1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density of Unit Cells

  1. Nov 18, 2013 #1

    Qube

    User Avatar
    Gold Member

    (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.

    Questions:

    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

    Qube

    User Avatar
    Gold Member

    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 :)!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Density of Unit Cells
  1. Unit cells (Replies: 2)

Loading...