Question

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the commonly used value (experimentally obtained) of 7.87 g/cm3.

Answer 1

Answer:

Explanation:

The density of the unit cell of a material, Iron in this case, has to be approximately equal  with its experimental value of 7.87 g/cm³.

The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.

For a BCC crystal structure the length of the side of the cube is given by:

a = 4r/√3

where a is the atomic radius of Iron

first we will convert this radius to cm since we want the density in g/cm³:

0.124 nm x  1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm

the volume of the cubic cell is:

v = a³ =  ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³

The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:

BCC = 2 atoms / unit cell       ( 1/8 from the 8 corners + 1 in the center)

m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms  x 55.85 g/mol

   = 1.85 x 10⁻²² g

Therefore,

d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³

An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.