I found different values of Avogadro constant in different places. So what is the correct value?

$\pu{6.0221367*10^{23}}$ or
$\pu{6.02214129*10^{23}}$ or
$\pu{6.0221415*10^{23}}$ or anything else?


Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold!

$$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

It is unlikely this value will change within our lifetimes, which is exciting in its own way!

For historical reasons, I'll leave my previous answer below:

As of 2015, the latest data for the Avogadro constant is from 2014. According to CODATA, the most accurate value is:

$$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$

The relative uncertainty in the measurement is thus only 12 parts per billion!

Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light. This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic $\ce{^{28}Si}$ monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel.

All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry.

Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was:

$$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$

This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science!

| improve this answer | |
  • 7
    $\begingroup$ Personally, I like memorizing Avogadro's constant as $6.022\times 10^{23}\ \mathrm{mol^{-1}}$, because it nicely matches the fundamental charge $1.6022\times 10^{-19}\ \mathrm{C}$ and I recall that $22^2=484$, which reminds me of the Faraday constant $96484\ \mathrm{C\ mol^{-1}}$ (actually $96485\ \mathrm{C\ mol^{-1}}$ is a better approximation). Precision to spare for calculations, and pretty mathematical patterns to help my poor memory! $\endgroup$ – Nicolau Saker Neto Jun 4 '15 at 16:39
  • $\begingroup$ +1 especially for link to Veritasium. If physicists really wanted to make things easier for everyone, they would get rid of both the Faraday constant and Avogadro's constant, defining a new "Faragadro constant" as being precisely equal to say $10^21$. That would be both the number of particles in a "new mole" (we need another name for that). Coloumbs and amperes would be made obsolete and current and charge would be counted in "new moles per second" or "new moles". Volts would be made obsolete, we could just say "kJ per new mole" instead. $\endgroup$ – Curt F. Jun 4 '15 at 19:03
  • 1
    $\begingroup$ +1, very good answer. For what concerns the redefinition of the kilogram, this is part of the ongoing redefinition of the SI (don't hold your breath, it'll be in 2018 probably). A glimpse of the so-called new SI can be found on this BIPM webpage where there is a draft of the new SI brochure and other information. The new unit definitions are implicit, from the given, exact, values of the constants (the values given in the draft are not definitive). $\endgroup$ – Massimo Ortolano Jun 4 '15 at 20:53
  • 2
    $\begingroup$ @Equinox If Avogadro's constant is changed to an exact value by definition (which seems likely this year or the next), I'll be sure to perform the last edit ever to this answer with the details! $\endgroup$ – Nicolau Saker Neto Mar 4 '18 at 21:38
  • 2
    $\begingroup$ @NicolauSakerNeto "If Avogadro's constant is changed to an exact value by definition" seems to have happened prior to that comment of yours ;) See Felipe's answer below. $\endgroup$ – Gaurang Tandon Jul 25 '18 at 16:27

A new definition of Avogadro's constant will replace the current one soon (emphasis mine):

The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly $\pu{6.02214076 \times 10^{23}}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, N$_\text{A}$, when expressed in mol$^{-1}$, and is called the Avogadro number.

This makes Avogadro number a fixed integer. The new definition was published on 8 January 2018 as an IUPAC Recommendation in Pure and Applied Chemistry, which is available online.

It has yet to be approved by CGPM (expected November 2018), but "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019" (thanks R.M.!).

| improve this answer | |
  • 2
    $\begingroup$ It was published as an IUPAC Recommendation on January 8th 2018. It has yet to be approved by CGPM (expected Nov 2018), and as your article states: "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019". $\endgroup$ – R.M. Jul 25 '18 at 23:29
  • $\begingroup$ Thanks for correcting it, I changed the text to reflect that. $\endgroup$ – Felipe S. S. Schneider Jul 26 '18 at 0:35

Avogadro's number has changed over time because it depends on how we standardize the atomic mass unit. Formerly, oxygen was used as a standard with a value of 16, and from a comment in another post I recall that physicists used 16 specifically for oxygen-16 (making natural oxygen slightly heavier than 16). When they went to the carbon 12 = 12 standard naturally occurring oxygen dropped below 16, so Avogadro's number represented less mass of oxygen (and everything else) and it dropped accordingly.

| improve this answer | |

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.