Question

The crystal structure of an element has fcc lattice. If the edge length of the crystal is \(4\ \text{\AA}\), what is the atomic weight \((g\ mol^{-1})\) of the element, if the density of the crystal is \(11.21\ g\ cm^{-3}\)? \[ (N_A=6.023\times10^{23}\ mol^{-1}) \]

• A. 63.5
• B. 85.5
• C. 108.0
• D. 197.0

Hint:

For crystal density problems, always use \[ \rho=\frac{ZM}{N_Aa^3} \] where \(Z=4\) for fcc, \(Z=2\) for bcc and \(Z=1\) for simple cubic lattices.

Answer 1

The Correct Option is C

Step 1: Think in terms of the mass of one unit cell.
For an fcc lattice, \(Z=4\) atoms sit inside every unit cell. If we work out the mass of that single cell and how many moles of atoms it holds, dividing the two gives the molar mass directly.

Step 2: Find the volume and mass of the unit cell.
The edge length is \(a=4\ \text{\AA}=4\times10^{-8}\ cm\), so \[ a^3=(4\times10^{-8})^3=6.4\times10^{-23}\ cm^3 \] Mass of one unit cell = density times volume: \[ m=11.21\times6.4\times10^{-23}=7.174\times10^{-22}\ g \]

Step 3: Find the moles of atoms held in that unit cell.
Since a unit cell has 4 atoms, \[ n=\frac{Z}{N_A}=\frac{4}{6.023\times10^{23}}=6.64\times10^{-24}\ mol \]

Step 4: Divide mass by moles to get the molar mass.
\[ M=\frac{m}{n}=\frac{7.174\times10^{-22}}{6.64\times10^{-24}}\approx108\ g\ mol^{-1} \] This matches option (3).
\[ \boxed{108.0} \]