Go to the body-centered cubic problems
Go to the 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 / (4/3) π = [(1.26 x 10-23 cm3) (3)] / [(4) (3.14159)] = 3.008 x 10-24 cm3r = 1.4435 x 10-8 cm = 1.44 Å
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 mol0.098071 mol times 6.022 x 1023 atoms/mol = 5.9058 x 1022 atoms
2) Part b:
4.08 Å = 4.08 x 10-8 cm1 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/mol78.074 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 = 3.998There 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
Problem #9: The radius of gold is 144 pm, and the density is 19.32 g/cm3. Does gold crystallize in a face-centered cubic structure or a body-centered cubic structure?
Solution:
1) I will assume the unit cell is face-centered cubic. I will use that assumption and the atomic radii to calculate the volume of the cell. From there, I will use the fact that there are 4 atoms of gold in the unit cell to determine the density. The final step will be to compare it to the 19.32 value. Here is one face of a face-centered cubic unit cell:
2) Across the face of the unit cell, there are 4 radii of gold, hence 576 pm. Using the Pythagorean Theorem, we determine the edge length of the unit cell:
d2 + d2 = 5762d = 407.2935 pm
3) Let us convert the pm to cm:
4407.2935 pm times (1 cm / 1010 pm) = 4.072935 x 10-8 cm
4) The volume of the unit cell:
(4.072935 x 10-8 cm)3 = 6.75651 x 10-23 cm3
5) The mass of 4 gold atoms:
(196.96655 g/mol divided by 6.022 x 1023 atoms/mol) times 4 atoms = 1.308313 x 10-21 g
6) Let's see what density results:
1.308313 x 10-21 g / 6.75651 x 10-23 cm3 = 19.36 g/cm3We conclude that gold crystallizes fcc because we were able to reproduce the known density of gold.
7) Let's do the bcc calculation (which we know will give us the wrong answer). Here's an image showing what to do with the Pythagorean Theorem:
8) The rest of the calculation with minimal comment:
d2 + (d√2)2 = 5762d = 332.55 pm
332.55 pm = 3.3255 x 10-8 cm
(3.3255 x 10-10 cm)3 = 3.6776 x 10-23 cm3
There are two atoms in a body-centered cubic.
(197 g/mol divided by 6.022 x 1023 atoms/mol) times 2 atoms = 6.5427 x 10-22 g
6.5427 x 10-22 g / 3.6776 x 10-23 cm^3 = 17.79 g/cm^3
Gold does not crystallize bcc because bcc does not reproduce the known density of gold.
Problem #10: Avogadro's number has been determined by about 20 different methods. In one approach, the spacing between ions in an ionic substance is determined by using X-ray diffraction. X-ray diffraction of sodium chloride have shown that the distance between adjacent Na+ and Cl¯ ions is 2.819 x 10-8 cm. The density of solid NaCl is 2.165 g/cm3. By calculating the molar mass to four significant figures, you can determine Avogadro's number. What value do you obtain?
Solution:
1) Imagine a cube with 4 Na and 4 Cl at adjacent vertices. I'll call it the reference cube.
There's going to be a twist and it involves how many Na and Cl are in the cube. Think about it before the reveal in the last step.
2) Determine the volume of the cube:
(2.819 x 10-8 cm)3 = 2.2402 x 10-23 cm3
3) Calculate the mass of NaCl inside the cube:
2.2402 x 10-23 cm3 times 2.165 g/cm3 = 4.85 x 10-23 g
4) The molar mass divided by the mass inside the cube equals Avogadro's Number. Here's where the twist comes into play.
Each Na and each Cl at a vertex of the reference cube is shared by a total of 8 cubes. (You may verify this on your own.) At any one vertex, there is 1/8 of a Na atom and 1/8 of a Cl atom inside the reference cube.With the reference cube having 4 vertices of Na and 4 vertices of Cl, this means there is a total of 1/2 of a Na atom and 1/2 of a Cl atom inside the reference cube.
The "molar mass" of (1/2)NaCl is half of 58.443.
29.2215 g/mol divided by 4.85 x 10-23 g = 6.025 x 1023 mol-1
Problem #11: 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, platinum has a density of 21.45 g/cm3 and a unit cell side length a of 3.93 Å. What is the atomic radius of platinum?
Solution:
1) Determine the volume of the unit cell:
(3.93 x 10¯8 cm)3 = 6.07 x 10¯23 cm3Note that I converted from Å to cm. That's because of the density.
2) Determine the mass of Pt in one unit cell:
6.07 x 10¯23 cm3 times 21.45 g/cm3 = 1.302 x 10¯21 g
3) Determine number of Pt atoms in the given mass:
195.078 g/mol divided by 6.022 x 10^23 atom/mol = 3.2394 x 10¯22 g/atom1.302 x 10¯21 g divided by 3.2394 x 10¯22 g/atom = 4 atoms
I did the above calculations in order to determine if the unit cell was face-centered or body-centered. The answer of 4 atoms in the unit cell tells me that it is face-centered. I now know what to do to determine the atomic radius. To do so, I will use the Pythagorean Theorem. (See Problem #9 for an image illustrating a face-centered cubic.)
4) Solve the Pythagorean Theorem:
d2 + d2 = (4r)2 (where r is the radius of the Pt atom. Four of them span the hypotenuse.)3.932 + 3.932 = 16r2
r2 = 30.8898 / 16 = 1.9306125
r = 1.39 Å
Problem #12: The density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, (a) determine how many formula units of TlCl there are in a unit cell. Based on your answer for the number of formula units of TlCl(s) in a unit cell, (b) how is the unit cell of TlCl(s) likely to be structured?
Solution to (a):
1) Get the volume in cm:
(385 pm) (100 cm / 1012 pm) = 3.85 x 10¯8 cm
2) Calculate the volume of the unit cell:
(3.85 x 10¯8 cm)3 = 5.7066625 x 10¯23 cm3
3) Calculate the mass of TlCl in one unit cell:
(7.00 g/cm3) (5.7066625 x 10¯23 cm3) = 3.99466375 x 10¯22 g
4) Determine how many moles of TlCl are in the unit cell:
3.99466375 x 10¯22 g / 239.833 g/mol = 1.6656022107 x 10¯24 mol
5) Formula units of TlCl in the unit cell:
(1.6656022107 x 10¯24 mol) (6.022 x 1023 mol¯1) = 1.00
Solution to (b):
Body-centered cubic has two atoms per unit cell and a formula unit of TlCl has two atoms. One atom, say the chloride ion, occupies the center of the unit cell, while 1/8th of eight Tl+ ions are present at each of the 8 vertices of the cube.Face-centered cubic has 4 atoms per unit cell. The distribution of TlCl formula units into an fcc cell does not work.
Go to the body-centered cubic problems
