Problem 39 - Chemistry: The Central Science in SI Units, Global Edition

Question

Aluminum metal crystallizes in a face-centered cubic unit cell. (a) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, \(a,\) from the atomic radius of aluminum (143 pm). (d) Calculate the density of aluminum metal.

Step-by-Step Solution

Answer

(a) There are 4 aluminum atoms in a face-centered cubic unit cell. (b) The coordination number of each aluminum atom is 12. (c) The edge length of the unit cell is approximately 404.95 pm. (d) The density of aluminum metal is approximately 2.70 g/cm³.

Step 1: (a) Finding the number of aluminum atoms in a unit cell.

In a face-centered cubic (FCC) unit cell, there are eight atoms at the corners and six atoms on the face centers. Each corner atom is shared by eight adjacent unit cells, so they contribute 1/8 of an atom to each cell. Since there are eight corners, they contribute a total of 8*(1/8) = 1 atom. Each face-centered atom is shared by two adjacent unit cells, so they contribute 1/2 of an atom to each cell. Since there are six faces, they contribute a total of 6*(1/2) = 3 atoms. Therefore, there are #(1+3)=4# aluminum atoms in a face-centered cubic unit cell.

Step 2: (b) Finding the coordination number of each aluminum atom.

In an FCC unit cell, each atom is in contact with 12 other atoms. This is due to the fact that each atom: - touches 4 atoms along the edges of the same face - touches 4 atoms on the adjacent face in front - touches 4 atoms on the adjacent face behind So, the coordination number of each aluminum atom in an FCC unit cell is 12.

Step 3: (c) Estimating the length of the unit cell edge, \(a\), from the atomic radius of aluminum (143 pm).

In an FCC unit cell, the diagonal of each face can be related to the atomic radius (r) and edge length (a) as follows: \(4r = \sqrt{2a^2}\) Given the atomic radius of aluminum, r = 143 pm, we can find the length of the unit cell edge. \(4(143 \,\text{pm}) = \sqrt{2a^2}\) Solve for \(a\): \(a = \frac{\sqrt{2(4)(143\,\text{pm})^2}}{4}\) \(a \approx 404.95\,\text{pm}\) So, the edge length of the unit cell is approximately 404.95 pm.

Step 4: (d) Calculating the density of aluminum metal.

To find the density of aluminum, we can use the formula: \(\text{Density} = \frac{\text{mass of atoms in the unit cell}}{\text{volume of the unit cell}}\) First, we need to find the mass of Aluminum atoms. Aluminum has an atomic mass of 26.98 g/mol. In one unit cell, there are 4 aluminum atoms, so the mass of atoms in the unit cell is: \(\text{Mass} = 4 \times \frac{26.98\, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.794 \times 10^{-22}\, \text{g}\) Now, we can calculate the volume of the unit cell: \(\text{Volume} = a^3 = (404.95\, \text{pm})^3 = 6.644 \times 10^{-23} \,\text{cm}^3\) Now, we can find the density: \(\text{Density} = \frac{1.794 \times 10^{-22}\, \text{g}}{6.644 \times 10^{-23} \,\text{cm}^3} \approx 2.70\, \text{g/cm}^3\) So, the density of aluminum metal is approximately 2.70 g/cm³.

Key Concepts

Face-Centered Cubic (FCC)

Aluminum crystallizes in a structure known as the face-centered cubic (FCC) unit cell. This structure features atoms positioned at each of the cube's corners and an additional atom centered on each of the six faces.
In total, a face-centered cubic (FCC) unit cell houses four aluminum atoms.

• The corner atoms, which are shared among eight unit cells, contribute 1/8 of an atom per unit cell. There are eight corners, which collectively contribute 1 atom per unit cell.
• The face-centered atoms, each shared between two unit cells, contribute 1/2 of an atom per unit cell. With six face-centered atoms total, they contribute 3 atoms per unit cell.

Altogether, this arrangement offers a highly efficient packing of atoms, ensuring a dense and stable metallic configuration.

Coordination Number

The coordination number refers to the number of nearest neighbor atoms surrounding a given atom within the crystal structure. For aluminum in a face-centered cubic (FCC) crystal structure, this number is notably high due to the compact arrangement of atoms.
Each aluminum atom in an FCC unit cell contacts 12 adjacent atoms. This high coordination number results from the substantial contact each atom has with its neighbors:

• Four atoms lie along the same face.
• Four atoms are situated on the face in front of the given atom.
• Another four are found on the face directly behind it.

Such a high coordination number contributes to the strength and durability of aluminum, making it a suitable material for various industrial applications.

Atomic Radius

The atomic radius is a fundamental property of an element, offering insights into the size of its atoms. For aluminum, this atomic radius is 143 picometers (pm). The unique geometric attributes of an FCC unit cell provide a way to relate this atomic radius to the edge length of the cell.
By considering the diagonal of the face in an FCC structure, you can establish a relationship as follows:\[4r = \sqrt{2a^2}\]In this equation, \(r\) represents the atomic radius, and \(a\) is the unit cell edge length. For aluminum, substituting the known atomic radius, you can solve for the edge length \(a\):\[a = \frac{\sqrt{2(4)(143\,\text{pm})^2}}{4}\]Upon solving, the edge length \(a\) is approximately 404.95 picometers. This calculation links the microscopic atomic dimensions directly to the larger-scale measurements of crystal structures.

Density Calculation

The density of a material is calculated by dividing the mass of atoms in a unit cell by its volume. This calculation is vital as it helps predict the weight and structural attributes of materials. For aluminum's face-centered cubic (FCC) unit cell, here's how you proceed:
First, compute the mass of the atoms within a unit cell. Aluminum's atomic mass is 26.98 g/mol. Given there are four atoms within the FCC unit cell, the mass is determined by:\[\text{Mass} = 4 \times \frac{26.98\, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.794 \times 10^{-22}\, \text{g}\]Now, calculate the unit cell's volume using the edge length determined earlier:\[\text{Volume} = a^3 = (404.95\, \text{pm})^3 = 6.644 \times 10^{-23} \,\text{cm}^3\]Finally, use these to calculate the density:\[\text{Density} = \frac{1.794 \times 10^{-22}\, \text{g}}{6.644 \times 10^{-23} \,\text{cm}^3} \approx 2.70\, \text{g/cm}^3\]This computation reveals an aluminum density of approximately 2.70 g/cm³, aligning with empirical observations and underscoring the efficient packing and arrangement of atoms within an FCC structure.