Question

Aluminum metal crystallizes with a face-centered cubic unit cell. The volume of the cell is \(0.0662 \mathrm{nm}^{3}\). (a) What is the atomic radius of aluminum in \(\mathrm{cm}\) ? (b) What is the volume of a single aluminum atom? (c) What is the density of a single aluminum atom? (d) In face-centered cubic cell packing, the fraction of empty space is \(26.0 \%\). When this is factored in, what is the calculated density of aluminum?

Short answer

Question: Calculate the density of aluminum, including empty space, given that a unit cell of the face-centered cubic structure has a volume of 0.0662 nm³, the atomic weight of aluminum is 26.98 g/mol, and 26% of the structure is empty. Answer: The density of aluminum, including empty space, is \(\rho_{Al}\). Follow the steps in the provided solution to calculate the atomic radius, volume of an aluminum atom, density of an aluminum atom, and finally the density of aluminum including empty space.

Step-by-step solution

Calculate the side length of a unit cell

Given the volume of the unit cell, find the side length by calculating the cube root of the volume: \(a = \sqrt[3]{V}\). The volume of the unit cell is \(0.0662 \mathrm{nm}^3\). Convert this to cm: \(V(\mathrm{cm}^3)=0.0662 \times (10^{-7})^3\). Now find the side length (a).

Determine the atomic radius of aluminum

In a face-centered cubic unit cell, the diagonal of the unit cell corresponds to 4 times the atomic radius (r). Using the Pythagorean theorem in 3D, we can find the relationship between side length (a) and atomic radius (r): \(\sqrt{a^2 + a^2 + a^2} = 4r \Rightarrow r=\frac{\sqrt[2]{3}}{4}a\). Now substitute the value of 'a' found in step 1 and calculate the atomic radius.

Calculate the volume of a single aluminum atom

The aluminum atom can be approximated by a sphere. Use the formula for the volume of a sphere: \(V_{atom} = \frac{4}{3}\pi r^3\). Substitute the value of atomic radius (r) from step 2 and find the volume of an aluminum atom.

Determine the density of a single aluminum atom

Density is defined as mass per unit volume: \(\rho_{atom} = \frac{m_{atom}}{V_{atom}}\), where \(\rho_{atom}\) is the atomic density, \(m_{atom}\) is the mass of an aluminum atom, and \(V_{atom}\) is the volume of an aluminum atom. To find the mass of an aluminum atom, first find its molar mass (approximately 26.98 g/mol) and divide by Avogadro's number: \(m_{atom}=\frac{26.98}{6.022\times10^{23}}\) g. Now, substitute the values of \(m_{atom}\) and \(V_{atom}\) (from step 3) to calculate the density.

Calculate the density of aluminum including empty space

Given that the empty space in the face-centered cubic cell packing is \(26\%\), the fraction of filled space will be \(74\%\), or \(0.74\). To find the calculated density, we multiply the density of a single aluminum atom (found in step 4) by the fraction of filled space: \(\rho_{Al} = \rho_{atom} \times 0.74\). Calculating this value gives the final answer for the density of aluminum including empty space.