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#### Atome

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##### Atomic Radius of an Aluminium Atom
« on: December 15, 2007, 01:51:49 PM »
Hello everyone,

I am having some trouble with the following activity.

My work is shown below.

Thank you very much!

---

1. Based on your previous determination of the thickness of a piece of Aluminum foil, you will need the measurements and the calculated thickness of the foil (among other things) to determine the radius of one aluminium atom.

Assume that an aluminium atom is spherical. To make it easier, think of the sphere sitting inside a cube. Stack the cubes. State your answer for the radius of an Al atom in picometers.

Dimension of Al foil = 10.1 cm x 10.1 cm x 1.5E-3 cm

Volume of foil = 0.153 cm^3

Mass of foil = 0.42 g

Work:

Since 1 mol = 26.98 g, this foil would be 0.01557 mol.

There would be 0.01557 x 6.02E23 = 9.3714E21 atoms in this piece of foil.

Therefore, one atom would have a volume of 1.633E-23 cm^3.

Since the volume of a sphere is (4/3)pi(r^3), the radius of one Al atom would be 1.57E-8 cm = 15.7 pm.

Thank you again!
« Last Edit: December 15, 2007, 02:37:45 PM by Atome »

#### Yggdrasil

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #1 on: December 15, 2007, 02:01:11 PM »
It's 0.01557 mol of aluminum.

#### Atome

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #2 on: December 15, 2007, 02:31:53 PM »

I am still not getting the right answer. Could you please see if there are any other mistakes?

#### Sev

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #3 on: December 15, 2007, 07:11:23 PM »
Quote
Therefore, one atom would have a volume of 1.633E-23 cm^3.

This is volume of imaginary cube around atom.  so (1.633*10-23)1/3 will be diameter of atom.

#### AWK

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #4 on: December 16, 2007, 05:16:06 AM »
Spheres does not use the whole volume of crystal.
In the case of simple cubic structure it is only slightly above 50 %.
AWK

#### ARGOS++

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #5 on: December 16, 2007, 08:37:08 AM »

Dear Atome;

1.57 *10-8 cm   =   1.57 * 10-10 m   =   157 pm

All other numeric I did not test, as other have done it.

Good Luck!
ARGOS++

#### ARGOS++

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #6 on: December 16, 2007, 02:27:12 PM »
Dear Atome;

And if you correct your calculations by the factor for “Regular Sphere Packing”:

157 pm * (0.74048)1/3  =  157 pm * 0.9047 = 142.05 pm
I think that’s pretty close for the precision you were taking.

For “Regular Sphere Packing” see under:     "Sphere Packing

I hope it may be of help to you.

Good Luck!
ARGOS++

« Last Edit: December 16, 2007, 04:47:52 PM by ARGOS++ »

#### AWK

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #7 on: December 17, 2007, 01:14:34 AM »
Dear Argos.
0.9047 is a coefficient for two dimensional circles packing. For spheres in three dimensional space this coefficient equals to 74 % (ccp), 68 % for bcc and 52 % for simple cubic packing.
« Last Edit: December 17, 2007, 04:15:09 AM by AWK »
AWK

#### ARGOS++

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #8 on: December 17, 2007, 04:19:13 AM »

Dear Mr AWK;

SORRY!, - If I only partially agree with you!
(And in the part I agree with you, you are telling exactly the same as I!)

• A.)   The Packing Density for the two dimensional case PV(2) is ~0.9069 and not (PV(3))1/3 = ~0.9047, as I told.
• B.)   With PV(3) = VA / VU, with VA = Volume of the Atom, VU = Volume of the Unit Cell, and VA = k * (rA)3 it must follow  by rearrangement for the proportionality:
rA : rU      ------> rA = rU * (PV(3))1/3
But that's exactly the formula I have used.
Sorry, that 0.9047 is coming close to 0.9069, but there is a whole dimension between them.

Good Luck!
ARGOS++

#### AWK

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #9 on: December 17, 2007, 10:36:31 AM »
Dear Argos
You are right, but for cubic close packing. The problem concerns simple cubic packing, which is much less dense.
AWK

#### ARGOS++

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #10 on: December 17, 2007, 03:47:05 PM »
Dear Mr AWK;

I was not able to detect during writing my posting (Replay #8), that you edited your Replay #7 in the same moment in such a significant way, that my answer is not really fitting your posting anymore.

In case we really imply that the descriptions must mean the identical as the “Simple Cubic Packing” we have to end in, conform to Sev (Posting #3), that VA consumes the same Volume as VU and that must result in:
VU = (2rA)3  --->  rA = ½ * VU1/3  = 0.5 * (1.632247 * 10-23)1/3 = 126.835 pm

- with the final conclusion that the result of the textbook is not congruent to the calculation parameters of the same book!

So the only thing that can solve this dilemma, at least in my opinion, is that the parameters for the “Cubic Close Packing” must be used, what results in rA = 142.365 pm.
From my very little knowledge about crystallography this is not a real antagonism.

But anyway:
It seems that we finally have to accept the one or the other, as I don’t see any other solution.

Good Luck!
ARGOS++

#### AWK

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #11 on: December 18, 2007, 01:21:46 AM »
I agree, the problem posted by Atome is an odd one since it does not concern a real structure of Al.
Itn this case it would be much better to use "some element" instead of "Al", but of course there is not excluded someone somewhere will obtain this form of Al.
AWK

#### ARGOS++

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##### Re: Atomic Radius of an Aluminium Atom
« Reply #12 on: December 18, 2007, 09:15:54 AM »

Dear Mr AWK;

Thank you!   ─    And:  Merry  Christmas!

Good Luck!
ARGOS++