Deriving Avogadro's Number without using "mol"

[Moondog]

TL;DR

Defining the quantity of a mol while using a mol of something (electrons) to divide into is like using the word being defined in the definition.

Frustratingly, everything I read about deriving Avogadro's number uses the word "mole" somewhere in the explanation. Per Scientific American, for example, Robert Millikan divided the charge on a mole of electrons by the charge on a single electron to obtain a value of Avogadro’s number of 6.02214154 x 1023 particles per mole. How did Millikan know how many electrons were in that mole to begin with?

 

[Dale]

everything I read about deriving Avogadro's number uses the word "mole" somewhere in the explanation

Of course it does. Avogadro’s number is the number of particles per mole. There is simply no possible way to talk about Avogadro’s number without talking about moles.

 

[ergospherical]

You can say that in 12g of carbon-12 there are Avogadro's number of carbon atoms.

 

[Dale]

You can say that in 12g of carbon-12 there are Avogadro's number of carbon atoms.

Which is, not coincidentally, one mole of carbon atoms.

 

[ergospherical]

Not exactly any more, but in practice yes (anywhere outside of the système international HQ).

 

[Ibix]

The Wiki article on Avogadro's number points to the one on Faraday's constant which says this was initially measured by measuring the mass of silver electroplated onto something by a known charge.

 

[Borek]

Historically mole was first defined as 16 grams of oxygen-16. You take 16 grams, you divide it by O-16 atom mass and you know mole contains about 6.02e23 elementary entities. This number was called Avogadro's number, but its value depended and how accurately we could measure 16 grams of substance and how accurately we could determine mass of an atom. That's where the value used by Millikan came from.

Later the mole definition was changed to 12 grams of carbon-12, which was a bit easier to realize in practice, but suffered from the same accuracy problems, and Avogadro's number was still something determined experimentally. Once we got deeper and required higher accuracy it became obvious this definition is inconvenient as well. Final solution was to make the things other way around - define that Avogadro's number is exactly some number (you can google exact value), and the mole is this number of elementary entities. That removes ambiguities and inaccuracies.

 

[Moondog]

Of course it does. Avogadro’s number is the number of particles per mole. There is simply no possible way to talk about Avogadro’s number without talking about moles.

My question is being misunderstood (my fault). How can Milikan talk about a MOL of electrons before the mol had been quantified at 6.022 e23? Put another way, what was Milikan's understanding of how many electrons were in a mol of electrons?

 

[Moondog]

Of course it does. Avogadro’s number is the number of particles per mole. There is simply no possible way to talk about Avogadro’s number without talking about moles.

What is the number of particles in a mol of electrons? When Milikan divided the mol of electrons by the electron charge, how did he get 6.022e23?

 

[Dale]

What is the number of particles in a mol of electrons?

The number of particles in a mol of anything is Avogadro’s number.

How can Milikan talk about a MOL of electrons before the mol had been quantified at 6.022 e23? Put another way, what was Milikan's understanding of how many electrons were in a mol of electrons?

The mol had been quantified well before Millikan. I think that in his day Avogadro’s number was defined as the number of atoms in 16 g of oxygen, but I could be wrong on that. As always one mol of any substance was Avogadro’s number of particles of that substance.

 

[Ibix]

My question is being misunderstood (my fault). How can Milikan talk about a MOL of electrons before the mol had been quantified at 6.022 e23? Put another way, what was Milikan's understanding of how many electrons were in a mol of electrons?

That was the point of the Faraday constant reference. They knew the charge a mole of electrons had, so they knew when they had a mole of electrons even if they didn't know Avogadro's number.

 

[Borek]

My question is being misunderstood (my fault). How can Milikan talk about a MOL of electrons before the mol had been quantified at 6.022 e23? Put another way, what was Milikan's understanding of how many electrons were in a mol of electrons?

Have you read my post?

 

[robphy]

When I need the mass of a proton (or mass of the hydrogen atom) approximately,
I use the Avogadro number.

1/(avogadro) is approximately the proton-mass in grams.

