Misunderstanding in converting atomic weight to kg (s.state physics)?

1. Jul 12, 2012

naftali

1. The problem statement, all variables and given/known data

Two questions :
1. find the (mass) density of Sodium. Atomic weight: 22.9898. Crystal structure: bcc. a=4.23 angstrom (crystal constant).
2. Given the (mass) density of Aluminum : 2.7 g/cm^3. Atomic weight: 26.982. Find the number of atoms density (# of atoms per $cm^{3}$).

2. Relevant equations

These questions looks trivial, but my answers are slightly different from the given solutions.
I think that this might be because I have a mistake in the conversion from atomic weight to the weight of an atom.
To my understanding, the atomic weight is the weight in grams of Avogadro number ($6.022\times10^{23}$) of atoms.

3. The attempt at a solution

Therefore I get:
1. The weight of an atom is : $\frac{22.9898}{6.022\times10^{23}}=3.818\times10^{-23} gram$. And since in bcc there are 2 atoms in unit cell we have density of:
$\frac{2\times3.818\times10^{-23}}{{(4.23\times10^{-8})}^{3}}=1.009 g/cm^{3}$
But the given solution is $0.97 g/cm^{3}$

2. The number of atoms density (# of atoms per $cm^{3}$), is density/atom's weight. Atom's weight = $\frac{26.982}{6.022\times10^{23}}=4.481\times10^{-23} gram$. Therefore, the atomic density is : $\frac{2.7}{4.481\times10^{-23}}=6.025\times10^{23} atoms/cm^{3}$.
But the given solution is $6.028 \times10^{23} atoms/cm^{3}$.

Where is my mistake?
Thanks,
Naftali

2. Jul 13, 2012

CWatters

You have correctly calculated the weight of an atom.

I think you need to check how you have calculated the volume of a cell. See..

http://www.southampton.ac.uk/~engmats/xtal/crystal/crystal.html [Broken]

Sorry for editing this a few times.

Last edited by a moderator: May 6, 2017
3. Jul 13, 2012

CWatters

From one corner of the bcc to the diagonally opposite corner is four time the radius of an atom (r+2r+r = 4r). If the length of a side is "a" then

a2+a2+a2=(4r)2

a=SQURT(4r2/3)

Wikipedia says the radius is 1.86 Angstroms so

a=4.295 x 10-8m

Thats different from your value of 4.23

Then density is

= (2 x 3.818 x 10-23)/(4.295 x 10-8)3
= 0.964g/cm3

4. Jul 13, 2012

CWatters

5. Jul 13, 2012

nasu

The lattice constant is given in the problem (a=4.23 A). No need to calculate it. And the volume of the BCC cell is a^3. You don't need the atomic radius for this problem.

You did the calculations correctly for both parts. If the numbers are a little off then maybe they used different values for Avogadro number or something like this. Or your calculator is a little off.
On the other hand, if you compare with values from literature, then the given lattice constant may be a little off.

PS. I did the calculations with the values given (4.23 A, 22.9898) and NA=6.022 10^23 and I've got density of 0.96g/cm^3.
So after all, it may be a problem with your computation process.

Last edited: Jul 13, 2012
6. Jul 15, 2012

naftali

Thanks you all for the answers.

nasu - sorry, despite my tries I don't get this result (and I tried another calculator..), I will be thankful if you can show your calculation process.

Thanks

7. Jul 15, 2012