Question

a metallic object with atoms in a face-centered cubic unit cell with an edge length of 392 pm has a density of 21.45 g/cm^3. calculate the atomic mass and radius of the metal. identify the metal.

Answer 1

The metal = Platinum(Pt)

Further explanation

FCC : Face centered cubic : the unit structures of the unit cell((other than BCC, HCP)

In FCC ⇒ atoms on every corner and side (total 4 atoms)

The length of the face diagonal (b) = 4 times the atomic radius  (r)

Formula :

$\tt b=4r$

$\tt a=r\sqrt{8}$

$\tt V=16r\sqrt{2}=a^3$

$\tt \rho=\dfrac{n.Ar}{V.No}$

Ar = atomic mass, g/mol

No = Avogadro number = 6.02.10²³

n = number of atoms

V = volume (cm³)

a = side length  

b = diagonal of the side surface  

r = atomic radius

a= 392 pm=3.92 x 10⁻⁸ cm

ρ = 21.45 g/cm³

$\tt V=a^3\\\\V=(3.92\times 10^{-8})^3\\\\V=6.024\times 10^{-23}~cm^3$

The atomic mass :

$\tt 21.45~g/cm^3=\dfrac{4\times Ar}{6.024\times 10^{-23}\times 6.02\times 10^{23}}\\\\Ar=194.47\approx 195~g/mol$

radius :

$\tt a=r\sqrt{8}\\\\3.92\times 10^{-8}=r\sqrt{8}\\\\r=1.386\times 10^{-8}~cm$

Element with atomic mass 195 g/mol and radius = 139 pm (1.386x10⁻⁸cm) = Platinum(Pt)