Calcium crystallizes in a face-centered cubic unit cell at room temperature that has an edge length of \(558.8 \mathrm{pm}\). (a) Calculate the atomic radius of a calcium atom. (b) Calculate the density of Ca metal at this temperature.

In a face-centered cubic unit cell, the atoms are located at the corners and face centers of the cube. The relationship between the atomic radius (r) and the edge length (a) can be determined by using simple geometry. For a face-centered cubic structure, the diagonal of the cube can be defined in terms of atomic radius: \[Diagonal = 4r\] Moreover, the diagonal of the cube can also be defined using Pythagorean theorem: \[Diagonal = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2}\] Now, equating both expressions for the diagonal and solving for the atomic radius: \[4r = \sqrt{3a^2}\] \[r = \frac{\sqrt{3a^2}}{4}\] Given the edge length a = 558.8 pm, we can calculate the atomic radius: \[r = \frac{\sqrt{3(558.8 \mathrm{pm})^2}}{4}\]

Calculate r using the equation found in step 1: \[r \approx \frac{\sqrt{3(558.8 \mathrm{pm})^2}}{4} = 192.5 \mathrm{pm}\] The atomic radius of a calcium atom is approximately 192.5 pm.

To calculate the density of calcium metal, we need the mass of one unit cell and the volume of the unit cell. - The mass of one unit cell can be found using the atomic mass of calcium, the Avogadro's number, and the number of atoms in a face-centered cubic unit cell (which is equal to 4). \[Mass_{unit \thinspace cell} = \frac{4 \times Atomic \thinspace Mass_{Ca}}{Avogadro's \thinspace Number}\] - The volume of the unit cell can be calculated using the cube edge length (a): \[Volume_{unit \thinspace cell} = a^3\] Finally, the density can be computed as follows: \[Density = \frac{Mass_{unit \thinspace cell}}{Volume_{unit \thinspace cell}}\]

The atomic mass of calcium (Ca) is 40.08 g/mol, and the Avogadro's number is \(6.022 \times 10^{23}\) atoms/mol. Calculate the mass of one unit cell: \[Mass_{unit \thinspace cell} = \frac{4 \times 40.08 \thinspace g/mol}{6.022 \times 10^{23}\thinspace atoms/mol} \approx 2.672 \times 10^{-22} \thinspace g\]

Calculate the volume of the unit cell using the cube edge length a = 558.8 pm (note that 1 pm = \(10^{-12}\) m): \[Volume_{unit \thinspace cell} = (558.8 \times 10^{-12} \thinspace m)^3 = 1.74 \times 10^{-28} \thinspace m^3\]

Calculate the density of Ca metal using the mass and volume of the unit cell: \[Density = \frac{2.672 \times 10^{-22} \thinspace g}{1.74 \times 10^{-28} \thinspace m^3} \approx 1.54 \thinspace g/cm^3\] (Don't forget to convert the units from g/m³ to g/cm³ with the conversion factor \(10^6\)) The density of Ca metal at room temperature is approximately 1.54 g/cm³.