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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 cm3r = 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 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
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