Talk:Avogadro constant: Difference between revisions

Content deleted Content added

Inline

Revision as of 19:15, 24 January 2006

The following statement is incorrect:

Such an atom consists of 6 protons, 6 neutrons and 6 electrons, and N_A is therefore equal to 12 grams divided by the sum of the masses of a proton, neutron and electron.

While an atom of Carbon-12 does include the various subatomic particles mentioned above, it also has some mass due to the nuclear binding energy. Therefore, the sum of the masses of 6 free protons, neutrons and electrons will be different from that of an atom of Carbon-12. -- User:Matt Stoker

Ah yes, that makes sense. I originally added the above sentence, and I was a little worried that the numbers didn't come out quite right... :-)

But now you've got me thinking. The example on mole unit adds molecular masses like this: atomic mass of carbon is 12, atomic mass of hydrogen is 1, so molecular mass of C₂H₆ is 2*12+6 = 30. This is incorrect, isn't it? It neglects the mass in the binding energy. AxelBoldt

Strictly speaking, you are correct that the chemical binding energy contributes mass. However, this mass is very small compared to the mass due to the nuclear binding energy and so can usually be neglected. For example, the chemical binding energy in a hydrogen molecule contributes a mass of ~2.4E-9 g/mol. By contrast the nuclear binding energy in Deuterium contributes a mass of ~0.0029 g/mol (over six orders of magnitude greater than the chemical binding energy). For the purpose of defining Avogadros Number to arbitrary precision, both nuclear binding energy and chemical binding energy must be accounted for and I would assume the definition refers to 0.012 kg of free Carbon-12 atoms (ie. no chemical bonding), since graphite or diamond would have a slightly different mass. However, for most other applications the chemical binding energy contribution to mass can be neglected, so the example under mole unit should be fine. --User:Matt Stoker

I see, thanks. This is interesting stuff, too interesting to be hidden away in Talk:. It would be nice if we had this information either in mole unit or Avogadro's number.

Also, am I correct in assuming that free protons have a higher mass than nuclear ones? AxelBoldt

This last question is not as simple to answer as it sounds, since there is no way to independently determine the mass of a proton bound into a nucleus. However, it is true that the atomic mass of an atom is always less than the sum for a corresponding number of free electrons, protons and neutrons. This mass deficiency corresponds to the nuclear binding energy (E=mc^2) of the atom and represents the amount of energy that would have to be added to break the atom into its component subatomic particles. If you plot the mass per nucleon (neutrons and protons) vs. number of nucleons for all of the isotopes, you will notice that it initially decreases, reaches a minimum at 56 nucleons, then gradually increases. (This is why fusion is possible for light elements and fission for heavy elements). --User:Matt Stoker

It's a Good Thing that such discussions take place and result in encyclopedic material. However, I think that the section Connection to the mass of protons and neutrons has become a little unwieldy (like its title). Perhaps we could shorten this to maybe two or three well chosen sentences?

—Herbee 01:04, 2004 Apr 3 (UTC)

just a detail about the name: after the change, it could now seem that it was discovered by Amedeo Avogadro, while it is instead named this way only in honour of him (and after his death). Wasn'it more correct before?

Ok, I'll put something in to that effect. AxelBoldt

True or false??

True or false: it has been proven that Avogadro's number is not an integer. 66.245.8.219 23:16, 13 Sep 2004 (UTC)

Currently it isn't an integer, however it would be possible to get an integer if we redefined the kilogram. This might sound stupid, but in fact it would be pretty reasonable to redefine the kilogram in a way that would give you an integer.

Disagree. Why "currently" not an integer? Currently, the recommended value is 6.0221415×10^{23 which is an integer! And, if it is defined as "the number of atoms...", then it is also an integer (since "atom" means "not divisible" ;-). <nowiki></nowiki>— [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:01, 24 January 2006 (UTC)}[reply]

Avogadro's number is essentially just a conversion factor between the microscopic mass system (atomic mass units or Daltons) and the kilogram system. The microscopic mass system is based on the mass of Carbon-12, while the kilogram system is currently based on the mass of a particular "standard" lump of metal in France. So naturally there's no clean integer conversion relationship between the two. However, it wouldn't be totally outlandish to redefine the kilogram in terms of some particular number of atoms, rather than an arbitrary lump. If the atoms picked were Carbon-12, then you would end up with an integer relationship (because of the nuclear binding mass/energy, it would have to be Carbon-12, see above).

