# Is it possible for me to derive Avogadro's number?

To my understanding the mole is the unit used to translate between mass on the atomic level and mass in the macro level, defined as the number of atoms in $12$ grams of Carbon-$12$, which apparently turns out to be $\pu{6.0221409e23}$ atoms, and this was defined as Avogadro's Number.

How was this number derived? Is it something I can derive myself? If I had $12$ grams of Carbon-$12$, how would I be able to deduce that it had $\pu{6.0221409e23}$ atoms?

• It would be hard, very hard. I'll point out that there is a gigantic different in stating that $N_A \approx 6.022 \times 10^{23}$ and stating that $N_A = 6.0221409\times 10^{23}$. Determining a value to eight significant figures requires extremely good experimental technique. – MaxW Jul 13 '18 at 3:29
• @MaxW How did Avogadro do it in his time, to whatever degree of precision that he did? – user51819 Jul 13 '18 at 3:48
• This information is readily available on the web in sources such as Wikipedia. en.wikipedia.org/wiki/Avogadro_constant, "By dividing the charge on a mole of electrons by the charge on a single electron..." Look up Millikan oild rop experiment for charge of e-, Faraday and electrolysis for atomic mass/charge ratio etc. – DrMoishe Pippik Jul 13 '18 at 4:36
• @DrMoishePippik What is interesting to me is that apparently the oil drop experiment gave an answer that was technically wrong / inaccurate, but we still used it anyway to derive all these other numbers? – user51819 Jul 13 '18 at 4:43
• It's been refined over time. As @MaxW states, original values had a very large margin of error. It's like Galileo using a chandelier as a clock... instrumentation has improved since then. – DrMoishe Pippik Jul 13 '18 at 5:09

As per the answer to this question on Chemistry SE it might become the other way round. The Avogadro number will be absolute (defined, with 0 error) and then the kilogram will be redefined as a function of the number of atoms in a monocrystalline perfect $\ce{^{28}Si}$ sphere.