Question 10.9: Using Unit-Cell Dimensions to Calculate the Density of a Met......
Using Unit-Cell Dimensions to Calculate the Density of a Metal
Nickel has a face-centered cubic unit cell with a length of 352.4 pm along an edge. What is the density of nickel in g/cm³ ?
STRATEGY
Density is mass divided by volume. The mass of a single unit cell can be calculated by counting the number of atoms in the cell and multiplying by the mass of a single atom. The volume of a single cubic unit cell with edge d is d³ = (3.524 × 10^{-8} cm)³ = 4.376 × 10^{-23} cm³ .
Each of the eight corner atoms in a face-centered cubic unit cell is shared by eight unit cells, so that only 1/8 × 8 = 1 atom belongs to a single cell. In addition, each of the six face atoms is shared by two unit cells, so that 1/2 × 6 = 3 atoms belong to a single cell. Thus, a single cell has 1 corner atom and 3 face atoms, for a total of 4, and each atom has a mass equal to the molar mass of nickel (58.69 g/mol) divided by Avogadro’s number (6.022 × 10^{23} atoms/mol). We can now calculate the density:
\text { Density }=\frac{\text { Mass }}{\text { Volume }}=\frac{(4 \text { atoms })\left(\frac{58.69 \frac{\mathrm{g}}{\mathrm{mol}}}{6.022 \times 10^{23} \frac{\text { atoms }}{\mathrm{mol}}}\right)}{4.376 \times 10^{-23} \mathrm{~cm}^3}=8.909 \mathrm{~g} / \mathrm{cm}^3Te calculated density of nickel is 8.909 g/cm³ . (The measured value is 8.90 g/cm³ .)