Solid State: Application Problems (3)

Question

A metal crystallizes in a face-centered cubic (FCC) lattice with edge length $a = 4.0 \text{ Å}$ and density $\rho = 8.96 \text{ g/cm}^3$. Calculate the atomic mass of the metal. ($N_A = 6.022 \times 10^{23}$.)

Solution — Step by Step

Final answer: Atomic mass $\approx \mathbf{86.4 \text{ g/mol}}$ (close to strontium, $87.6$, or yttrium, $88.9$ — likely strontium given FCC structure).

Why This Works

The unit cell is a microscopic packet of the crystal. Knowing the cell contents ($Z$) and dimensions ($a$), we can reverse-engineer atomic-scale properties from bulk density. This same equation lets us find $a$ from $M$, density from $a$ and $M$, and so on — it’s a four-quantity relation where any three determine the fourth.

For FCC: $Z = 4$. For BCC: $Z = 2$. For simple cubic: $Z = 1$. Memorize.

Alternative Method

Compute mass per unit cell directly: $\rho \times V_{cell} = 8.96 \times 6.4 \times 10^{-23} = 5.74 \times 10^{-22} \text{ g}$. Then $M = (\text{mass per cell}/Z) \times N_A = (5.74 \times 10^{-22}/4) \times 6.022 \times 10^{23} \approx 86.4 \text{ g/mol}$.

Common Mistake