 

[Moondog]

When I need the mass of a proton (or mass of the hydrogen atom) approximately,
I use the Avogadro number.

1/(avogadro) is approximately the proton-mass in grams.

Robphy: How did Millikan know what a "mol of electrons" was before he had quantified Avogadro's number? It is said he determined Avogadro's number by dividing a mol of electrons by the charge of an electron to get 6.022e23. Please tell me what number is equal to a mol of electrons?

 

[Moondog]

That was the point of the Faraday constant reference. They knew the charge a mole of electrons had, so they knew when they had a mole of electrons even if they didn't know Avogadro's number.

Ibix: What is a mole of electrons? Quantify it for me if you can.

 

[Dale]

@Moondog you are asking questions that have already been answered in this thread

 

[Moondog]

@Moondog you are asking questions that have already been answered in this thread

I guess I missed it. What is a mol of electrons? If it's 6.022e23 then why is Millikan doing the experiment since he would already know what a mol is?

 

[robphy]

It might be good to go to the source.

R. A. Millikan, “On the elementary electrical charge and the Avogadro constant,” Phys. Rev. 2, 109–143 (Aug. 1913).
https://journals.aps.org/pr/pdf/10.1103/PhysRev.2.109
https://authors.library.caltech.edu/6438/
https://history.aip.org/exhibits/gap/PDF/millikan.pdf

Milikan wrote on p. 109:

The experiments herewith reported were undertaken with the view of introducing certain improvements into the oil-drop method of determining e and N and thus obtaining a higher accuracy than had before been possible in the evaluation of these most fundamental constants.Millikan wrote on p.140:

When correction is made for the fact that the numbers in Table XX. were obtained on the basis of the assumption $\eta_{23} = .0001825$, instead of $\eta_{24} = .0001824$ (see sec.2) the final mean value of $e^{2/3}$ obtained from the first 23 drops is $61.085\times 10^{-8}$. This corresponds to
$$e = 4.774 \times 10^{-10}\ \mbox{electrostatic units}.$$

Since the value of the Faraday constant has now been fixed virtually by international agreement at 9,650 absolute electromagnetic units and since this is the number N of molecules in a gram molecule times the elementary electrical charge, we have
$$\begin{align*}
N\times 4.774\times 10^{-10} &=9,650\times 2.9990 \times 10^{10}\\
\therefore N &= 6.062 \times 10^{23}
\end{align*}
$$

For more details, you should consult the original paper linked above.

(I'm not up on the history of this topic... I just went to the source and scanned for N.)

 

[Dale]

What is a mol of electrons? If it's 6.022e23 then why is Millikan doing the experiment since he would already know what a mol is?

As I already told you, a mol of anything is Avogadro’s number of particles. In Millikan’s day Avogadro’s number was the number of atoms in 16 g of oxygen 16, which is a quantity that has to be measured.

 

[Ibix]

Ibix: What is a mole of electrons? Quantify it for me if you can.

The point is that he doesn't need to know the value of Avogadro's number to know that he has one mole of electrons. Faraday's constant tells him that one mole of electrons has a certain charge - when he's accumulated some known fraction of that charge he has that fraction of a mole.

Faraday's constant was derived without knowing the value of Avogadro's number because Faraday understood how many electrons he had to supply to deposit one atom of silver. When he'd deposited 108g of silver, 1 mol of silver, by definition he'd supplied that many moles of electrons. So he knew the charge of one mole of electrons without knowing how many electrons that was.

Then the maths in the quote @robphy provided gets you the rest of the way, I think.

 

[Moondog]

It might be good to go to the source.

R. A. Millikan, “On the elementary electrical charge and the Avogadro constant,” Phys. Rev. 2, 109–143 (Aug. 1913).
https://journals.aps.org/pr/pdf/10.1103/PhysRev.2.109
https://authors.library.caltech.edu/6438/
https://history.aip.org/exhibits/gap/PDF/millikan.pdf

For more details, you should consult the original paper linked above.

