Go to some body-centered cubic problems
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Here are the problems:
Problem #1: Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.023 g/cm^{3}. Calculate the atomic radius of palladium. |
Problem #2: Nickel crystallizes in a face-centered cubic lattice. If the density of the metal is 8.908 g/cm^{3}, what is the unit cell edge length in pm? |
Problem #3: Nickel has a face-centered cubic structure with an edge length of 352.4 picometers. What is the density? |
Problem #4: Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium. |
Problem #5: Krypton crystallizes with a face-centered cubic unit cell of edge 559 pm. (a) What is the density of solid krypton? (b) What is the atomic radius of krypton? (c) What is the volume of one krypton atom? |
Problem #6: You are given a small bar of an unknown metal. You find the density of the metal to be 11.5 g/cm^{3}. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.06 x 10^{-10} m. Find the gram-atomic weight of this metal and tentatively identify it. |
Problem #7: A metal crystallizes in a face-centered cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm^{3}. What is this metal? |
Problem #8: The density of an unknown metal is 2.64 g/cm^{3} and its atomic radius is 0.215 nm. It has a face-centered cubic lattice. Determine its atomic weight. |
Problem #9: Metallic silver crystallizes in a face-centered cubic lattice with L as the length of one edge of the unit cube. What is the center-to-center distance between nearest silver atoms? |
Problem #10: Iridium has a face centered cubic unit cell with an edge length of 383.3 pm. The density of iridium is 22.61 g/cm^{3}. Use these data to calculate a value for Avogadro's Number. |
Problem #1: Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.023 g/cm^{3}. Calculate the atomic radius of palladium.
Solution:
1) Calculate the average mass of one atom of Pd:
106.42 g mol¯^{1} ÷ 6.022 x 10^{23} atoms mol¯^{1} = 1.767187 x 10¯^{22} g/atom
2) Calculate the mass of the 4 palladium atoms in the face-centered cubic unit cell:
1.767187 x 10¯^{22} g/atom times 4 atoms/unit cell = 7.068748 x 10¯^{22} g/unit cell
3) Use density to get the volume of the unit cell:
7.068748 x 10¯^{22} g ÷ 12.023 g/cm^{3} = 5.8793545 x 10¯^{23} cm^{3}
4) Determine the edge length of the unit cell:
[cube root of] 5.8793545 x 10¯^{23} cm^{3} = 3.88845 x 10¯^{8} cm
5) Determine the atomic radius:
Remember that a face-centered unit cell has an atom in the middle of each face of the cube. The square represents one face of a face-centered cube:
Here is the same view, with 'd' representing the side of the cube and '4r' representing the 4 atomic radii across the face diagonal.
Using the Pythagorean Theorem, we find:
d^{2} + d^{2} = (4r)^{2}
2d^{2} = 16r^{2}
r^{2} = d^{2} ÷ 8
r = d ÷ 2(√2)
r = 1.3748 x 10¯^{8} cm
You may wish to convert the cm value to picometers, the most common measurement used in reporting atomic radii. Try it before looking at the solution to the next problem.
Problem #2: Nickel crystallizes in a face-centered cubic lattice. If the density of the metal is 8.908 g/cm^{3}, what is the unit cell edge length in pm?
Solution:
This problem is like the one above, it just stops short of determining the atomic radius.
1) Calculate the average mass of one atom of Ni:
58.6934 g mol¯^{1} ÷ 6.022 x 10^{23} atoms mol¯^{1} = 9.746496 x 10¯^{23} g/atom
2) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell:
9.746496 x 10¯^{23} g/atom times 4 atoms/unit cell = 3.898598 x 10¯^{22} g/unit cell
3) Use density to get the volume of the unit cell:
3.898598 x 10¯^{22} g ÷ 8.908 g/cm^{3} = 4.376514 x 10¯^{23} cm^{3}
4) Determine the edge length of the unit cell:
[cube root of] 4.376514 x 10¯^{23} cm^{3} = 3.524 x 10¯^{8} cm
5) Convert cm to pm:
cm = 10¯^{2} m; pm = 10¯^{12} m.Consequently, there are 10^{10} pm/cm
(3.524 x 10¯^{8} cm) (10^{10} pm/cm) = 352.4 pm
Problem #3: Nickel has a face-centered cubic structure with an edge length of 352.4 picometers. What is the density?