Essentially, this discussion shows that whether or not Avogadro's number is an integer isn't a very interesting question; Avogadro's number is simply a conversion factor, not some fundamental physical value, so whatever number it happens to be is just not that significant. --Chinasaur 01:12, 14 Sep 2004 (UTC)

Amp

I thought remembered this. Is it woorth adding?

A coulomb is Avagadros Number of electrons (6.023x10**23). AMPERE - The unit used for measuring the quantity of an electrical current flow.

One ampere represent a flow of one coulomb per second.--Jirate 01:48, 14 Sep 2004 (UTC)

No, if you check the Coulomb article you'll see that's incorrect. --Chinasaur 02:38, 14 Sep 2004 (UTC)

but not according to * http://fig.cox.miami.edu/~lfarmer/BIL265/BIL2001/neuron/tsld010.htm it is. It also corrisponds with my memory though the SI defintion of an amp invloves wires of negligable xsection and infinite length in vacuum.

No, the parenthetical statement on that slide is definitely wrong or confused. The rest of the slide is right though: F is equal to elementary charge (charge on one electron) times Avogadro's number. This is probably what you are thinking of. As you can see, F is not 1C, but rather ~96500C. --Chinasaur 01:07, 16 Sep 2004 (UTC)

I 'll go for that but their are several places on the WWW that are wrong. Not just my memory.--Jirate 12:39, 16 Sep 2004 (UTC)

A's # is just a convertion factor then?

In chemistry class, we always looked upon Avogadro's number as an almost magical physical universal cosmological constant. Yet I think this type of thinking could lead to a severe misunderstanding of subject (I didn't argue with the professor because I wanted good grades).

Isn't the number simply the number of amu's in a gram? That means that it is no more than a convertion factor, like 2.45 cm = 2 inch, or 1 pound = 4.45 newtons. There is nothing "special" about the gram; it's just an arbitrary unit, as is the amu (albeit less arbitrary).

So then, if you consider avogadro's number a "physical constant," that you might as well consider 2.45 and 4.45 "physical constants" too. The magical avogradro's number is in no way even close to being in the league of true constants, such as the gravitational constant, or the cosmological constant, or planck's constant, or the fine structure constant etc.

This artical does not reflect that. It seemes we are elevation this number to a place it does not belong. GWC Autumn 57 2004 13.10 EST

That's right, Avogadro's number is just as "thingys in an arbitary wossname" type of number, caused by the fact that the original definition of the kilogram had something to do with the density of water. There's no real reason why the mass unit should not be defined in some other way involving the other fundamental units and dimensionless constants: in fact, there are a number of efforts to do exactly that. -- Anon 18:36, 17 Nov 2004 (UTC)

You will find this discussed fairly exhaustively two headers above. I'm starting to get the impression this misconception should be addressed more explicitly in the article proper. --Chinasaur

I've refactored the article to introduce N_A as a scaling factor, rather than as a derived value from the mole. -- Karada 11:21, 2 Dec 2004 (UTC)

What symble does A constant have

Just, im currently at school, and weve been told its L, not N_A, although i expect its just a school thing... tooto 17:59, 3 Jan 2005 (UTC)

Both (check IUPAC's Goldbook: Goldbook). Though L may be used as a Loschmidt constant which is No. of molecules in 1 m³ (at normal p,T conditions), so it better to use N_A which cannot be taken for anything else. AWM~ads 23:38, 25 Mar 2005 (UTC)

Mathematical constant???

It's said on the page that $N_{A}$ is a mathematical constant. I think the author meant to say dimensionless constant, perhaps? "Unlike physical constants, mathematical constants are defined independently of any physical measurement." (from mathematical constant). --207.216.107.80 18:54, 27 Feb 2005 (UTC)

Not mathematical but physical/chemical constant AWM~ads 23:38, 25 Mar 2005 (UTC)

It is not dimensionless. Its units are inverse moles which depend eventually on the kilogram. If the kilogram were twice as large, Avogadro's number would also double. Pdn 06:02, 18 July 2005 (UTC)[reply]

Move Avogadro's number to Avogadro's constant

Avogadro's number is only a historical name/old terminology (though still widely used). The only legal one is Avogadro's constant. The reason: it is not a number (dimensionless) but has quite clearly a dimension [mol^-1]. The same situation is with Loschmidt constant (historically called Loschmidt number - it is quoted in Loschmidt's biogram). Avogadro's constant should be left as a REDIR to Avogadro's number (same with L.no.).

