Aluminum metal crystallizes in a face centered cubic lattice. If each Al atom has a radius of 1.43 A what is the density of aluminum metal?
I'm not sure how to go about this. I know that eventually I will use V=l^3, but I'm not sure how to get there. I also know that there are 4 atoms in a face-centered unit cell, but that's it.
I think that the crystal lattice has little to do with it at this point. By saying that the Al atom has a volume, they just want you to divide the mass of an atom by its volume. Without more information on the lattice spacings in 3D, it's hard to say how much space an atom truly occupies.
Yes, the crystal lattice can be used to calculate the density. The theory is that the density of a unit cell is the same as the density of the macroscopic unit.
So the fcc has 4 atoms/unit cell. That's right. The mass of a unit cell then is
4 * 26.98/6.02E23 = mass unit cell.
For a fcc, a(2)^1/2 = 4r.
You know the radius is 1.43 A, I would convert that to cm (1.43E-8 cm), substitute and solve for a which is the length of the unit cell in cm. Cube that to find the volume of the unit cell in cc.
Then density = mass unit cell/volume unit cell. I get about 2.70 g/mL.
To determine the density of aluminum metal, we need to calculate its volume and mass. Here are the steps to calculate the density of aluminum metal:
1. Determine the unit cell length (l):
- In a face-centered cubic (FCC) lattice, there are four atoms at the corners of a cube, and one additional atom at the center of each face.
- The distance between the center of the cube and the corners is equal to the edge length (l).
- Since the atoms touch along the edge, the diagonal of the face-centered cube is equal to 4 times the atomic radius.
- Therefore, l = 4 * atomic radius.
2. Calculate the volume of the unit cell (V):
- In an FCC lattice, the unit cell volume (V) is given by V = l^3.
3. Find the mass of the unit cell:
- Each unit cell contains four aluminum atoms.
- To calculate the mass, we need to know the molar mass of aluminum (Al) and the atomic mass unit (amu).
4. Calculate the density (ρ):
- Density (ρ) is defined as mass divided by volume: ρ = mass / V.
Let's go through the calculations using the given information and formulae:
Given:
Atomic radius of aluminum (Al) = 1.43 Å
1. Calculate the unit cell length (l):
l = 4 * atomic radius
l = 4 * 1.43 Å
l = 5.72 Å
2. Calculate the volume of the unit cell (V):
V = l^3
V = (5.72 Å)^3
V = 184.32 Å^3
3. Determine the mass of the unit cell:
- The molar mass of aluminum (Al) is 26.98 g/mol.
- The atomic mass unit (amu) is defined as 1/12th the mass of a carbon-12 atom, which is approximately 1.66 x 10^-24 grams.
Mass of one aluminum atom = 26.98 g/mol / Avogadro's constant (6.022 x 10^23)
Mass of one aluminum atom = 4.48 x 10^-23 g
Mass of one unit cell = 4 atoms * mass of one aluminum atom
Mass of one unit cell = 4 * (4.48 x 10^-23 g)
Mass of one unit cell = 1.79 x 10^-22 g
4. Calculate the density (ρ):
Density = mass / V
Density = (1.79 x 10^-22 g) / (184.32 Å^3)
Now, to convert the units to a more convenient form:
- Convert grams (g) to kilograms (kg): Divide by 1000.
- Convert cubed angstroms (Å^3) to cubic meters (m^3): Multiply by (10^-10)^3.
- The final density unit will be kg/m^3.
Therefore, the final result will be the density of aluminum metal in kg/m^3.