Talk:Avogadro constant

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The following statement is incorrect:

Such an atom consists of 6 protons, 6 neutrons and 6 electrons, and NA 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

[edit] 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 6.0221415×1023 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>&mdash; [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:01, 24 January 2006 (UTC)
Currently the value stated on the page is a plus or minus value. That is, an experimentally determined number. Keep in mind, when you see a number expressed in scientific notation, especially when it has many decimal places and a very high power of ten, is probably not precise. The trailing digits are likely to be non-zero; however, we do not know (or are not concerned) about what they are. Avagadro's number is not, as far as has been shown yet, an integer. Unless, of course, as stated above, we redefine the kilogram. However, the kilogram is defined, and there is no reason to think Avagadro's number is an integer.Vijay 05:24, 13 March 2006 (UTC)
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)

[edit] 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)

[edit] 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
NIST lists Avogadro's number as a "fundamental physical constant", as does the Institute of Chemistry at the Free University of Berlin. Are they wrong? Jeff Connelly

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)

[edit] What symble does A constant have

Just, im currently at school, and weve been told its L, not NA, 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 NA which cannot be taken for anything else. AWM~ads 23:38, 25 Mar 2005 (UTC)

[edit] Mathematical constant???

It's said on the page that NA 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)

[edit] 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):

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

[edit] 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

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)

Boy, you are too young. In my times (the early nineteenth century), the standard was hydrogen. ;-) Itub 21:34, 21 March 2006 (UTC)

[edit] 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 :)
Unfortunately, physicists do not agree on a single definition for "fundamental constant". NIST and John Baez use the term not only to refer to dimensionless constants but also dimensioned ones, which are not independent of units. It appears the book Just Six Numbers: The Deep Forces that Shape the Universe takes "fundamental constant" to mean dimensionless and independent of units. I have updated fundamental physical constants to reflect this ambiguity. Jeff Connelly
The atomic mass unit (u) is defined as one twelfth of the mass of an unbound atom of the carbon-12 nuclide and the Avogadro constant is one gram divided by one (u). This means the Avogadro constant is as fundamental as the mass of an unbound atom of the carbon-12 nuclide, since both properties can be calculated immeadiately from each other. But, what is so special about the carbon-12 nuclide ? --84.59.139.74 12:41, 8 September 2007 (UTC)

[edit] redundant

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? — Omegatron 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)
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>&mdash; [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:13, 24 January 2006 (UTC)

[edit] 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)

[edit] 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>&mdash; [[User:MFH|MFH]]: [[User talk:MFH|Talk]] 19:15, 24 January 2006 (UTC)

I found a few things. Apparently, CODATA had estimated the number at 6.023 X 1023 in 1973, I think their first publishing of standardized values. They didn't revise until 1986, at which time it became 6.022, although not quite the same in the digits after "2" as it is today. The previous values are available on their website back to 1986, but 1973 data is no longer available. 2006 is their 5th revision since 1973 of the various constants. Here is the link to the 1986 report:

http://physics.nist.gov/Document/codata86.pdf

In the text, they explain that the large change is "a direct consequence of the 7.8 ppm decrease from 1973 to 1986 in the quantity KV and the high correlation between KV and the calculated values of e, h, me, NA, and F." NA, is the symbol for Avogadro's number.

There was also an interesting abstract I ran across on the history of the number. I only read the abstract, but it looks like the fulltext might explain this change, and it also talks about the entire history of its calculation. Its called, "History and progress in the accurate determination of the Avogadro constant" by Peter Becker in 2001 Rep. Prog. Phys.

http://www.iop.org/EJ/abstract/0034-4885/64/12/206

I would guess than anyone who had chemistry before 1990 was likely to have the old number because of the lifecycle of the textbooks. I know I took it in 1985 and 1987-88 and I was always told 6.023 in both high school and college. I too wondered when and why they changed it.

