Question

Could you please explain in depth how you come to the answer on this?

At room temperature, iron exists as a body centered cubic (BCC) crystal? Estimate the density (mass per unit volume, kg/m3) of iron assuming each unit cell contains the mass of two atoms of iron. Note: the atomic radius of iron is 1.24 A and its atomic mass is 55.84 g/mol; 1A 10-10m; 1mol 6.032x1023.

I understand that the atomic mass, m = 55.85 g/mole and the radius, r = 1.25 a ---- r = 1.24 x 10 ^-8

But for BCC how do you find the atomic number? I know it is Z = 2. but how do you find it and what does it signify?

Also, for BCC how do you find r? I know it is ((sqrt 3)*9) / 4 and a = (u*r) / sqrt 3. But where do you find this?

Thanks! :)

Answer 1

density (d) = z times M/a^3 times N_A Z = number of atoms per unit cell (for bcc Z = 2) M = Molar mass in g mol^- a = edge length in cm d = Density (gcm ^-3) N_A = Avogadro number 6.023 times 10^23 molecules per moles radius (r) = squareroot 3 a /4 p a = 4 /squareroot 3 r = 1.24A ^0: M = 55.848 g mol^-1 d = 2 times 55.84 x10^-3 kg mol ^-1 /(4 times 1.24 times 10^-10 m/ squareroot 3)^3 times 6.023 times 10^23 = 7896.98 kgm ^-3 = 7.9 cm^-3