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The definition of atomic mass says that it is the mass of one atom of the element with respect to 1|12 of the mass of carbon-12 atom.

My question is that why there is need to compare the atomic mass with carbon-12 or any other element , why not just we measure the mass of 1 atom of an element and call it as the atomic mass.

Please do not mark this question as duplicate because I have searched the web but didn't find any appropriate answer.

Any help is appreciated.

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  • $\begingroup$ Please read this reference with historical point of view. $\endgroup$ – Mathew Mahindaratne Apr 14 '18 at 9:46
  • $\begingroup$ Not sure if I get this question right but if I call the mass of 1 atom hydrogen 1 atomic mass and I also call the mass of one atom helium 1 atomic mass and also 1 atom fluorine is one atomic mass then one atom mass isn't defined at all because every atom weights one atomic mass while in reality they all have a different weight. $\endgroup$ – DSVA Apr 14 '18 at 10:34
  • $\begingroup$ @DSVA I think he's just asking why is there's amu at all, and not just in SI unit. OP should put it more clear and not taunt me with supposed unduplicability of the question ;) $\endgroup$ – Mithoron Apr 14 '18 at 16:52
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the mass of carbon-12 is used as it is just a very abundant isotope. we then compare each of the masses to this one as it allows to have a comparison for all the other masses. if we did each element individually it would acts as if each one had a different scale system so it wouldn't be very useful.

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  • $\begingroup$ We could perfectly well give masses in SI units, why would you think there would be different scales? Also OP isn't asking why C-12 is used... $\endgroup$ – Mithoron Apr 14 '18 at 16:42
  • $\begingroup$ each element would have a slightly different value for each of an atomic mass of one, as you are comparing each element to its won control so there is no standardisation across all elements. but by using carbon-12 the control is the same for all elements so we can then compare all elements. $\endgroup$ – H.Linkhorn Apr 15 '18 at 14:17

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