Check the IUPAC's Compedium of chemical Terminology Goldbook

See NIST CODATA (BTW it is cited in this article):

Avogadro's constant: NIST CODATA http://physics.nist.gov/cgi-bin/cuu/Value?na%7Csearch_for=Avogadro
Loschmidt constant: http://physics.nist.gov/cgi-bin/cuu/Results?search_for=Loschmidt

AWM~ads 23:38, 25 Mar 2005 (UTC)

Additional Physical Relations

I thought the old, pre-carbon AMU standard was set to 1/16th of oxygen-16, not to hydrogen-1. --66.81.79.104 04:01, 18 July 2005 (UTC) AKA User:Arkuat[reply]

I agree. When I was a boy (1940's) I learned some atomic weights and Hydrogen was 1.008 Pdn 06:03, 18 July 2005 (UTC)[reply]

Fundamental Constant?

quote: "The value of Avogadro's number depends on the definition of the mole, which depends on the definition of the kilogram. Both definitions, especially that of the kilogram, are arbitrary: the kilogram system is currently based on the mass of a particular "standard" bar of metal in France. Clearly, this means that the value of Avogadro's number is less fundamental than other physical constants in the sense that there is no physical reason for its particular value. However, Avogadro's number is still a fundamental constant: all constants depend on the units used and on the definition of the units, and therefore, such a dependence does not exclude that a constant can be called fundamental."

Isn't that just plain nonsense? Avogadro's number is just a conversion factor. What's so fundamental about that? And the claim that "all constants depend on the units used" is just plain wrong. For example the really fundamental Fine_structure_constant is dimensionless.

I'm not entire sure about the exact definition of a "fundamental constant", so I won't edit this now. But I hope someone can look into it.

Well, surely at least it should match with Fundamental physical constant, which says it should be "independent of systems of units". By that standard, Avogadro's number is not fundamental. I think a good operative definition of a fundamental constant is whether space aliens can be expected to know the same numerical value as we do.

I edited the paragraph now. I hope it's correct now :)

redudant

Does the first paragraph have to say "defined as the number of ~~carbon-12~~ atoms in 12 grams (0.012 kg) of carbon-12"? Or is it like "the number of carbon-12 atoms in a typical sample of carbon that may not be entirely carbon-12" or something like that? User:Omegatron/sig 03:37, 5 October 2005 (UTC)

It's probable slightly redundant, BUT not in the way you are surgesting, the Avogadro's number/Constant, is ment to be a constant so it can be used as a conversion factor... basicaly the sample is ment to be "number of atoms in 12 grams of pure carbon-12" (therefore they should all be caron-12 atoms).

note that if it were to be measured using, say using Carbon-14 then you would need to use 14 grams to work out the ratio. more generaly

"y grams of a molecule whose relative molecular mass is y, will contain 6.022 x 10^23 molecules." Carbon-12 is normaly used because measurments can be (traditionaly) carried out more acuratly on it.tooto 18:43, 13 October 2005 (UTC)[reply]

this latter "definition" is really redundant (or rather useless), since "molecular mass" is defined (via 1 mole) in terms of Avogadro's number. <nowiki></nowiki>— [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:13, 24 January 2006 (UTC)[reply]

reversion 23 Oct 2005

I reverted the article tonight because of a copyvio from

http://www.moleday.org/htdocs/avogadro.html

Ian Cairns 21:51, 23 October 2005 (UTC)[reply]

mysterious 6 023 x 10 23

Strangely enough, I always have remembered N as 6.023 x 10^23, and this value is present on many many places (even in the introduction of this article, in contradiction with the value given later (I'll fix this right now)). What is the reason thereof ?

I couldn't find any info about this number's (precision) history, but I suppose we know it to 3 decimal places since quite some time. (And I also exclude a change of convention/definition that would induce such an important discrepancy.) So, is it just the (false) mnemonic effect of 6023 x 1023 that makes up the popularity of this error? <nowiki></nowiki>— [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:15, 24 January 2006 (UTC)[reply]