Some hypothetical metal has the simple cubic crystal structure. If its atomic weight is 70.4 g/mol and the atomic radius is 0.126 nm, compute its density in g/cc.
To compute the density of a metal with a simple cubic crystal structure, we need to know the number of atoms per unit cell. In a simple cubic lattice, there is only one atom per unit cell located at each corner.
To find the volume of a unit cell, we can calculate the cube of the side length. In a simple cubic lattice, the side length can be determined using the atomic radius. Since the metal has a simple cubic crystal structure, the side length of the unit cell (a) is twice the atomic radius (r):
a = 2r = 2 * 0.126 nm = 0.252 nm
Now, we need to convert the side length to cm because density is typically measured in g/cm³:
0.252 nm * (1 cm / 10^-7 nm) = 2.52 * 10^-6 cm
Next, we can calculate the volume of the unit cell:
Volume = a³ = (2.52 * 10^-6 cm)³ = 1.588 * 10^-17 cm³
Since there is only one atom per unit cell in a simple cubic lattice, the density can be calculated using the atomic weight (MW) and the volume (V):
Density = (MW / NA) / V
Where NA is Avogadro's number (6.022 × 10²³ mol⁻¹).
Density = (70.4 g/mol) / [(6.022 × 10²³ mol⁻¹) * (1.588 * 10^-17 cm³)]
Now, let's calculate the density:
Density = 70.4 g / [(6.022 × 10²³) * (1.588 * 10^-17)] g/cm³
Therefore, the density of the metal is approximately 8.41 g/cm³.