Answer to Question #144779 in General Chemistry for Joshua

Question

A metal has a density of 10.22 g/cm^3, atomic weight of 95.94 g/mole and atomic radius of 0.136 nm. Is it BCC or FCC?

Answer 1

Density ($\rho$) = 10.22$g/cm^3$

Atomic weight (A) = 95.94g/mol

Atomic radius (R) = 0.136nm = 1.36 × $10^{-8}$cm

In order to determine whether metal has a FCC or a BCC crystal structure we need to compute the density for each of the crystal structures.

FCC ---->

$\rho = \dfrac{nA}{(2\sqrt{2}R)^3N_A}$

$\rho = \dfrac{4×95.94}{(2\sqrt{2}×1.36×10^{-8})^3× 6.02× 10^{23}}$

$= 11.19g/cm^3$

BCC ----->

$\rho = \dfrac{nA}{(\frac{4}{\sqrt{3}}R)^3N_A}$

$\rho = \dfrac{2×95.94}{(\frac{4}{\sqrt{3}}×1.36×10^{-8})^3× 6.02×10^{23}}$

$= 10.22g/cm^3$

The density derived from the BBC formula is the value we're provided with in the problem. Therefore, the metal has a BCC crystal structure, and not a FCC structure.