Hhoblit (talk) 01:31, 28 November 2007 (UTC)

In front of me I have a mechanical engineering text book for a senior level class.
Energy Systems Third Edition. Pourmovahed, et al. John Wiley & Sons. NJ. 2008.
On page 183: "At standard temperature and pressure (STP), 1 kmol of any gas (such as hydrogen) occupies 22.41 liters and consists of 6.022 x 1023 molecules (Avogadro's number, Na).
Apparently Avogadro's number can be either 6.022 x 1023 or 6.022 x 1020, but not 6.023 x 1023.
75.45.79.31 (talk) 06:08, 8 February 2008 (UTC)mont3904


I am sorry, but what are you talking about? The above is about those of us who took chemistry when the official number was still being taught as 6.023 X 1023, or before the textbooks changed it to the updated number (6.022 X 1023). --Hhoblit (talk) 19:55, 4 March 2008 (UTC)

[edit] "rule of thumb"

The article states

"Another common sense application shows that without determining the actual weight of a substance, a good rule of thumb to use is that a cubic centimeter of solid matter contains about 10^24 atoms [4]"

Is this really a "good rule of thumb"? The density of iron is 7874 kg/(m^3). This would mean 1 cc of iron has a mass of 7.874 g and would be 0.141 moles, or 8.5 x 10^22 atoms, two orders of magnitude lower than that "rule of thumb".

Gold has a density of 19300 kg/(m^3), or 19.3 g/mL. That means 1mL gold contains only 0.0980 moles, or 5.9 x 10^22 atoms; and gold is very dense. Pt's density of 21090 kg/(m^3) and the most dense element, iridium (22650 kg/(m^3)) put them in the same range, still 2 orders of magnitude less than the suggested "rule of thumb" in atomic count.

If you refer to the wiki page on density, http://en.wikipedia.org/wiki/Density simple calculation will show none of these "representative substances" have densities even beginning to approach one where you would expect to find 10 moles of a substance in 1 cc. I don't know if this was a typographical error that was not corrected, or a simple misinformation issue, but perhaps it should be fixed? Tomteboda 05:16, 10 July 2006 (UTC)

Yes, it's pretty wrong. 10^22 would be more accurate. Even for water the correct number would be 3 * 10^22. (Of course, dense elements are dense more due to the mass of the atom than due to the number of atoms per cm^3, which explains why they don't go far from this range.) Itub 11:37, 16 June 2006 (UTC)
I agree that 10^24 seems too high, this implies that the majority of solids will lie in the range 5.0 x 10^23 to 5.0 x 10^24. By the example given above, however, both iron and gold approximate to 10^23; i.e. one order of magnitude lower. The difficulty comes when dealing with non-elemental solids e.g. polymers: how are we to estimate the number of atoms in 1cm^3 of polyethylene? And what about natural substances such as wood or stone?
However, 10^24 is a convenient number if you want to take its cube root. For example, if you assume that a solid has a cubic packing with the 10^24 rule, the linear density would be about 10^8 / cm... Itub 13:52, 18 June 2006 (UTC)
Wouldn't 100 x 10^21 be equally as convenient for a solid? Then the linear density would run around 5 x 10^7, which is a perfectly reasonable number. Tomteboda 05:16, 10 July 2006 (UTC)
On a pedantic note what do henrys (L) have to do with density? The SI symbol for litre (UK spelling) is l not L.
In 1979 the BIPM adopted the capital "L" to designate litres (I'll give you the spelling) to avoid confusion of this often-used unit with the numeral 1. In SI units currently the henry is abbreviated H. (reference BIPM) Tomteboda 05:16, 10 July 2006 (UTC)

My own "rule of thumb" is to state that Avogadro's Constant is the aggregate molecular weight of one gram of anything — with the possible exception of neutronium. Josh-Levin@ieee.org 01:49, 10 March 2007 (UTC)

[edit] Energy does not 'have' mass.

Saying "energy = mass" is not the same as saying energy HAS mass. So, I think the text that reads (which means that all energy has mass), the released energy has mass should be struck, and replaced with something like the lost mass from the formation of a Carbon-12 atom is released as energy, typically in the form of heat at leaving it like that. Objections? -- Ch'marr 18:53, 11 July 2006 (UTC)

[edit] Request regarding electrons in "Connection to masses of protons and neutrons"

The section Connection to masses of protons and neutrons has an informative discussion regarding the "mass defect". However electrons in Carbon-12 are regarded as having a negligible impact on the calculation. I didn't find that comment entirely convincing. For instance I recall the factoid that a 70 kilogram person contains about an ounce of electrons; while small, I don't find this insignificant given the decimal accuracy of the calculations on the page. Would anyone care to spruce up this section with a deeper commentary about the part played by electrons? Kaimiddleton 22:25, 11 July 2006 (UTC)

