Question

Compute the theoretical density of CaF2 (in g/cm^3), which has the fluorite structure. There are 4 CaF2 units per unit cell; a=2[r(Ca^2+) + r(F^1-)]/3; and V=(2a)3. Round and report the answer to 3 significant figures in the format: 1.23 g/cm^3

Answer #1

Theoretical density = n(A_{Ca} +
2*A_{F})/(Vc*N_{A})

CaF2 = Ca2+ + 2F-

n = number of atoms / unit cell = 4

A_{Ca} = atomic mass of Calcium = 40.08 g/mol

A_{F} = atomic mass of Calcium = 19 g/mol

N_{A} = Avogadro number = 6.023 x 10^23 atoms/mol

The sub cell edge length

a = (2*r_{Ca2+} + 2*r_{F-})
/(3^{0.5})

= (2*0.100 + 2*0.133) /(3^{0.5})

= 0.269 nm x 10^-7 cm/nm

= 2.69 x 10^-8 cm

Volume of unit cell

Vc = (2a)^{3}

= ( 2*2.69 x 10^-8 cm )^{3}

= 1.56 x 10^-22 cm3

Put all the values in the formula

Theoretical density = 4*(40.08 + 2*19)/(1.56 x 10^-22*6.023 x 10^23)

= 3.32 g/cm3

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