Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its theoretical density?

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07-09-2017
Physics

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Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its theoretical density?

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meerkat18 meerkat18 · Answer 1 · 2017-09-19T11:19:05+02:00

The equation to be used is written as:

ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol

r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³

BladeRunner212 BladeRunner212 · Answer 2 · 2019-07-25T09:18:13+02:00

21.46 g/cm³

Further explanation

The problem of this calculation is the main part of the subject about the structure of crystalline solids.

This problem asks that we calculate the theoretical density of platinum (Pt).

The general formula of theoretical density is [tex]\boxed{ \ \rho = \frac{(number \ of \ atoms/unit \ cell)(atomic \ mass)}{(volume \ of \ unit \ cell)(Avogadro's \ number)} \ }[/tex]

by using symbols, i.e.,

[tex]\boxed{ \ \rho = \frac{(n)(A)}{(V_C)(N_A)}\ }[/tex]

Given data:

The atomic radius of 0.1387 nm, so [tex]\boxed{R = 0.1387 \ nm \times \Big( \frac{10^{-9} \ m}{1 \ nm} \Big) \times \Big( \frac{10^2 \ cm}{1 \ m} \Big)}[/tex]

Thus, [tex]\boxed{ \ R = 0.1387 \times 10^{-7} \ cm \ = 1.387 \times 10^{-8} \ cm \ }[/tex]

The atomic weight of 195.08 g/mol

Since platinum has the FCC crystal structure, n = 4 atoms, and [tex]\boxed{ \ V_C = (2R \sqrt{2})^3 \ }[/tex]

Let's calculate the volume of unit cells.

[tex]\boxed{ \ V_C = (2(1.387 \times 10^{-8}) \sqrt{2})^3 \ }[/tex]

[tex]\boxed{ \ V_C = (1.387 \times 10^{-8})^3 \times 16\sqrt{2} \ }[/tex]

[tex]\boxed{ \ V_C = 6.038 \times 10^{-23} \ cm^3 \ }[/tex]

Substitute all data into the general formula.

[tex]\boxed{ \ \rho = \frac{(4 \ atoms/unit \ cell)(195.08 \ g/mol)}{(6.038 \times 10^{-23} \ cm^3/unit \ cell)(6.023 \times 10^{23} \ atoms/mol)} \ }[/tex]

Thus, the theoretical density of platinum is 21.46 g/cm³ (rounded to 2 decimal places).

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Keywords: platinum, Pt, the FCC crystal structure, atomic radius, weight, mass, theoretical density, volume, face-centered cubic