Worksheet - Unit Cell Problems - AP level

Go to some body-centered cubic problems

Go to some face-centered cubic problems

Return to the Liquids & Solids menu


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 to a:

(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 to b:

(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 to c:

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 cm3

r = 1.4 x 10-8 cm = 144 pm


Problem #4: 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, gold has a density of 19.32 g/cm3 and a unit cell side length of 4.08 Å. (1 Å = 1 x 10-8 cm.)

(a) How many gold atoms are in exactly 1 cm3?
(b) How many unit cells are in exactly 1 cm3?
(c) How many gold atoms are there per unit cell?
(d) The atoms/unit cell suggests that gold packs as a (i) simple, (ii) body-centered or (iii) face-centered unit cell.

Solution:

1) Part a:

19.32 g / 197.0 g/mol = 0.098071 mol

0.098071 mol times 6.022 x 1023 atoms/mol = 5.9058 x 1022 atoms

2) Part b:

4.08 Å = 4.08 x 10-8 cm

1 cm divided by 4.08 x 10-8 cm = 24509804 (this is how many 4.08 Å segments in 1 cm)

24509804 cubed = 1.47238 x 1022 unit cells

3) Part c:

5.9058 x 1022 atoms / 1.47238 x 1022 unit cells = 4 atom/unit cell

4) Part d:

face-centered

Problem #5: A metal nitride has a nitrogen atom at each corner and a metal atom at each edge. Which is the empirical formula for this nitride?

a. Ba3N2
b. Na3N
c. AlN
d. Ti3N4

Solution:

A cube has eight corners and an atom at a corner is in eight different cubes; therefore 1/8 of an atom at each corner of a given cube. So:

1/8 times 8 = 1 total nitrogen atom in each cube

A cube has 12 edges and each edge is in 4 different cubes, so there is 1/4 of an atom in each individual cube. So:

1/4 times 12 = 3 total metal atoms in each cube

The only choice to fit the above criteria is answer choice b, Na3N.


Problem #6: Calcium fluoride crystallizes with a cubic lattice. The unit cell has an edge of 546.26 pm and has a density of 3.180 g/cm3. How many formula units must there be per unit cell?

Solution:

1) Convert pm to cm:

546.26 pm times (1 cm / 1010 pm) = 5.4626 x 10-8 cm

2) Determine volume of unit cell:

(5.4626 x 10-8 cm)3 = 1.63 x 10-22 cm3

3) Determine mass of CaF2 in unit cell:

3.180 g/cm3 times 1.63 x 10-22 cm3 = 5.1835 x 10-22 g

4) Determine mass of one formula unit of CaF2:

formula weight of CaF2 = 78.074 g/mol

78.074 g/mol divided by 6.022 x 1023 formula units / mole = 1.2965 x 10-22 g

5) Determine number of formula units in one unit cell:

5.1835 x 10-22 g divided by 1.2965 x 10-22 g = 3.998

There are 4 formula units of CaF2 per unit cell


Problem #7: Tungsten has an atomic radius of 137 pm and crystallizes in a cubic unit cell having an edge length d = 316 pm. What type of cubic unit cell does tungsten crystallize in?

Solution:

Let us assume the cell is face-centered. If this is the case, then this relationship should hold true:

d2 + d2 = (4r)2

where r = the radius.

Please see a small discussion of this in problem #1 here.

Using 316 pm for d and 548 pm for 4r, we have this:

3162 + 3162 ?=? 5482

We find 199712 for the left and 300304 for the right, so the idea that tungsten is fcc fails.

For body-centered, please see problem #2 here for this equation:

d2 + (d√2)2 = (4r)2

3d2 = (4r)2

3(3162) ?=? 5482

299568 = 300304

Due to the fact that these numbers are roughly equivalent, we can conclude that tungsten is being body-centered cubic.


Problem #8: What is the formula of the compound that crystallizes with Ba2+ ions occupying one-half of the cubic holes in a simple cubic arrangement of fluoride ions?

Solution:

We must consider two cubic unit cells, one with the barium ion and one without (this gives us one-half of the cubic holes filled).

Cell 1: 8 F atoms at the 8 vertices. Since each vertex is in a total of 8 cells, we have 1 F atom in the unit cell. The cubic hole in the middle of the cell is empty.

Cell 2: 8 F atoms at the 8 vertices. Since each vertex is in a total of 8 cells, we have 1 F atom in the unit cell. The cubic hole in the middle of the cell has a barium in it.

Total for the two cells: one Ba and two F

Formula: BaF2


Go to some body-centered cubic problems

Go to some face-centered cubic problems

Return to the Liquids & Solids menu