to be fixed and obtains the same uncertainty as mu, equal to 1/12 of the mass of NA carbon-12 atoms. $\endgroup$ – Sheldon Nov 21 '18 at 10:54
  • 1
    $\begingroup$ My take is that a mole is exactly $N_A$, so a mole of $^{12}C$ isn't exactly 12g, instead it's 12.0000000g. $\endgroup$ – PM 2Ring Nov 21 '18 at 11:06
  • $\begingroup$ I assume it's as close to 12 grams within current experimental error. But still seems weird. Like why even bother defining $N_A\;$ as an exact number ??? $\endgroup$ – Sheldon Nov 21 '18 at 16:08

The status of the mass of a mole of $^{12}\rm C$ in the revised SI is basically identical to that of the magnetic vacuum permeability $\mu_0$, which is explored in this and this question. The SI revision does not consist of a re-definition of the kilogram in isolation $-$ instead, the full set of the seven base units get redefined, including in particular the mole, which is no longer defined in terms of a fixed value of the molar mass of carbon-12 and which is instead tied to a fixed value of the Avogadro constant. For the full details, see the over-arching Q&A What are the proposed realizations in the New SI for the kilogram, ampere, kelvin and mole? (as well as this table in the Wikipedia page to see how all the relevant units and constants change status), but the short answer to your question

Then how close would the mass of $N_A$ atoms of Carbon-12 be to 12 grams?

is that it will be close, but not exactly equal, and it will be an experimentally-determined quantity with a finite uncertainty in the future.

This is spelled out explicitly in the Resolution that implements the revision (which will presumably be named as Resolution 1 of the CGPM 2018), and particularly in its appendices, which contain an explicit definition of the mole,

The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly $6.022 140 76 \times 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, when expressed in the unit $\rm mol^{–1}$ and is called the Avogadro number.

The amount of substance, symbol $n$, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles.

as well as an explicit comment on the status of the molar mass of carbon-12,

the molar mass of carbon 12, $M(^{12}\mathrm C)$, is equal to $0.012 \:\rm kg \: mol^{–1}$ within a relative standard uncertainty equal to that of the recommended value of $N_Ah$ at the time this Resolution was adopted, namely $4.5 \times 10^{–10}$, and that in the future its value will be determined experimentally.

The appearance of the Planck constant $h$ on the relative uncertainty of $M(^{12}\mathrm C)$ after the change is directly tied to the fact that the relative uncertainty in the mass of the International Prototype Kilogram immediately after the change is exactly equal to the current relative uncertainty in $h$ (because of the way the switch is carried out, and because the second and the meter don't have significant changes).

  • $\begingroup$ Thanks for your answer. The uncertainty of $4.5 \times 10^{-10}$ seems very impressive; Wikipedia gives an uncrtainly for the Kibble balance measurement of Plank's constant as about $9 \times 10^{-9}$. $\endgroup$ – Sheldon Nov 24 '18 at 13:57

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.