Aluminum (atomic mass 26.98 g/mol) crystallizes in a face-centered cubic unit cell. in addition, aluminum has an atomic radius of 143 pm. what is the density of aluminum?
The correct answer and explanation is :
To calculate the density of aluminum, we use the following formula:
[
\text{Density} = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}
]
Step 1: Determine the number of atoms per unit cell
Aluminum crystallizes in a face-centered cubic (FCC) unit cell. The number of atoms per unit cell in an FCC structure is 4.
Step 2: Calculate the mass of the unit cell
The atomic mass of aluminum is 26.98 g/mol, and Avogadro’s number is:
[
N_A = 6.022 \times 10^{23} \text{ atoms/mol}
]
The mass of one aluminum atom:
[
\frac{26.98 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} = 4.48 \times 10^{-23} \text{ g}
]
Since the unit cell contains 4 atoms:
[
\text{Mass of unit cell} = 4 \times (4.48 \times 10^{-23} \text{ g})
]
[
= 1.792 \times 10^{-22} \text{ g}
]
Step 3: Calculate the volume of the unit cell
In an FCC structure, the edge length (a) is related to the atomic radius (( r )) by:
[
a = \frac{4r}{\sqrt{2}}
]
Given ( r = 143 ) pm ( = 143 \times 10^{-12} ) m:
[
a = \frac{4 \times (143 \times 10^{-12} \text{ m})}{\sqrt{2}}
]
[
= \frac{572 \times 10^{-12}}{1.414}
]
[
= 4.05 \times 10^{-10} \text{ m}
]
The volume of the unit cell:
[
V = a^3 = (4.05 \times 10^{-10} \text{ m})^3
]
[
= 6.64 \times 10^{-29} \text{ m}^3
]
Step 4: Calculate the density
[
\text{Density} = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}
]
[
= \frac{1.792 \times 10^{-22} \text{ g}}{6.64 \times 10^{-29} \text{ m}^3}
]
[
= 2.70 \times 10^3 \text{ kg/m}^3
]
[
= 2.70 \text{ g/cm}^3
]
Answer:
The density of aluminum is ( 2.70 ) g/cm³.
Here is the 3D diagram of the face-centered cubic (FCC) unit cell structure of aluminum. It illustrates the atomic arrangement, including the atomic radius (143 pm) and edge length (a).