- #1

Qube

Gold Member

- 468

- 1

(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.

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)

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.

Last edited: