Chromium metal crystallizes as a body-centered cubic lattice. If the atomic radius of Cr is 1.25 angstroms, what is the density of Cr metal in g/cm3?



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Posted on (Answer #1)

To determine the density of the unit cell, we need to find the mass of the unit cell and the volume of the unit cell.  A body centered cubic unit cell contains a total of 2 atoms; one in the middle (i.e. the body centered) and there are 8 at each corner each of which contributes 1/8 to the number of atoms. 

We need to find the mass of the unit cell which we've already determined contains two atoms.  We need to convert atoms to moles to grams to find the mass of the two atoms.

2 atoms/cell * (1 mole/6.022*10^23 atoms) * (52.00 g/mole) = 1.73 x 10^-22 g

Now, we need to find the volume of the unit cell.  Since we see that the density is in terms of cm^3, the first thing to do would be to convert Angstroms to cm.

1.25 A (1 m / 1 * 10^10 A) * (100 cm / 1 m) = 1.25 x 10^-8 cm

The edge length of a body centered unit cell is 4r/(sqrt3) where r is the radius of the atom.  The edge length of this unit cell is

4(1.25x10^-8)/(sqrt3) = 2.89 x 10^-8 cm

To find the volume, we can cube the edge length to get

2.41 x 10^-23 cm^3

Now that we know the mass and the volume, we can find the density

1.73 x 10^-22 g / 2.41 x 10^-23 cm^3

= 7.17 g/cm^3

Compared to the density value given in the link below (7.19 g/cm^3), this is a reasonable answer.



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