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 experimental value. You will find a number of very useful values of most elements inside of the front cover of the text book.

Answer 1

Answer:

d = 7.89 g/cm³

experimental value = 7.87 g/cm³

% differrence = 0.28

Explanation:

The density of a material is the relation betwEen the mass and the volume it occupies, thus the density of the unit cell compared to its experimental value should agree reasonably well.

to solve this question lets calculate the density of the unit cell and then compare it with the experimental value.

For a body centered cubic unit cell the volume is the edge length, a, cubed.

In the bcc crystal this edge is related to the radius of the atom  as

a= 4r/√3

r = (0.124 nm)( 1 x10⁻⁷ cm/nm) = 1.24 x 10⁻⁸ cm

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

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

To calculate the mass of the cell we have to count the number of atoms per cell:

corners: 8 atoms x 1/8 = 1

center of the cube = 1

So we have 2 atoms of iron per unit cell , and the mass m is:

m = (2 atoms/unit cell )(1mol/6.022 x 10²³ atoms)(55.845 g/mol)

m = 1.85 x 10⁻²² g/unit cell

Finally the density will be:

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

            = 7.89 g /cm³

The experimental value from the literature is 7.87 g /cm³

The percent difference is:

(7.89 - 7.87) / 7.87 = 0.28 %