Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many formula units of TlCl there are in a unit cell.

**Answer 1**

see the attached pic. There is 1 formula unit per cell. Which means 1 net Tl atom and 1 net Cl atom per cell. The structure will then probably be bcc

https://www.e-education.psu.edu/matse81/sites/www….

with either the Tl in the center or the Cl in the center of the cell and 8 corner atoms of the other with 1/8 of each atom in the unit cell.

**Answer 2**

385 pm = 3.85 x 10^-8 cm

(3.85 x 10^-8 cm)^3 = 5.7066625 x 10^-23 cm^3 (volume of unit cell)

(7.00 g/cm^3) (5.7066625 x 10^-23 cm^3) = 3.99466375 x 10^-22 g (mass of TlCl in unit cell)

3.99466375 x 10^-22 g / 239.833 g/mol = 1.6656022107 x 10^-24 mol (moles of TlCl in unit cell)

(1.6656022107 x 10^-24 mol) (6.022 x 10^23 mol^-1) = 1.00 (formula units of TlCl in unit cell)

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I’m a bit out of my comfort zone on the second question but I feel comfortable with telling you body-centered cubic. I’m not 100% sure, so do some checking before you commit to bcc.