Gold (Au) has a face-centered cubic structure with a unit cell edge length of 407.8 pm.What is the calculated value of the density of gold based on this information?

Question

Quizplus · Accepted Answer

In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell. The atomic mass of gold (Au) is approximately 197 g/mol. The volume of the unit cell can be calculated using the edge length (a) as $V = a^3$, where $a = 407.8 \, \text{pm} = 407.8 \times 10^{-10} \, \text{cm}$. The density ($\rho$) can be calculated using the formula $\rho = \frac{Z \times M}{N_A \times V}$, where $Z = 4$ (number of atoms per unit cell for FCC), $M = 197 \, \text{g/mol}$ (molar mass of gold), $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's number), and $V$ is the volume of the unit cell in cm$^3$. Plugging in the values, the calculated density closely matches the given option for 19.30 g/cm$^3$, which is a known density of gold.

Gold (Au) Has a Face-Centered Cubic Structure with a Unit