The mass of the electron is not really negligible. However, it can be neglected as a first approximation, in the same way that the mass of the proton is approximately the same as the mass of the neutron. The initial part of the paragraph as it is is qualitative, or rather semi-quantitative. It could be rewritten in a more precise way, but given my lack of inspiration today I just clarified a bit exactly how negligible the electron is as a first approximation. Itub 23:56, 11 July 2006 (UTC)

[edit] Mass of a hydrogen atom

In the section Additional physical constants the following line is incorrect (I hope): "In the 19th century physicists measured the mass of one atom of hydrogen to be about 6.02214199×10-23 grams. " If this were true, the hydrogen atom in question would be about 3 times heavier than a carbon atom ~ 12/(6.022×1023) gram, as stated somewhere else in he article.

[edit] Brain Teaser

As a substitute Chemistry teacher, I tried to "break the ice" with the students by posting, on the blackboard, "DCIIEXXI" and ask them what that means. Only the brightest students in one of the best schools in the city figured out that this is Avogadro's constant rendered into Roman Numerals, with the "E" being borrowed from computer notation and standing for "times ten to the power of".

Actually, this is 602 x 1021, because Roman Numerals don't have a decimal point. Josh-Levin@ieee.org 02:58, 25 February 2007 (UTC)

[edit] History Formula

The number density, n(0), is given by p/kT surely? Not RT/p as stated. RT/p is the inverse of the molar density I think. Spinningspark 09:50, 3 March 2007 (UTC)

May I revise the article?

If you think you know what you are doing, why not Spinningspark 17:21, 13 June 2007 (UTC)

I also think that n_\circ = \frac{RT}{p} \, is a mistake.
According to the article Loschmidt constant, n_\circ = \frac{N}{V} = \frac{N_A}{V_m} \,, and n_\circ = \frac{p}{k_B T} \,.
The gas constant R = N_A k_B \,, therefore k_B = \frac{R}{N_A} \,, and thus n_\circ = \frac{p}{\left ( \frac{R}{N_A} \right ) T} = \frac{p N_A}{R T} \,.
Ben 15:32, 8 September 2007 (UTC)


Hello, I think the formula of NA which uses the n0 is wrong. It gives a different result. Can someone check and fix this please. —Preceding unsigned comment added by 87.162.215.200 (talk) 08:44, 10 September 2007 (UTC)


Sorry to undo your edit Virginia Fried Chicken, but I don't think that was correct.

P = n(0)k(b)T

combines with,

R = N(a)k(b)

to give,

N(a) = n(0)RT/p as stated

it cannot be right that N(a) = n(0)P/[k(b)T] since,

P/k(b)T = n(0) and N(a) would then be n(0) squared.

Perhaps you where trying to get an expression in terms of the boltzmann constant k(b) rather than the gas constant R ? In that case you could use;

p = n(m)RT

where n(m) is the molar density rather than the number density, then,

R = N(a)k(b)

yields,

N(a) = P/[n(m)k(b)T]

sorry for not putting this in proper math format Spinningspark (talk) 18:03, 8 December 2007 (UTC)

[edit] Silicon unit cell equation

The article currently states that the formula used to calculate NA based on the length of the unit cell of silicon is:

N_A = \frac{4M}{a^3\rho} \,

Shouldn't this be

N_A = \frac{8M}{a^3\rho} \,

considering the fact that the unit cell of silicon contains 8 atoms?

Ben 18:05, 10 September 2007 (UTC)

You are right. Putting in the values for silicon (density = 2.33 g/cm3, atomic weight = 28.0855 and a = 5.431 angstrom) yields a value of NA roughly half the correct value. I will correct the formula and add some references to this data.

Spinningspark (talk) 14:48, 9 December 2007 (UTC)

[edit] Standardization of the mole - rewrite?

When I apply the algorithm described in, "When the mole entered the International System of Units (SI) in 1971 as the base unit of amount of substance, the definitions were interchanged: what had previously been a number became a physical constant with the unit of reciprocal moles (mol−1).", to the pre1960 definitions given, I get that now;

Avogadro number = one Avogadro number of atoms, molecules or other entities.mol−1

a mole = the number of atoms in 12 grams of carbon-12, that is as a dimensionless quantity

But:

and

  • "Avogadro number(post1971) = one Avogadro number(pre1971).mol−1" might be more clearly communicated. SmithBlue (talk) 13:00, 19 January 2008 (UTC)