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[Jordi Fabre]

Kulbir tried to publish the message below but it was rejected by our anti spam system due the email address. I corrected it and published it on his behalf.
Please remember the rules to publish web or email addresses in this forum: https://www.mineral-forum.com/message-board/viewtopic.php?t=74

Hello,
I am working on nanocrystalline materials. I want to find that how can i find total no. of atoms present in nanocrystalline materials if only size is known.(rutile of 4nm size is having 1068 atoms?)
Thanks for any help.
kulbir
kulbir28(at)yahoo(dot)com

[Cesar M. Salvan]

It is not really easy. If the material is actually crystalline, simply by knowing the volume of the unit cell or the lattice parameters. If you know the number of the atoms in the unit cell... you can calculate the total atoms in the nanocrystal with simple aritmethics.

But is necessary to know the porosity of the nanocrystal and the amount of amorphous material present. All crystalline nanomaterials have a relatively high percentage of amorphous material. You could stimate the number of atoms usign the unit cell, but in practice this number is frequently under the real number if the amorphous material percentage is high or over the real number if the material has high porosity.

I do not know if this explanation respond to your question...

[Pete Modreski]

Kulbir,
Assuming (putting aside the complications Cesar mentioned; that there is no porosity and that the nanoparticle has the ideal density of the crystalline mineral) the size and shape of the particle give you the volume; multiplying by the density gives the mass; dividing by the formula weight gives the number of moles; and multiplying by the number of atoms (3 in this case) per forumula unit and by avogadro's number, gives the total number of atoms.

For your example of a 4 nm rultile particle; assuming it is a cube 4 x 4 x 4 nm (an appropriate geometrical correction would apply if it were a sphere), and that rutile has the ideal density of 4.25 g/cm3, and with the gram-formula-weight of TiO2 being 79.88,

V = (4 x 10-9 m)3 = 64 x 10-27 m3 = 6.4 x 10-26 m3 = 6.4 x 10-20 cm3

mass = 6.4 x 10-20 cm3 x 4.25 g/cm3 = 27.2 x 10-20 g = 2.72 x 10-19 grams

no. of moles = 2.72 x 10-19 g / 79.88 g/mole = .03405 x 10-19 = 3.405 x 10-21 moles

no. of TiO2 molecules = 3.405 x 10-21 moles x 6.02214 x 10(23) molecules/mole
= 20.505 x 10(2) = 2050 molecules

no. of total atoms = 2050 x 3 = 6150 atoms

(You or someone else can double-check the math here and make sure I haven't made any errors!)

Sincerely, Pete

[Les Presmyk]

It all looks good to me. Besides, Pete, you need to get a life. But, it is all very interesting and way beyond my mineralogical knowledge.

[Pete Modreski]

Hey, Les, it's good mental exercise, a good chemistry refresher, maybe it will be helpful to someone, and it doesn't get you all dirty like climbing around in oxidized lead mines. : )

Pete

[Les Presmyk]

Or, breathing in all that nasty lead oxide dust.

[Cesar M. Salvan]

We must take in mind that Kulbir are working on nanocrystalline materials. He must know the crystal structure of the material, evaluate the presence of distortions, the values of porosity and the amount of amorphous materlal

The ideal density, used for the estequiometric calculation of Pete, are calculated using the unit cell parameters, and the unit cell parameters and structure are the starting point Kulbir shoud use.
People working on nanomaterials must describe the material from a structural point of view: usually they are dealing with new phases or materials. Even if you are working in relatively well known phases, (for example, now I am working on a special property of maghemite nanoparticles, absent in the non-nanoparticullate material) the special properties of nanomaterials forces you to avoid the undergraduate estequiometric calculation and to be pretty "molecular".

In the example of rutile (now, I will present a back-envelope calculation and I use my memory, so I could be mistaken in some point. Unfortunately I have not time for elaborate discussion) :

rutile could be considered a distorted HCP (hexagonal close packed) structure.

The unit cell, tetragonal, contains 15 atoms. In the crystal we have 6 atoms per unit cell, taking into account the edge sharing and the presence of octahedral holes in crystal structure. I can not describe it in detail...I do not remember!

The volume of unit cell is approx 58 cubic angstroms (again, I do not remember the exact number).

If our ideal crystalline particle is a cube of 4 nm edge, and we assume that all the particle is filled with rutile unit cells (almost impossible using native tetragonal cells...), we have a volume of about 64.000 cubic angstroms.

So, in the particle we have 1103 unit cells. 1103x6=6620 atoms.

If the interested people find the exact unit cell volume, you will have the exact number of atoms in the ideal particle. Of course, the exact atoms count must be close to the estequiometric calculation of Pete, since the ideal density is obtained using the same starting point: the unit cell structure and parameters.
And, I said earlier, Kulbir should use this starting point and calculate all the parameters of his nanoparticles.

[Claus]

Kulbir,

Are your particles small enough to be accelerated by a mass spectrometer without disintegrating? I do not know as I have only used MS on organic molecules, so pl. let me know if you find a solution.

If so, you will be able to find the actual mass of each particle (i.e., you will get a distribution function for the particle masses) irrespective of structure and shape and you do not have to make any assumptions.

All the best

Claus

[kulbir]

Hey guys,

Thank you so much for such a great discussion. I don't know whether u will get my reply or not. Its been three years when I asked this question. And didn't got any notification for answers (that's possibly because of some mistake by me only).
I had used what Pete has suggested at that time.

Thanks again.

Kulbir Kaur.