relating Density to Crystal Structure Aluminum crystallizes with a face-centered cubic unit cell. The radius of an aluminum atom is 143 pm. Calculate the density of solid crystalline aluminum in g/cm3.

Question

relating Density to Crystal Structure

Aluminum crystallizes with a face-centered cubic unit cell. The radius of an aluminum atom is 143 pm. Calculate the density of solid crystalline aluminum in g/cm³.

SORT You are given the radius of an aluminum atom and its crystal structure. You are asked to find the density of solid aluminum.

GIVEN r = 143 pm, face-centered cubic
FIND d

STRATEGIZE The conceptual plan is based on the definition of density. Since the unit cell has the physical properties of the entire crystal, find the mass and volume of the unit cell and use these to calculate its density.

CONCEPTUAL PLAN
d = m/V
m = mass of unit cell
= number of atoms in unit cell ×mass of each atom
V = volume of unit cell
= (edge length)³

SOLVE Begin by finding the mass of the unit cell. Obtain the mass of an aluminum atom from its molar mass. Since the face centered cubic unit cell contains four atoms per unit cell, multi ply the mass of aluminum by 4 to get the mass of a unit cell.

Next, calculate the edge length (l) of the unit cell (in m) from the atomic radius of aluminum. For the face-centered cubic structure, l = 2[latex]\sqrt{2}r[/latex].

Calculate the volume of the unit cell (in cm) by converting the edge length to cm and cubing the edge length. (Use centimeters to report the density in units of g/cm³).

Finally, calculate the density by dividing the mass of the unit cell by the volume of the unit cell.

Accepted Answer

m(Al atm) = 26.98[latex]\frac{ g}{\cancel{mol}} imes\frac{1 \cancel{mol}}{6.022 imes 10^{23} atoms}[/latex]
= 4.480 ×10[latex]^{-23}\frac{ g}{atom}[/latex]
m(unit cell) = 4[latex]\cancel{ atom} (4.480 imes 10^{-23} \frac{g}{\cancel{atom}} )[/latex]
= 1.792 × 10[latex]^{-22}[/latex] g

l = 2[latex]\sqrt{2}[/latex] r
= 2[latex]\sqrt{2}[/latex] (143 pm)
= 2[latex]\sqrt{2} (143 imes 10^{-12} m)[/latex]
= 4.0[latex]\underline{4}5 imes 10^{-10}[/latex] m

V = l³
= [latex](4.0\underline{4}5 imes 10^{-10} \cancel{m} imes\frac{1 cm}{10^{-2} \cancel{m} })[/latex]³
= 6.6[latex]\underline{1}8 × 10^{-23} cm^{3}[/latex]

d = [latex]\frac{m}{V} = \frac{1.792 imes 10^{-22} g}{6.6\underline{1}8 imes 10^{-23} cm^{3}}[/latex]
= 2.71 g/cm³

CHECK The units of the answer are correct. The magnitude of the answer is reasonable because the density is greater than 1 g/cm³(as you would expect for metals), but still not too high (aluminum is a low-density metal).

Question 11.6: relating Density to Crystal Structure Aluminum crystallizes ...

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