Question

Iron oxide (FeO) has the rock salt crystal structure and a density of 5.70 g/cm3. The atomic weights of iron and oxygen are 55.85 g/mol and 16.00 g/mol, respectively. (a) Determine the unit cell edge length

Answer 1

Answer:

The unit cell edge length is  $4.3\times 10^{-8} m$

Explanation:

NaCl has FCC structure. and so is the FeO.

Number of atom in unit cell of FCC (Z) = 4

Density of FeO= $5.70 g/cm^3$

Edge length = a

Atomic mass of FeO(M) = 55.85 g/mol +16.00 g/mol = 71.85 g/mol

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$5.70 g/cm3=\frac{4\times 71.85 g/mol}{6.022\times 10^{23} mol^{-1}\times a^{3}}$

$a =  4.3\times 10^{-8} m$

The unit cell edge length is  $4.3\times 10^{-8} m$