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Qube
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(Skip to the bottom for my questions). I'm just sounding off my thoughts above.
Find the density of aluminum, which crystallizes in a face-centered cubic unit cell. The atomic radius is 143 pm.
edge length of FCC is (4/sqrt2)r.
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.
Homework Statement
Find the density of aluminum, which crystallizes in a face-centered cubic unit cell. The atomic radius is 143 pm.
Homework Equations
edge length of FCC is (4/sqrt2)r.
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.
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