This problem is the exact reverse of problem #2. (See problem 5a below for an example set of calculations.)
Solution:
1) Convert pm to cm
2) Calculate the volume of the unit cell
3) Calculate the average mass of one atom of Ni
4) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell
5) Calculate the density (value from step 4 divided by value from step 2)
Problem #4: Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium.
Solution:
1) Convert pm to cm:
197 pm x (1 cm/10^{10} pm) = 1.97 x 10¯^{8} cm
2) Determine the edge length of the unit cell:
Use the Pythagorean Theorem (see problem #1 for a discussion):r = d ÷ 2(√2)
1.97 x 10¯^{8} cm = d ÷ 2(√2)
d = 5.572 x 10¯^{8} cm
3) Determine the volume of the unit cell:
(5.572 x 10¯^{8} cm)^{3} = 1.730 x 10¯^{22} cm^{3}
4) Determine mass of 4 atoms of Ca in a unit cell (cubic closest packed is the same as face-centered cubic):
40.08 g/mol divided by 6.022 x 10^{23} atoms/mol = 6.6556 x 10¯^{23} g/atom6.6556 x 10¯^{23} g/atom times 4 atoms = 2.66224 x 10¯^{22} g
5) Determine density:
2.66224 x 10¯^{22} g divided by 1.730 x 10¯^{22} cm^{3} = 1.54 g/cm^{3}
Problem #5: Krypton crystallizes with a face-centered cubic unit cell of edge 559 pm.
a) What is the density of solid krypton?
b) What is the atomic radius of krypton?
c) What is the volume of one krypton atom?
d) What percentage of the unit cell is empty space if each atom is treated as a hard sphere?
Solution to a:
1) Convert pm to cm:
559 pm x (1 cm/10^{10} pm) = 559 x 10¯^{10} cm = 5.59 x 10¯^{8} cm
2) Calculate the volume of the unit cell:
(5.59 x 10¯^{8} cm)^{3} = 1.7468 x 10¯^{22} cm^{3}
3) Calculate the average mass of one atom of Kr:
83.798 g mol¯^{1} divided by 6.022 x 10^{23} atoms mol¯^{1} = 1.39153 x 10¯^{22} g
4) Calculate the mass of the 4 krypton atoms in the face-centered cubic unit cell:
1.39153 x 10¯^{22} g times 4 = 5.566 x 10¯^{22} g
5) Calculate the density (value from step 4 divided by value from step 2):
5.566 x 10¯^{22} g / 1.7468 x 10¯^{22} cm^{3} = 3.19 g/cm^{3}
Solution to b:
Use the Pythagorean Theorem (see problem #1 for a discussion):r = d ÷ 2(√2)
r = 5.59 x 10¯^{8} cm ÷ 2(√2)
r = 1.98 x 10¯^{8} cm
Solution to c:
V = (4/3) π r^{3}V = (4/3) (3.14159) (1.98 x 10¯^{8} cm)^{3}
V = 3.23 x 10¯^{23} cm^{3}
Solution to d:
1) Calculate the volume of the 4 atoms in the unit cell:
3.23 x 10¯^{23} cm^{3} times 4 = 1.29 x 10¯^{22} cm^{3}
2) Calculate volume of cell not filled with Kr:
1.7468 x 10¯^{22} cm^{3} minus 1.29 x 10¯^{22} cm^{3} = 4.568 x 10¯^{23} cm^{3}
3) Calculate % empty space:
4.568 x 10¯^{23} cm^{3} divided by 1.7468 x 10¯^{22} cm^{3} = 0.261526%
Problem #6: You are given a small bar of an unknown metal. You find the density of the metal to be 11.5 g/cm^{3}. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.06 x 10^{-10} m. Find the gram-atomic weight of this metal and tentatively identify it.
Solution:
1) Convert meters to cm:
4.06 x 10^{-10} m = 4.06 x 10^{-8} cm
2) Determine the volume of the unit cube:
(4.06 x 10^{-8} cm)^{3} = 6.69234 x 10^{-23} cm^{3}
3) Determine the mass of the metal in the unit cube:
11.5 g/cm^{3} times 6.69234 x 10^{-23} cm^{3} = 7.696193 x 10^{-22} g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.696193 x 10^{-22} g is to 4 atoms as x grams is to 6.022 x 10^{23} atomsx = 116 g/mol (to three sig figs)
This weight is close to that of indium.