(I'm not up on the history of this topic... I just went to the source and scanned for N.)

Robphy: Thanks so much for providing the link to R.A. Millikan's original work. It will take time to plow through. Avogadro's constant should equal roughly the mass of the number of protons and neutrons in 12 grams of C-12, yet it doesn't add up. Any idea why not? (Note: I assume neutron mass counts since we specify Carbon-12.)

 

[robphy]

Robphy: Thanks so much for providing the link to R.A. Millikan's original work. It will take time to plow through. Avogadro's constant should equal roughly the mass of the number of protons and neutrons in 12 grams of C-12, yet it doesn't add up. Any idea why not?

What is the ratio between what you calculated and what you expected to get?
Please show your work.

 

[Moondog]

What is the ratio between what you calculated and what you expected to get?
Please show your work.

P(mass) = 1.67262e-24g x 6 protons = 1.0035738e-23 g
N(mass) = 1.67492e-24g x 6 neutrons = 1.0049571e-23g
Total C(mass) = 2.00853e-23g
2.00853e-23g x 12g(C) = 2.410236e-22g (Mass of 12 grams of C-12)

 

[Dale]

Avogadro's constant should equal roughly the mass of the number of protons and neutrons in 12 grams of C-12

Not quite. Avogadro’s constant equals 12 g divided by the mass of a single carbon-12 atom. The mass of a single carbon-12 atom will be roughly equal to the mass of 6 protons and 6 neutrons.

2.00853e-23g x 12g(C) = 2.410236e-22g (Mass of 12 grams of C-12)

This is in units of $\mathrm{g^2}$. Avogadro’s number is not a mass-squared.

 

[Moondog]

Not quite. Avogadro’s constant equals 12 g divided by the mass of a single carbon-12 atom. The mass of a single carbon-12 atom will be roughly equal to the mass of 6 protons and 6 neutrons.

This is in units of $\mathrm{g^2}$. Avogadro’s number is not a mass-squared.

Dale: I don't see any mass squared in my above. That said, I see my mistake: I needed to have divided 12 not multiplied by 12. (whoops!) Thanks for the help.

 

[Dale]

: I don't see any mass squared in my above.

It was in your calculation that I directly quoted. On the left hand side of the equal sign you had grams times grams, that is grams squared.

 

[Moondog]

It was in your calculation that I directly quoted. On the left hand side of the equal sign you had grams times grams, that is grams squared.

Roger that, a careless mistake. Thanks. I appreciate all of your time and attention. I'm now able to follow the effort to standardize the kilogram in terms of Avogadro's number and Planck's constant.

 

[Mike S.]

The confusingness of how moles are presented is right up there with my pet peeves about how chemistry is taught. I mean, they hit people with the most absurd nonsense - teach them two different terms (mole and Avogadro's number), call it a "unit of quantity", otherwise known as a number, and treat it as an SI unit, as if it were not possible to convert it to a number and totally take it out of the equation. And to top it off, they hit them with moronically worded problems meant to confuse the living daylights out of people, like training them to write "1 mole = 6.02E+23 molecules" to answer repeated calculations. NO! Students need to understand that 1 mole = 6.02E+23, period, no units after that number, and that a mole only equals some number of molecules if you're talking about a mole of molecules, which is multiplied by that number; otherwise it could be atoms or peas. I bet 10% of the people who veered away from STEM majors were lost to this tomfoolery.

Oh, and anyway, this number is simply the number of carbon-12 atoms in 12 grams of carbon, a ratio chosen to make 12 daltons / atomic mass units (roughly 12 nucleons) match up precisely with 12 grams.

 

[Dale]

treat it as an SI unit,

It is an SI unit, according to the BIPM which is the only authoritative body on the definition of the SI.

You have reasonable objections to the BIPM’s decision, but their decision is authoritative and the mol is in fact an SI unit with its own dimension.

 

[Mike S.]

The average learner is far enough away from whoever BIPM is that they can probably figure out how to do chemistry before they have to worry about what those people think. :)