Question

The density of solid Cr is 7.15 g/cm3. What is the volume per atom of Cr?As a solid, Cr adopts a body-centered cubic unit cell. What is the volume of a unit cell of this metal?What is the edge length of a unit cell of Cr?

Answer 1

In a body-centered cubic solid, each unit cell contains 2 full atoms of the compound - 1/8*8 of the atom at each corner (=1) and one full atom at the centre (=1). The mass of one chromium atom is 52 amu = 8.63*10^-23 g. The mass of two chromium atoms is twice that, 1.73*10^-22 g.

Since the density must be the same, we can say (V is volume of the unit cell):

7.15 g/cm^3 = (1.73*10^-22 g)/V --> V = 2.42*10^-23 cm^3
The edge length of a unit cell is the cube root of the volume, which is 2.89*10^-8 cm or 289 pm.

Answer 2

I understand how to get the volume for a unit cell of the metal but how do we get the simple volume per ATOM of Cr?

Answer 3

To find the volume of an atom of Cr, you can't just divide the volume of the unit cell in half since there is some empty space in addition to the actual atoms. Therefore you would first have to find the proportion of the unit cell volume that is occupied by Cr atoms (Vatom/Vcell). The solution only looks long and difficult, but I promise it isn't :P

Call the edge length "a", the face diagonal "f" (that's the diagonal across one face), and the body diagonal "b" (that's the diagonal through the unit cell, from one vertex on say the bottom right of one face to the top left of the opposite face). You know from the Pythagorean theorem that:

\[f^2 = a^2 + a^2 = 2a^2\]\[b^2 = f^2 + a^2\]and by substitution:\[b^2 = 2a^2 + a^2 = 3a^2\]
The atoms along the body diagonal (the one full atom in the centre and the two 1/8 atoms in each corner along the diagonal) "touch" each other, so the body diagonal is also equal to four times the radius of the atom (b = 4R). Substituting this into the above equation gives:

\[16R^2 = 3a^2\] and if you isolate "a" you get:\[a = 4/(\sqrt3R)\]
Now we can figure out the proportion of the cube occupied by actual atoms:

\[V _{atoms}/V_{cell}=(4/3piR^3)/a^3\]and by substituting in the above expression for a we get:\[V _{atoms}/V_{cell}=(4/3piR^3)/(4/(\sqrt3R))^3\]
You'll notice that "R" reduces out, and if you do the math you get about 68%.

*breathes*

We know from the original solution that Vcell = 2.42*10^-23 cm^3, so 68% of that, or 1.65*10^-23 cm^3, is occupied by actual atoms. There are two Cr atoms in the cell, so just divide that volume by two to find the volume of one atom, and you should get 8.23*10^-24 cm^3.

Hope that helps! if anything is unclear let me know.

Answer 4

how do you get the number of unit cells that are present per cubic cm of Cr?