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

  1. Jul 12, 2012 #1
    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 [itex]cm^{3}[/itex]).

    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 ([itex]6.022\times10^{23}[/itex]) of atoms.

    3. The attempt at a solution

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

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

    Where is my mistake?
  2. jcsd
  3. Jul 13, 2012 #2


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    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
  4. Jul 13, 2012 #3


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    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



    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
  5. Jul 13, 2012 #4


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  6. Jul 13, 2012 #5
    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
  7. Jul 15, 2012 #6
    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.

  8. Jul 15, 2012 #7
    Sorry, my bad. I suspect I used 6.22 instead of 6.022 in Avogadro number.
    You are right, with the numbers given in the problem, you get a value of 1.00 if rounded to 3 digits. This is a little larger than the values measured by macroscopic techniques, which are below 1 (sodium floats on water).
    It does not necessarily mean that the given values are wrong.
    The so called x-ray density (calculated above) may happen to be larger than the actual density of the crystal. And even more so for polycrystalline materials. Defects, vacancies, pores and other factors may all contribute to reduce the density.
  9. Jul 16, 2012 #8
    Thanks anyway..
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