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Density of Unit Cells

  1. Nov 18, 2013 #1


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    (Skip to the bottom for my questions). I'm just sounding off my thoughts above.

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

    Find the density of aluminum, which crystallizes in a face-centered cubic unit cell. The atomic radius is 143 pm.

    2. Relevant equations

    edge length of FCC is (4/sqrt2)r.

    3. The attempt at a solution

    Okay. I need to express density in grams per centimeter cubed so here goes.

    1) There are 4 atoms per face-centered cubic unit cell. Therefore there are 4 aluminum atoms. What is the mass of these atoms?

    The mass of one mole of Al is 26.98 grams. 26.98 grams divided by the number of atoms in a mole gives me the mass of each individual atom. 26.98 / (6.022 * 10^23) = 4.480 * 10^-23 grams per atom of Al.

    Multiply by 4 and I have the mass of one unit cell of Al (1.79 * 10^-22 grams per unit cell).

    2) I have mass. Now I need to find the volume.

    10^12 pm = 1 m. Therefore 10^10 pm = 1 cm.

    I can cube both sides and now I have 10^30 pm^3 = 1 cm^3. There we go. A measure of volume.

    3) Volume of the unit cell.

    Using the formula given above, the edge length of a unit cell is 404.46 pm, or 4.04 * 10^-8 cm.

    Therefore the volume of a unit cell is simply the edge length (given above) cubed. Or 6.61 * 10^-23 cm^3.

    4) We have grams and we have cm^3. We can divide them and get the (simplified) final result.

    (1.79 * 10^-22 grams per unit cell) / (6.61 * 10^-23 cm^3) = 2.705 g/cm^3.


    1) I know my answer is correct. The second step was an intermediary step; I wasn't familiar with working with picometers. Is there, however, a simplified formula for the densities of the various unit cells?

    2) How come dividing molar mass / moles = mass of one atom of the element?

    g / moles = molar mass right?

    (g/moles) / moles = g/moles^2 right?

    Intuitively, however, dividing molar mass by moles seems to work. The mass of Avagadro's number of atoms divided by Avagadro's number is ... the mass of one atom.
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 19, 2013 #2
    I suggest you not to look for it. The way you approached the problem is fine. It's always best to use the basic definition that density is mass/volume.

    I don't get this. You agree that molar mass is mass of one mole? How many atoms are there in a mole? If there are N atoms, what is the mass of single atom?
  4. Nov 19, 2013 #3


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    Thanks! I think my problem was not realizing the correct units for Avogadro's number, which are atoms/mole.

    I mistakenly thought the units for A's number was imply moles. If that were the case, then molar mass (units of which are grams/mole) divided by A's number would yield:

    (g/mole) / mole = g/moles^2

    But that's simply not the case.

    The units of A's number are atoms / mole.

    So grams / mole (molar mass) divided by atoms / mole is

    grams / mole * (mole / atoms) = grams/atoms.

    There we go :)!
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