Titanium has an HCP crystal structure and a density of \(4.51 \mathrm{g} / \mathrm{cm}^{3}\) (a) What is the volume of its unit cell in cubic meters? (b) If the \(c / a\) ratio is 1.58 , compute the values of \(c\) and \(a.\)

In an HCP crystal structure, there is a lattice point at the center of each of the hexagonal faces and at the vertices of the unit cell. To find the volume of an HCP unit cell, we can use the following formula: \(V = \frac{\sqrt{2}}{2} a^2 c\) where \(V\) is the volume of the unit cell, \(a\) is the side length of the hexagonal face, and \(c\) is the height of the unit cell.

If we count the atoms at the vertices, there are 12 \(\times\) \(\frac{1}{6}\) (as each vertex atom is shared by 6 other unit cells) and 6 \(\times\) \(\frac{1}{2}\) (on top and bottom face) total one atom contribution per unit cell. Additionally, there are three more atoms completely inside the unit cell (in an HCP structure). Therefore, the total number of atoms per unit cell is: \(N_{atoms} = 1 + 3 = 4\)

To find the molar mass of titanium, we can use the periodic table, where the atomic mass of titanium (Ti) is approximately 47.87 g/mol.

Now that we have the density (\(ρ\)), molar mass (\(M\)), and number of atoms (\(N_{atoms}\)) per unit cell, we can use the following formula to calculate the volume (\(V\)) of the unit cell: \(ρ = \frac{N_{atoms}M}{V \times N_A}\) where \(N_A\) is the Avogadro's number (\(6.022 \times 10^{23} \frac{1}{\mathrm{mol}}\)). Rearranging the formula to solve for \(V\): \(V = \frac{N_{atoms}M}{ρ \times N_A}\) Plug in the known values: \(V = \frac{4 \times 47.87 \frac{\mathrm{g}}{\mathrm{mol}}}{4.51 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 6.022 \times 10^{23} \frac{1}{\mathrm{mol}}}\) \(V = 5.595 \times 10^{-23} \mathrm{cm}^3\) To convert this to cubic meters: \(V = 5.595 \times 10^{-23} \mathrm{cm}^3 \times (\frac{1 \mathrm{m}}{100 \mathrm{cm}})^3 = 5.595 \times 10^{-29} \mathrm{m}^3\) So, the volume of the unit cell is \(5.595 \times 10^{-29} \mathrm{m}^3\).

We are given the ratio \(c/a = 1.58\). Since we now know the volume of the unit cell, we can use the formula from Step 1 to solve for \(c\) and \(a\): \(V = \frac{\sqrt{2}}{2} a^2 c\) Plug in the known values: \(5.595 \times 10^{-29} \mathrm{m}^3 = \frac{\sqrt{2}}{2} a^2 (1.58a)\) Rearrange the equation to solve for \(a\): \(a^3 = \frac{5.595 \times 10^{-29} \mathrm{m}^3 \times 2}{\sqrt{2} \times 1.58} = 2.503 \times 10^{-29} \mathrm{m}^3\) Taking the cube root: \(a = \sqrt[3]{2.503 \times 10^{-29} \mathrm{m}^3} = 2.920 \times 10^{-10} \mathrm{m}\) Now that we have the value of \(a\), we can calculate the value of \(c\) using the given ratio: \(c = 1.58a = 1.58 \times 2.920 \times 10^{-10} \mathrm{m} = 4.613 \times 10^{-10} \mathrm{m}\) The values of \(c\) and \(a\) are \(4.613 \times 10^{-10} \mathrm{m}\) and \(2.920 \times 10^{-10} \mathrm{m}\), respectively.