# Understanding the reasoning behind the mole

## Homework Statement

My book says that if the ratio of the masses of the samples is the same as the ratio of the masses of the components then the two samples contain the same number of components. That part makes sense. But how do we know that the same number of components has to be 6.02 x 10^23?

Let's compare natural aluminium (average mass of 12.01 amu) and natural antimony (average mass of 26.98 amu). The overall sample mass of each is 12.01 g and 26.98 g so because the ratio of the sample masses (26.98 g/12.01 g) is equal to the ratio of the masses of the components involved (26.98 amu/12.01 amu) they have the same number of components. Now what is the next step to determining the number of components involved? (I know the rule states that it's 6.02 x 10^23, but I want to know the reasoning behind why)

epenguin
Homework Helper
Gold Member

## Homework Statement

My book says that if the ratio of the masses of the samples is the same as the ratio of the masses of the components then the two samples contain the same number of components. That part makes sense. But how do we know that the same number of components has to be 6.02 x 10^23?

Let's compare natural aluminium (average mass of 12.01 amu) and natural antimony (average mass of 26.98 amu). The overall sample mass of each is 12.01 g and 26.98 g so because the ratio of the sample masses (26.98 g/12.01 g) is equal to the ratio of the masses of the components involved (26.98 amu/12.01 amu) they have the same number of components. Now what is the next step to determining the number of components involved? (I know the rule states that it's 6.02 x 10^23, but I want to know the reasoning behind why)

Sorry, no one can answer your question, we can't know what it is, you haven't formulated anything understandable. You talk vaguely of "the sample" and we don't know what it is and what facts are given about it. You give a figure of 26.98 related to antimony - that figure happens to be the atomic mass of aluminium so I think you have several confusions.

AlephZero
Homework Helper
The first accurate measurement was by measuring the electric charge of one mole of electrons, divided by the charge on one electron. Later measurement were done by comparing the mass of one electron and a mole of electrons, or by finding the number of atoms in 1kg of a single isotope of silicon.

All right epenguin I'll address what you said first. So in my scenario there are two samples of two different elements. One sample is of aluminium and the other sample is of americium. The reason why the numbers are wrong is because I changed the elements at the last moment and forgot to change the amu for them as well. Now that this is clarified I'll just rewrite the paragraph with the correct masses. Sorry for the confusion.

Let's compare natural aluminium (aluminium atoms have an average mass of 26.98 amu) and natural antimony (antimony atoms have an average mass of 121.8 amu). There are two samples, one of each element. The mass of the first sample (aluminium) is 26.8 g and the mass of the second sample (antimony) is 121.8 g. Because the ratio of the sample masses (26.98 g/121.8 g) is equal to the ratio of the masses of the components (atoms) involved (26.98 amu/121.8 amu) theoretically they should have the same number of components (atoms). Now what is the next step to determining the number of components involved? (I know the rule states that it's 6.02 x 10^23, but I want to know the reasoning behind why)

I know that it is something that is determined experimentally but what is the explanation for why 26.98 g of aluminium has exactly 1 mol of aluminium atoms, just because we determined it experimentally?

Thank you all once again

Borek
Mentor
what is the explanation for why 26.98 g of aluminium has exactly 1 mol of aluminium atoms

That's because we defined it in such a way.

We defined mole to be "the amount of any substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 grams of pure carbon-12". This definition means we can - knowing mass of C-12 atom - calculate the number of these entities. It happens to be 6.02x1023. It simply follows from the definition.

You may ask why it is that mass of C-12 is exactly 12 - well, that's again because we defined it this way. a.m.u. could have any value (in terms of how many grams is an a.m.u.), but we selected value that makes calculations easy.

epenguin
Homework Helper
Gold Member
Let's compare natural aluminium (aluminium atoms have an average mass of 26.98 amu) and natural antimony (antimony atoms have an average mass of 121.8 amu). There are two samples, one of each element. The mass of the first sample (aluminium) is 26.8 g and the mass of the second sample (antimony) is 121.8 g. Because the ratio of the sample masses (26.98 g/121.8 g) is equal to the ratio of the masses of the components (atoms) involved (26.98 amu/121.8 amu) theoretically they should have the same number of components (atoms). Now what is the next step to determining the number of components involved? (I know the rule states that it's 6.02 x 10^23, but I want to know the reasoning behind why)

Yes that's right now. Those two samples contain the same number of atoms, one of aluminium, the other of antimony. The way this was known originally had nothing to do with this 6.02 x 10^23 not known till a lot later - rather it was the chemical combining weight. Aluminium (Al) combines with oxygen ("forms an oxide") and does also with antimony (Sb) and the of Al combines with the same amount (grams) of oxygen as does 121.8 g of Sb. And then the same combining ratios will be found in other compounds e.g. Aluminium sulphate bowever much sulphate (SO4) is combined with 26.98 g Al, the same amount of sulfate is combined with 121.8 g of Sb. And these non-trivial and experimentally found consistencies of ratios are very naturally explained by simple atomic theory.

The actual amount of oxygen in aluminium oxide can be calculated from the molecular formula Al2O3 and the atomic masses so 26.98 X 2 g Al combines with 16 X 3 g O and so does 121.8 g of Sb.
Etc. for other compounds.

So traditional chemistry, and most chemical laboratory calculations today, are about chemical combining ratios. They would be exactly the same calculations with the same exact numbers if the absolute masses of the atoms or the Avogadro number were a thousand or a million times bigger or smaller.

About the Avogadro number I haven't time this morning but others have said it, I want to come back later too.

Thank you all for your answers, a combination of Borek's and ePenguin's answer really cleared my confusion.