# Question #fde10

Dec 4, 2017

$221.30 g m o {l}^{- 1}$

#### Explanation:

1. Calculate the volume of the unit cell

Convert pm to cm
1 pm = $1 \cdot {10}^{- 10} c m$
419 pm = $4.19 \cdot {10}^{- 8} c m$

Volume of unit cell = ${\left(4.19 \cdot {10}^{- 8} c m\right)}^{3}$
Volume of unit cell = $7.36 \cdot {10}^{- 23} c {m}^{3}$

2. Determine the mass of the metal in the unit cell
Mass of metal in unit cell = $20 g c {m}^{- 3} \times 7.36 \cdot {10}^{- 23} c {m}^{3}$
Mass of metal in unit cell = $1.47 \cdot {10}^{- 21} g$

3. Determine number of atoms per unit cell

We have face-centered unit cell: Lattice points on the corner count for an 1/8 because they are shared with 8 other unit cells. Lattice points on the face count for 1/2 as they are shared with one other unit cell.

1/8 x 8 corners = 1 lattice point
1/2 x 6 faces = 3 lattice points
Total number of lattice points (atoms) per unit cell = 4

4. Calculate molar mass of unknown metal

Mass per atom = Total mass of metal in unit cell / number of atoms in unit cell
Mass per atom = $\frac{1.47 \cdot {10}^{- 21} g}{4}$
Mass per atom = $3.68 \cdot {10}^{- 22} g$

Use mass to determine molar mass by multiplying by Avogadro's number:

molar mass = $3.68 \cdot {10}^{- 22} g \times 6.022 \cdot {10}^{2} m o {l}^{- 1}$

molar mass = $221.30 g m o {l}^{- 1}$

Looking at the periodic table, this molar mass is the closest to radon.