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The mole is defined as the amount of a chemical substance which contains as many representative particles, e.g., atoms, molecules, ions, electrons, or photons, as there are atoms in 12 grams of carbon-12 (12C), the isotope of carbon with relative atomic mass 12 (from Wikipedia).

However, when I go to websites such as http://www.webqc.org/molecular-weight-of-C.html or http://www.convertunits.com/molarmass/Carbon and I ask what is the molar weight of Carbon, I get answers such as:

Molar mass of C is 12.01070 ± 0.00080 g/mol

When I asked on a forum why this isn't precisely 12, I was told that typically carbon is a mixture of isotopes which have relative atomic masses of 12, 13 and 14. With the ratio of each, the average value comes out to 12.0107.

But where does that number come from? Is it by definition? Is it by convention? Is it a practical number? Where do the error bars come from? For instance, I would expect to see different isotope distributions at different geographic locations (or what about the moon, asteroids, Jupiter, etc.). I tried googling for "standard carbon isotope ratio" but didn't find anything definitive. Of course then I start to wonder how molar weights of other elements are defined, with varying isotopic combinations, and I have the same question.

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    $\begingroup$ It is a number determined experimentally, hence the error bars. As for the abundance of different isotopes, these were mixed pretty thoroughly during the formation of the Solar System, though some minor variations do exist. $\endgroup$ – Ivan Neretin Apr 4 '17 at 18:07
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As you stated the molar mass is a weighted average based on isotopic abundance, specifically abundance on Earth. We can obtain the molar mass of carbon by averaging the mass of $\ce{^{12}C}$ (about $98.938\%$ abundance) and $\ce{^{13}C}$ (about $1.078\%$ abundance). $$(\mathrm{mass\space of\space \ce{^{12}C}})*(.9889)+(\mathrm{mass\space of\space \ce{^{13}C}})*(.0111)=(12)*(.9898)+(13.00335)*(.0111)=12.0206\text{g}$$ This deviates from your value above because the values I used for the abundance aren't very precise.

In terms of where these abundances comes from and how the error bars are determined, the short answer is that its complicated. The long answers is here in a technical report by IUPAC, particularly sections 1.2-1.4.

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    $\begingroup$ Thanks, that IUPAC report is exactly what I was looking for. $\endgroup$ – Purplejacket Apr 6 '17 at 0:45

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