Worksheet - Body-Centered Cubic Problems - AP level

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Here are the problems:
Problem #1: The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centered cubic. Tantalum has a density of 16.69 g/cm3 (a) calculate the mass of a tantalum atom. (b) Calculate the atomic weight of tantalum in g/mol.
Problem #2: Chromium crystallizes in a body-centered cubic structure. The unit cell volume is 2.583 x 10¯23 cm3. Determine the atomic radius of Cr in pm.
Problem #3: Barium has a radius of 224 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell?
Problem #4: Metallic potassium has a body-centered cubic structure. If the edge length of unit cell is 533 pm, calculate the radius of potassium atom.
Problem #5: Sodium has a density of 0.971 g/cm3 and crystallizes with a body-centered cubic unit cell. (a) What is the radius of a sodium atom? (b) What is the edge length of the cell? Give answers in picometers.
Problem #6: At a certain temperature and pressure an element has a simple body-centred cubic unit cell. The corresponding density is 4.253 g/cm3 and the atomic radius is 1.780 Å. Calculate the atomic mass (in amu) for this element.
Problem #7: Mo crystallizes in a body-centered cubic arrangement. Calculate the radius of one atom, given the density of Mo is 10.28 g /cm3.
Problem #8: see problem at end of file.


Problem #1: The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centered cubic. Tantalum has a density of 16.69 g/cm3.

(a) calculate the mass of a tantalum atom.
(b) Calculate the atomic weight of tantalum in g/mol.

Solution:

1) Convert pm to cm:

330.6 pm x 1 cm/1010 pm = 330.6 x 10¯10 cm = 3.306 x 10¯8 cm

2) Calculate the volume of the unit cell:

(3.306 x 10¯8 cm)3 = 3.6133 x 10¯23 cm3

3) Calculate mass of the 2 tantalum atoms in the body-centered cubic unit cell:

16.69 g/cm3 times 3.6133 x 10¯23 cm3 = 6.0307 x 10¯22 g

4) The mass of one atom of Ta:

6.0307 x 10¯22 g / 2 = 3.015 x 10¯22 g

5) The atomic weight of Ta in g/mol:

3.015 x 10¯22 g times 6.022 x 1023 mol¯1 = 181.6 g/mol

Problem #2: Chromium crystallizes in a body-centered cubic structure. The unit cell volume is 2.583 x 10¯23 cm3. Determine the atomic radius of Cr in pm.

Solution:

1) Determine the edge length of the unit cell:

[cube root of] 2.583 x 10¯23 cm3 = 2.956 x 10¯8 cm

2) Examine the following diagram:

The triangle we will use runs differently than the triangle used in fcc calculations.
d is the edge of the unit cell, however d√2 is NOT an edge of the unit cell. It is a
diagonal of a face of the unit cell. 4r is a body diagonal. Since it is a right triangle,
the Pythagorean Theorem works just fine.

We wish to determine the value of 4r, from which we will obtain r, the radius of the Cr atom. Using the Pythagorean Theorem, we find:

d2 + (d√2)2 = (4r)2

3d2 = (4r)2

3(2.956 x 10¯8 cm)2 = 16r2

r = 1.28 x 10¯8 cm

3) The conversion from cm to pm is left to the student.


Problem #3: Barium has a radius of 224 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell? (This is the reverse of problem #4.)

Solution:

1) Calculate the value for 4r (refer to the above diagram):

radius for barium = 224 pm

4r = 896 pm

2) Apply the Pythagorean Theorem:

d2 + (d√2)2 = (896)2

3d2 = 802816

d2 = 267605.3333. . .

d = 517 pm


Problem #4: Metallic potassium has a body-centered cubic structure. If the edge length of unit cell is 533 pm, calculate the radius of potassium atom. (This is the reverse of problem #3.)

Solution:

1) Solve the Pythagorean Theorem for r (with d = the edge length):

d2 + (d √2)2 = (4r)2

d2 + 2d2 = 16r2

3d2 = 16r2

r2 = 3d2 / 16

r = √3 (d / 4)

2) Solve the problem:

√3 (533 / 4)

r = 231 pm


Problem #5: Sodium has a density of 0.971 g/cm3 and crystallizes with a body-centered cubic unit cell. (a) What is the radius of a sodium atom? (b) What is the edge length of the cell? Give answers in picometers.

