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.
9 years ago
9 years ago
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.
1 year ago
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.