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Atomic Density question.

  1. Jul 4, 2010 #1
    Can anyone help.

    How do I calculate mass density of aluminium, given the atomic density of 6.04 X 10^22.

    I understand that Density = Mass/Volume. but the question confuses me.

  2. jcsd
  3. Jul 4, 2010 #2
    the mass density is equal to:

    \rho = \frac{\Delta m}{\Delta V}

    where [itex]\Delta m[/itex] is the mass contained in the volume [itex]\Delta V[/itex]. But, at the same time, the total mass is equal to:

    \Delta m = m_{0} \, \Delta N

    where [itex]m_{0}[/itex] is the mass of one particle (atom, molecule) and [itex]\Delta N[/itex] is the total number of particles in that volume. So, we can write:

    \rho = \frac{m_{0} \, \Delta N}{\Delta V} = m_{0} \, \frac{\Delta N}{\Delta V}

    But, by definition, the number density is:

    n_{0} \equiv \frac{\Delta N}{\Delta V}

    So, we have:

    \rho = m_{0} \, n_{0}

    So, all you need is to find the mass of one atom of aluminum. You will need to review the concept of atomic mass unit and relative atomic (molecular) mass.
  4. Jul 5, 2010 #3
    Thanks dickfore...

    so is this right?

    p = (6.04 X 10^22 / 6.023 X 10 ^23) x 26.98 = 2.7g
  5. Jul 5, 2010 #4
    in what volume are there [itex]6.04 \times 10^{22}[/itex] particles? You will need to divide with that volume to get a density. As it is now, your result has units of mass.
    Last edited: Jul 5, 2010
  6. Jul 5, 2010 #5
    Sorry,, I hope im not annoying you. Im getting mixed answers. Can you please work it out for me...

    Im gettng volume of 6.04 X 10^22 = 4.470 X10^-23, therefore the mass density equaling 3.645 X 10^46..


    (6.04 X 10^22 / 9.98 ) X 26.98 = 1.6g/cm^3

    Im lost
    Last edited: Jul 5, 2010
  7. Jul 5, 2010 #6
    no, i won't. In your original post, you hadn't specified a correct unit for the number density of Al atoms. This is why your mass density is in incorrect units. You should look back in your problem formulation again.
  8. Jul 5, 2010 #7
    Yes you're right,,, The question is the atomic density of 6.04 X 10^22 atoms per cm^3.
    I missed out the units.

    so would this mean, (6.04 X 10^22 cm^3 / 1.023 X 10^23) x 26.98 amu = 2.7 g/cm^3
    Last edited: Jul 5, 2010
  9. Jul 5, 2010 #8
    yes. look up density of aluminum and you compare with what you have found. However, I urge you to think about which one is easier to measure, the density of a bulk material or the number density of the atoms in the material?
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