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**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/cm^{3}.

(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/10^{10}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}cm^{3}

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

16.69 g/cm^{3}times 3.6133 x 10¯^{23}cm^{3}= 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 10^{23}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} cm^{3}. 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}cm^{3}= 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:

d^{2}+ (d√2)^{2}= (4r)^{2}3d

^{2}= (4r)^{2}3(2.956 x 10¯

^{8}cm)^{2}= 16r^{2}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 pm4r = 896 pm

2) Apply the Pythagorean Theorem:

d^{2}+ (d√2)^{2}= (896)^{2}3d

^{2}= 802816d

^{2}= 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):

d^{2}+ (d √2)^{2}= (4r)^{2}d

^{2}+ 2d^{2}= 16r^{2}3d

^{2}= 16r^{2}r

^{2}= 3d^{2}/ 16r = √3 (d / 4)

2) Solve the problem:

√3 (533 / 4)r = 231 pm

**Problem #5:** Sodium has a density of 0.971 g/cm^{3} 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 10^{23}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/cm^{3}= 7.863378 x 10^{-23}cm^{3}

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

[cube root of]7.863378 x 10^{-23}cm^{3}= 4.2842 x 10^{-8}cm

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

d^{2}+ (d√2)^{2}= (4r)^{2}3d

^{2}= 16r^{2}r

^{2}= 3(4.2842 x 10^{-8})^{2}/ 16r = 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/cm^{3} 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:

d^{2}+ (d√2)^{2}= (4r)^{2}3d

^{2}= 16r^{2}d

^{2}= (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}cm^{3}

4) Calcuate the mass inside the unit cell:

6.95 x 10^{-23}cm^{3}times 4.253 g/cm^{3}= 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 10^{23}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 /cm^{3}.

**Solution:**

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

95.96 g/mol divided by 6.022 x 10^{23}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/cm^{3}= 3.100175 x 10^{-23}cm^{3}

3) Determine the edge length:

[cube root of]3.100175 x 10^{-23}cm^{3}= 3.14144 x 10^{-8}cm

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

d^{2}+ (d√2)^{2}= (4r)^{2}3d

^{2}= 16r^{2}r

^{2}= 3(3.14144 x 10^{-8})^{2}/ 16r = 1.3603 x 10

^{-8}cm (or 136.0 pm, to four sig figs)

**Problem #8:** Sodium crystallizes in body-centered cubic system, and the edge of the unit cell is 430. pm. Calculate the dimensions of a cube that would contain one mole of Na.

**Solution:**

A cube that is bcc has two atoms per unit cell.6.022 x 10

^{23}atoms divided by 2 atoms/cell = 3.011 x 10^{23}cells required.430. pm = 4.30 x 10

^{-8}cm <--- I'm going to give the answer in cm^{3}rather than pm^{3}(4.30 x 10

^{-8}cm)^{3}= 7.95 x 10^{-23}cm^{3}<--- vol. of unit cell in cm^3(3.011 x 10

^{23}cell) (7.95 x 10^{-23}cm^{3}/cell) = 23.9 cm^{3}23.9 cm

^{3}would be a cube 2.88 cm on a side (2.88 being the cube root of 23.9)

**Problem #9:** Vanadium crystallizes with a body-centered unit cell. The radius of a vanadium atom is 131 pm. Calculate the density of vanadium. (in g/cm^{3})

**Solution:**

1) We are going to use the Pythagorean Theorem to determine the edge length of the unit cell. That edge length will give us the volume.

131 pm times ( 1 cm / 10^{10}pm) = 131 x 10^{-10}cm = 1.31 x 10^{-8}cmThe right triangle for Pythagorean Theorem is here. The image is in problem #2.

3d

^{2}= (4 * 1.31 x 10^{-8}cm)^{2}d

^{2}= (4 * 1.31 x 10^{-8}cm)^{2}/ 3d = 3.0253 x 10

^{-8}cm <--- this is the edge lengthCube the edge length to give the volume:

2.7689 x 10

^{-23}cm^{3}

2) We will use the average mass of one V atom and the two atoms in bcc to determine the mass of V inside the unit cell.

50.9415 g/mol divided by 6.022 x 10^{23}mol^{-1}= 8.459 x 10^{-23}g <--- average mass of one atom8.459 x 10

^{-23}g times 2 = 1.6918 x 10^{-22}g <--- mass of V in unit cell

3) Step 2 divided by step 1 gives the density.

1.6918 x 10^{-22}g / 2.7689 x 10^-23 cm^{3}= 6.11 g/cm^{3}

**Problem #10:** 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 Å^{3}?

(3.05 Å)^{3}= 28.372625 Å^{3}

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 Å^{3}?

(4/3) (3.141592654) (1.32)^{3}= 9.63343408 Å^{3}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 Å^{3}times 2) = 19.26816686 Å^{3}

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

19.26816686 Å^{3}/ 28.372625 Å^{3}= 0.67967.9%

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