Do we have that $\pu{1 g/mol} = \pu{1 amu}$ ?

Because we have, for the mass of an atom of carbon 12, call it $m(\ce{^12C})$, that

$$m(\ce{^12C}) = \pu{12 amu}$$

and furthermore

$$\pu{1 mol} \cdot m(\ce{^12C}) = \pu{12 g}$$


$$m(\ce{^12C}) = \pu{12 amu} = \pu{12 g/mol}$$

So finally we get that $\pu{1 g/mol} = \pu{1 amu}$ .

However, my chemistry teacher is telling me that those are two completely different things and that I am confused between the mass per atom and the mass per $6.022\cdot10^{23}$ atoms. I can't understand how, and this is really bugging me, so help is very appreciated.

Note that this requires the mole to be a number (or a "constant"), which may be where I'm wrong.


3 Answers 3


You are correct, but to make it a little more clear you can include the assumed "atom" in the denominator of amu:

$$ \begin{align} m_{\ce{C}^{12}} &= \pu{12amu atom^-1} \\ \\ m_{\ce{C}^{12}} &= \pu{12g mol^-1} \\ \\ \pu{12amu atom^-1} &= \pu{12g mol^-1} \\ \\ \pu{1amu atom^-1} &= \pu{1g mol^-1} \end{align} $$

In other words, the ratio of amu/atom is the same as the ratio of g/mol. The definitions of amu and moles were intentionally chosen to make that happen (I'm surprised your teacher didn't explain this, actually). This allows us to easily relate masses at the atomic scale to masses at the macroscopic scale.

To check this, look at the mass of an amu when converted to grams:

$\pu{1amu}= \pu{1.6605E-24 g}$

Now divide one gram by one mole:

$\pu{1g mol^-1}= \frac{\pu{1 g}}{\pu{6.022E23 atom}} = \pu{1.6605E-24 g atom^-1}$

It's the same number! Therefore:

$\pu{1g mol^-1}= \pu{ 1 amu atom^-1}$

  • $\begingroup$ Thanks for the clarification. Do you know know any reputable reference where this is stated? I searched but couldn't find any. $\endgroup$
    – Oliver
    Commented Jun 20, 2014 at 16:55
  • 1
    $\begingroup$ I think any general chemistry textbook should have it - I know that Tro's Chemistry, 3rd ed. has it on page 71. $\endgroup$
    – thomij
    Commented Jun 20, 2014 at 17:51
  • $\begingroup$ It should be noted that amu as a unit is deprecated. Now the unified atomic mass unit u should be used instead. $\endgroup$ Commented Feb 25, 2016 at 7:34
  • $\begingroup$ @Martin that is a good point, but I think that for the most part 'amu' is understood to be the unified atomic mass unit based on carbon-12, not oxygen-16. I at least have always seen 'amu' and not 'unified amu' in publications within the last couple of decades. $\endgroup$
    – thomij
    Commented Feb 25, 2016 at 16:21
  • 2
    $\begingroup$ @thomij I'm really glad you are still with us here, I missed you quality answers recently. The confusion is one of the worst, and it is all IUPAC/IUPAP's fault. They should have used something completely different. However, your argument does not lose any value through that change. $\endgroup$ Commented Feb 25, 2016 at 16:46

You need to be more careful with your units. The erroneous result is that you are equating a value in amu (a measure of mass, like grams) with a value in grams per mole (an invariant property of an element or compound, regardless of the amount you have).

  • $\begingroup$ I am equating them since they are both the mass of an atom of carbon, and I do believe that the mass of an atom of carbon is equal to itself. What is wrong with that? It is nothing exceptional to have units expressed in terms of others. $\endgroup$
    – Oliver
    Commented Jun 20, 2014 at 15:01
  • $\begingroup$ I support this statement, technically the comparison is not accurate, I would upvote this, if there were a little more explanation. $\endgroup$ Commented Feb 25, 2016 at 7:36
  • 1
    $\begingroup$ It's like comparing a speedometer and an odometer - they measure 2 different things. Saying a car has gone 100 miles is not the same as saying it has gone 100 miles per hour. $\endgroup$
    – iad22agp
    Commented Feb 25, 2016 at 19:34

There are two things that routinely mystify science students:

  1. anything to do with amount of substance (now to be called "chemical amount"), the mole, and the Avogadro constant (or the Avogadro number), and

  2. anything to do with the now-you-see-me-now-you-don't radian. Let me address the first.

If we have a general number of entities of kind X (e.g. X is the chemical symbol) represented by N(X), the corresponding chemical amount of X is denoted by n(X), which is an aggregate of N(X) entities. In symbols: n(X) = N(X) ent, where ent represents an amount of one entity (atom, molecule, ion, sub-atomic particle, . . .), i.e. the entity itself.

The Avogadro number is the (dimensionless) ratio of one gram to one "atomic mass unit" (now called dalton, Da): g/Da. One mole is an an Avogadro number of entities: mol = (g/Da) ent. Thus we have the important relationship: Da/ent = g/mol = kg/kmol, exactly. In other words, at the atomic level, the appropriate unit for amount-specific mass ("molar" mass) is dalton per entity--and, because of the mole definition as an Avogadro number of entities, dalton per entity is exactly equal to the macroscopic units gram per mole or kilogram per kilomole.

The critical problem is that IUPAC does not have a recognised symbol for one entity. It is sometimes (incorrectly) thought of as the (dimensionless) number one. In which case the "mole" is simply another name for the Avogadro number: "mol = g/Da". In this case we have the (incorrect) relationship: "Da = g/mol". Tables of "atomic weights" list the numerical values of atomic-scale masses in daltons--e.g. Ar(O) = ma(O)/Da = 16. The corresponding amount-specific mass is M(O) = 16 Da/ent; and this is (exactly) equal to 16 g/mol or 16 kg/kmol.


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