An unknown element crystallises in an FCC lattice. Its theoretical density is given as and the unit — Solid State Chemistry Question
Question
An unknown element crystallises in an FCC lattice. Its theoretical density is given as $10\text{ g cm}^{-3}$ and the unit cell edge length is exactly $400\text{ pm}$. What is the atomic mass of this element in $\text{g mol}^{-1}$? (Use $N_A = 6.0 \times 10^{23}\text{ mol}^{-1}$)
💡 Solution & Explanation
Using the density formula: $M = \frac{d \times N_A \times a^3}{Z}$. We know $Z=4$ for FCC, $d = 10\text{ g cm}^{-3}$, and $a = 400\text{ pm} = 4 \times 10^{-8}\text{ cm}$. $M = \frac{10 \times 6.0 \times 10^{23} \times (4 \times 10^{-8})^3}{4} = \frac{10 \times 6.0 \times 10^{23} \times 64 \times 10^{-24}}{4} = \frac{384}{4} = 96\text{ g mol}^{-1}$.