Problem #1: Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, sodium has a density of 0.968 g/cm3 and a unit cell side length (a) of 4.29 Å
a. How many sodium atoms are in 1 cm3?
b. How many unit cells are in 1 cm3?
c. How many sodium atoms are there per unit cell?
Solution:
1) Calculate the average mass of one atom of Na:
22.99 g mol¯1 ÷ 6.022 x 1023 atoms mol¯1 = 3.82 x 10¯23 g/atom
2) Determine atoms in 1 cm3:
0.968 g / 3.82 x 10¯23 g/atom = 2.54 x 1022 atoms in 1 cm3
3) Determine volume of the unit cell:
(4.29 x 10¯8 cm)3 = 7.89 x 10¯23 cm3
4) Determine number of unit cells in 1 cm3:
1 cm3 / 7.89 x 10¯23 cm3 = 1.27 x 1022 unit cells
5) Determine atoms per unit cell:
2.54 x 1022 atoms / 1.27 x 1022 unit cells = 2 atoms per unit cell
Problem #2: Metallic iron crystallizes in a type of cubic unit cell. The unit cell edge length is 287 pm. The density of iron is 7.87 g/cm3. How many iron atoms are there within one unit cell?
Solution:
1) Calculate the average mass of one atom of Fe:
55.845 g mol¯1 ÷ 6.022 x 1023 atoms mol¯1 = 9.2735 x 10¯23 g/atom
2) Determine atoms in 1 cm3:
7.87 g / 9.2735 x 10¯23 g/atom = 8.4866 x 1022 atoms in 1 cm3
3) Determine volume of the unit cell:
287 pm x (1 cm / 1010 pm) = 2.87 x 10¯8 cm
(2.87 x 10¯8 cm)3 = 2.364 x 10¯23 cm3
4) Determine number of unit cells in 1 cm3:
1 cm3 / 2.364 x 10¯23 cm3 = 4.23 x 1022 unit cells
5) Determine atoms per unit cell:
8.4866 x 1022 atoms / 4.23 x 1022 unit cells = 2 atoms per unit cell
Problem #3: (a) You are given a cube of silver metal that measures 1.015 cm on each edge. The density of silver is 10.49 g/cm3. How many atoms are in this cube?
Solution: (1.015 cm)3 x (10.49 g / cm3) x (1 mole Ag / 107.9 g) x (6.023 x 1023 atoms / 1 mole) = 6.12 x 1022 atoms Ag
(b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that 74% of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom.
Solution: (1.015 cm)3 x 0.74 = 0.774 cm3 filled by Ag atoms
0.774 cm3 / (6.12 x 1022 atoms) = 1.26 x 10-23 cm3 / Ag atom
(c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.
Solution: V = (4/3) π r3
r3 = V / 1.33 π = (1.26 x 10-23 cm3) / (1.33)(3.14) = 3.03 x 10-24 cm3r = 1.4 x 10-8 cm = 144 pm
Problem #4: Someday!