# Why does the Avogadro's number have the unit of reciprocal mole? [closed]

In my understanding,

$$1 \ mol \equiv N_A \ (atoms)$$

(Don't say it might not be atoms but molecules and also don't say what's it's not even a unit)

Then the unit should be $$N_A = 1 \ mol/atoms = 1 \ mol$$, not the reciprocal mole. My first question is what's wrong.

My second question is, we know that defining $$atoms$$ as being unit is absurd, because it's purely a number. And same applies to mole. There's not difference between them in the sense that they are dimensionless. But why is the mole a SI unit?

• The unit is $\pu{1 mol} = N_\mathrm{A} \text{ particles}$ . What is strange on that ? Note that there is the unitless Avogadro number and the Avogadro constant with dimension $\pu{mol^-1}$. Then $n [mol] \cdot N_\mathrm{A} [1/mol] = N$. The molar amount with 1 mol as the unit is useful concept, as many quantities relate rather to the number of particles than to their total mass. – Poutnik Nov 6 '20 at 8:00
• You may take any non-unit and call it a unit, and as long as it doesn't contradict anything, you'll be fine. That's why mole is a unit. – Ivan Neretin Nov 6 '20 at 8:01
• "we know that defining atoms as being unit is absurd, because it's purely a number." the same applies to dollars or what ever currency. It seems you are fine with more abstract units (length for instance) but have problem with countable hardware. – Alchimista Nov 6 '20 at 8:43
• Does this answer your question? What is the dimension of Avogadro's constant? – Mithoron Nov 6 '20 at 17:31

I realised where my confusion lies - so I'm leaving an answer so that other people learning chemistry for the first time are able to fully understand what's actually going on about Avogadro's number.

We have to make a clear distinction between Avogadro's number and Avogadro's constant.

The former is a unitless real number, defined as $$6.02214076\times 10^{23}$$.

The latter is a physical (or shall I call chemical?) constant, and thus has a unit. As @Poutnik and other people have commented on my question, the fact that $$mol$$ does not have a dimension doesn't mean it can't be a unit (the same applies to radian as a measure of angles); and that means there should exist a constant called the Avogadro's constant with the dimension $$mol^{-1}$$ so that the product of the two is dimensionless, i.e.

$$n \ [mol] \ \times N_A \ [mol^{-1}] = N \ [particles]$$

Here it's trivial that [particles] is not a unit, simply because we did not feel the need to (and thus chose not to) assign any unit, dimension, and whatever onto that thing called number of particles.

Again, although $$mol$$ is dimensionless it should still be treated as a unit so in order to cancel that out to produce a unitless number $$N$$ we can conclude that we need a number with value $$6.02214076\times 10^{23}$$ and unit $$mol^{-1}$$, or namely Avogadro's constant.

• dimension of the mole is mole, as of the kilogram is kilogram and of the metre is metre. Where is the problem ? – Poutnik Nov 6 '20 at 11:54
• You have solved your confusion but please see Poutnik comment. – Alchimista Nov 6 '20 at 15:47
• @Poutnik Yes you are right. I agree with that, and in my answer I was just emphasising that. I was originally confused about that. – curious Nov 6 '20 at 16:26