Problem #7: A metal crystallizes in a face-centered cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm^{3}. What is this metal?
Solution:
1) Convert nm to cm:
0.197 nm x (1 cm/10^{7} nm) = 1.97 x 10¯^{8} cm
2) Determine the edge length of the unit cell:
Use the Pythagorean Theorem (see problem #1 for a discussion):r = d ÷ 2(√2)
1.97 x 10¯^{8} cm = d ÷ 2(√2)
d = 5.572 x 10¯^{8} cm
3) Determine the volume of the unit cell:
(5.572 x 10¯^{8} cm)^{3} = 1.72995 x 10¯^{22} cm^{3}
4) Determine grams of metal in unit cell:
1.72995 x 10¯^{22} cm^{3} times 1.54 g/cm^{3} = 2.6641 x 10¯^{22} g
5) Determine atomic weight (based on 4 atoms per unit cell):
2.6641 x 10¯^{22} g is to 4 atoms as x grams is to 6.022 x 10^{23} atomsx = 40.11 g/mol
The metal is calcium.
Problem #8: The density of an unknown metal is 2.64 g/cm^{3} and its atomic radius is 0.215 nm. It has a face-centered cubic lattice. Determine its atomic weight.
Solution:
1) Convert nm to cm:
0.215 nm x (1 cm/10^{7} nm) = 2.15 x 10¯^{8} cm
2) Determine the edge length of the unit cell:
Use the Pythagorean Theorem (see problem #1 for a discussion):r = d ÷ 2(√2)
2.15 x 10¯^{8} cm = d ÷ 2(√2)
d = 6.08112 x 10¯^{8} cm
3) Determine the volume of the unit cell:
(6.08112 x 10¯^{8} cm)^{3} = 2.2488 x 10¯^{22} cm^{3}
4) Determine grams of metal in unit cell:
2.2488 x 10¯^{22} cm^{3} times 2.64 g/cm^{3} = 5.9368 x 10¯^{22} g
5) Determine atomic weight (based on 4 atoms per unit cell):
5.9368 x 10¯^{22} g is to 4 atoms as x grams is to 6.022 x 10^{23} atomsx = 89.4 g/mol
Problem #9: Metallic silver crystallizes in a face-centered cubic lattice with L as the length of one edge of the unit cube. What is the center-to-center distance between nearest silver atoms?
a) L/2
b) 2^{1/2} L
c) 2L
d) L/2^{1/2}
e) None of the above answers are valid.
Solution:
Call center-to-center distance = d. There are two of them on the face diagonal.
Therefore, by the Pythagorean Theorem:
L^{2} + L^{2} = (2d)^{2}2L^{2} = 4d^{2}
(L^{2} ) / 2 = d^{2}
L / 2^{1/2} = d
The answer is d
Problem #10: Iridium has a face centered cubic unit cell with an edge length of 383.3 pm. The density of iridium is 22.61 g/cm^{3}. Use these data to calculate a value for Avogadro's Number.
Solution:
1) Use the edge length to get the volume of the unit cell:
383.3 pm = 3.833 x 10¯^{8} cm(3.833 x 10¯^{8} cm)^{3} = 5.6314 x 10¯^{23} cm^{3}
2) Use the density to get the mass of Ir in the unit cell:
22.61 g/cm^{3} times 5.6314 x 10¯^{23} cm^{3} = 1.27326 x 10¯^{21} g
3) Use the atomic weight of Ir to determine how many moles of Ir are in the unit cell:
1.27326 x 10¯^{21} g divided by 192.217 g/mol = 6.624075 x 10¯^{24} mol
4) Use 4 atoms per face-centered unit cell to set up the following ratio and proportion:
4 atoms is to 6.624075 x 10¯^{24} mol as x is to 1.000 molx = 6.038 x 10^{23} atoms
For a different take on the solution to this problem, go here and take a look at the answer by m w.
Go to some body-centered cubic problems