Solution:

1) Determine mass of two atoms in a bcc cell:

22.99 g/mol divided by 6.022 x 1023 mol-1 = 3.81767 x 10-23 g (this is the average mass of one atom of Na)

3.81767 x 10-23 g times 2 = 7.63534 x 10-23 g

2) Determine the volume of the unit cell:

7.63534 x 10-23 g divided by 0.971 g/cm3 = 7.863378 x 10-23 cm3

3) Determine the edge length (the answer to (b)):

[cube root of]7.863378 x 10-23 cm3 = 4.2842 x 10-8 cm

4) Use the Pythagorean Theorem (refer to above diagram):

d2 + (d√2)2 = (4r)2

3d2 = 16r2

r2 = 3(4.2842 x 10-8)2 / 16

r = 1.855 x 10-8 cm

The radius of the sodium atom is 185.5 pm. The edge length is 428.4 pm. The manner of these conversions are left to the reader.


Problem #6: At a certain temperature and pressure an element has a simple body-centred cubic unit cell. The corresponding density is 4.253 g/cm3 and the atomic radius is 1.780 Å. Calculate the atomic mass (in amu) for this element.

Solution:

1) Convert 1.780 Å to cm:

1.780 Å = 1.780 x 10-8 cm

2) Use the Pythagorean Theorem to calculate d, the edge length of the unit cell:

d2 + (d√2)2 = (4r)2

3d2 = 16r2

d2 = (16/3) (1.780 x 10-8 cm)2

d = 4.11 x 10-8 cm

3) Calcuate the volume of the unit cell:

(4.11 x 10-8 cm)3 = 6.95 x 10-23 cm3

4) Calcuate the mass inside the unit cell:

6.95 x 10-23 cm3 times 4.253 g/cm3 = 2.95 x 10-22 g

Use a ratio and proportion to calculate the atomic mass:

2.95 x 10-22 g is to two atoms as 'x' is to 6.022 x 1023 mol-1

x = 88.95 g/mol (or 88.95 amu)


Problem #7: Mo crystallizes in a body-centered cubic arrangement. Calculate the radius of one atom, given the density of Mo is 10.28 g /cm3.

Solution:

1) Determine mass of two atoms in a bcc cell:

95.96 g/mol divided by 6.022 x 1023 mol-1 = 1.59349 x 10-22 g (this is the average mass of one atom of Mo)

1.59349 x 10-22 g times 2 = 3.18698 x 10-22 g

2) Determine the volume of the unit cell:

3.18698 x 10-22 g divided by 10.28 g/cm3 = 3.100175 x 10-23 cm3

3) Determine the edge length:

[cube root of]3.100175 x 10-23 cm3 = 3.14144 x 10-8 cm

4) Use the Pythagorean Theorem (refer to above diagram):

d2 + (d√2)2 = (4r)2

3d2 = 16r2

r2 = 3(3.14144 x 10-8)2 / 16

r = 1.3603 x 10-8 cm (or 136.0 pm, to four sig figs)


Problem #8: In modeling solid-state structures, atoms and ions are most often modeled as spheres. A structure built using spheres will have some empty space in it. A measure of the empty (also called void) space in a particular structure is the packing efficiency, defined as the volume occupied by the spheres divided by the total volume of the structure.

Given that a solid crystallizes in a body-centered cubic structure that is 3.05 Å on each side, please answer the following questions. (The ChemTeam formatted this question while in transit through the Panama Canal, Nov. 7, 2010.)

Solution:

a. How many atoms are there in each unit cell?

2

b. What is the volume of one unit cell in Å3?

(3.05 Å)3 = 28.372625 Å3

c. Assuming that the atoms are spheres and the radius of each sphere is 1.32 Å, what is the volume of one atom in Å3?

(4/3) (3.141592654) (1.32)3 = 9.63343408 Å3

I used the key for π on my calculator, so there were some internal digits in addition to that last 4 (which is actually rounded up from the internal digits).

d. Therefore, what volume of atoms are in one unit cell?

(9.63343408 Å3 times 2) = 19.26816686 Å3

e. Based on your results from parts b and d, what is the packing efficiency of the solid expressed as a percentage?

19.26816686 Å3 / 28.372625 Å3 = 0.679

67.9%


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