An element crystallizes in a face centered cubic lattice and has a density of 1.45 g cm-3 . the edge of its unit cell is 4.52 x 10-8cm.
a) how many atoms are in each unit cell?
b) what is the volume of a unit cell?
c) what is the mass of a unit cell?
d) calculate the approximate atomic mass of the element.

a) Four atoms.

b) 9.23 x 10⁻²³ cm³ .

c)  1.34 x 10⁻²² g .

d) 80.7 amu .

Explanation:

a) In the face-centered cubic cell, there are spheres (which represent atoms or molecules) at the center of each of the six faces of the cube, in addition to eight corner spheres. A face-centered cubic cell contains the equivalent of four complete spheres or atoms—three from  the six face-centered atoms and one from the eight shared corner spheres.

b)  The volume of a cube is , a being the edge length.

    Therefore V= (4.52 x 10⁻⁸ cm)³ = 9.23 x 10⁻²³ cm³

c) We know that δ= m ÷ v  

In the problem we are given the density and we just calculated the volume. We rearrange the equation to give:

m= δ × v = 9.23 × 10⁻²³ × 1.45 g/cm⁻³ = 1.34 x 10⁻²² g

d) We know the molar mass of carbon-12 is 12.00 g and there are 6.022 × 10²³ carbon-12 atoms in 1 mole of the substance; therefore,  the mass of one carbon-12 atom is given by :

12.00 g carbon-12 atoms ÷ 6.022 × 10²³ carbon-12 atoms= 1.993 x 10⁻²³ g

We can use this result to determine the relationship between atomic

mass units and grams. Because the mass of every carbon-12 atom is exactly 12 amu,  the number of atomic mass units equivalent to 1 gram is                                

amu/gram= 12 amu/1 carbon-12 atom × 1 carbon-12 atom/ 1.993 x 10⁻²³ g

amu/ gram= 6.022 x 10²³ amu/g

Thus, 1 g= 6.022 x 10²³ amu, and 1 amu = 1.661 x10⁻²⁴ g

To calculate the approximate atomic mass, we use the last conversion factor:

1.34 x 10⁻²² g × 1 amu/ 1.661 x10⁻²⁴ g = 80.7 amu

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