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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}) (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) (6.022 x 10^{23}mol¯^{1}) = 181.6 g/mol

**Problem #2a:** 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:

$\sqrt[3]{\mathrm{2.583\; x\; 10\xaf23cm3}}$ = 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$\sqrt{2}$)^{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 #2b:** Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 128 pm . Calculate the density of solid crystalline chromium in grams per cubic centimeter.

**Solution:**

1) Convert pm to cm:

(125 pm) (1 cm / 10^{10}pm) = 1.25 x 10¯^{8}cm

2) Use the Pythagorean theorem to calculate the unit cell edge length:

d^{2}+ (d$\sqrt{2}$)^{2}= (4r)^{2}3d

^{2}= (4r)^{2}3d

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

^{2}= (16) (1.25 x 10¯^{8}cm)^{2}d$\sqrt{3}$ = (4) (1.25 x 10¯

^{8}cm)d = [(4) (1.25 x 10¯

^{8}cm)] / $\sqrt{3}$d = 2.8868 x 10¯

^{8}cm

3) Calculate volume of the unit cell

(2.8868 x 10¯^{8}cm)^{3}= 2.4056 x 10¯^{23}cm^{3}

4) Determine mass of two atoms in body-centered unit cell:

51.996 g/mol / 6.022 x 10^{23}atoms/mol = 8.63434 x 10¯^{23}g/atom(2) (8.63434 x 10¯

^{23}g/atom) = 1.726868 x 10¯^{22}g

5) Determine the density:

1.726868 x 10¯^{22}g / 2.4056 x 10¯^{23}cm^{3}= 7.18 g/cm^{3}(to three sig figs)Book value is 7.15.

**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$\sqrt{2}$)^{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$\sqrt{2}$)^{2}= (4r)^{2}d

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

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

^{2}= 3d^{2}/ 16r = (d$\sqrt{3}$) / 4

2) Solve the problem:

r = (533$\sqrt{3}$) / 4r = 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?

**Solution:**

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

22.99 g/mol / 6.022 x 10^{23}mol¯^{1}= 3.81767 x 10¯^{23}g (this is the average mass of one atom of Na)(2) (3.81767 x 10¯

^{23}g) = 7.63534 x 10¯^{23}g

2) Determine the volume of the unit cell:

7.63534 x 10¯^{23}g / 0.971 g/cm^{3}= 7.863378 x 10¯^{23}cm^{3}

3) Determine the edge length, which is the answer to (b):

$\sqrt[3]{\mathrm{7.863378\; x\; 10\xaf23cm3}}$ = 4.2842 x 10¯^{8}cm

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

d^{2}+ (d$\sqrt{2}$)^{2}= (4r)^{2}3d

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

^{2}= 3(4.2842 x 10¯^{8})^{2}/ 16r = 1.855 x 10¯

^{8}cm

5) Typically, answers of this type are given in pm:

(a) The radius of the sodium atom is 185.5 pm.

(b) 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$\sqrt{2}$)^{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}) (4.253 g/cm^{3}= 2.95 x 10¯^{22}g)

5) 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 / 6.022 x 10^{23}mol¯^{1}= 1.59349 x 10¯^{22}g (this is the average mass of one atom of Mo)(2) (1.59349 x 10¯

^{22}g) = 3.18698 x 10¯^{22}g

2) Determine the volume of the unit cell:

3.18698 x 10¯^{22}g / 10.28 g/cm^{3}= 3.100175 x 10¯^{23}cm^{3}

3) Determine the edge length:

$\sqrt[3]{\mathrm{3.100175\; x\; 10\xaf23cm3}}$ = 3.14144 x 10¯^{8}cm

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

d^{2}+ (d$\sqrt{2}$)^{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 / 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) ( 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 / 6.022 x 10^{23}mol¯^{1}= 8.459 x 10¯^{23}g <--- average mass of one atom(2) (8.459 x 10¯

^{23}g) = 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:** Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm^{3}. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

**Solution:**

1) We need to determine the volume of one unit cell. I'll approach this in a dimensional analysis sorta way:

(volume/g) (g/mole) (mole/atoms) (atoms/cell) = (volume/cell)See how all the units cancel except for volume and cell. Once we have the volume of the cell, we can determine the edge lenth by taking the cube root of the volume.

I'll build it one calculation at a time.

2) (volume/g)

1.00 cm^{3}/4.50 g

3) (volume/g) (g/mole) <--- molar mass of Ti

(1.00 cm^{3}/4.50 g) (47.867 g/mol)

4) (volume/g) (g/mole) (mole/atoms) <--- Avogadro's Number

(1.00 cm^{3}/4.50 g) (47.867 g/mol) (1.00 mol/6.022 x 10^{23}atoms)

5) (volume/g) (g/mole) (mole/atoms) (atoms/cell)

(1.00 cm^{3}/4.50 g) (47.867 g/mol) (1.00 mol/6.022 x 10^{23}atoms) (2 atoms/cell) <--- because of body-centered cubic

6) Do the calculation for the volume of the unit cell. The answer is:

3.53275 x 10¯^{23}cm^{3}

7) The edge length is simply the cube root of the cell volume. The answer is:

3.28 x 10¯^{8}cm (to three sig figs)Comment: often the edge length is asked for in pm. The student is left to determine the conversion from cm to pm. The answer is 328 pm.

8) For the atomic radius, I will add some guard digits to the edge length (symbolized by 'd' in the Pythagorean theorem calculation I will use. By the way, remember that hint from the problem text? That's the thing that allows me to write 4r.

d^{2}+ (d$\sqrt{2}$)^{2}= (4r)^{2}(4r)

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

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

^{2}= 3d^{2}/ 16r = d$\sqrt{3}$ / 4

r = [(3.28124 x 10¯

^{8}cm) ($\sqrt{3}$)] / 4r = 1.42 x 10¯

^{8}cm = 142 pm

**Problem #11:** Aluminum B is a solid phase of aluminum still unknown to science. The only difference between it and ordinary aluminum is that Aluminum B forms a crystal with a bcc unit cell and a lattice constant a = 331 pm. Calculate the density of Aluminum B.

**Solution:**

1) Convert 331 pm to cm:

(331 pm) (10^{2}cm / 10^{12}pm) = 3.31 x 10¯^{8}cm

2) Determine the volume of the unit cell:

(3.31 x 10¯^{8}cm)^{3}= 3.6264691 x 10¯^{23}cm^{3}

3) Determine mass of Aluminum in the unit cell:

26.98 g/mol / 6.022 x 10^{23}atom/mol = 4.48 x 10¯^{23}g/atom(2 atom) (4.48 x 10¯

^{23}g/atom) = 8.96 x 10¯^{23}gRemember, two atoms in a bcc unit cell

4) Calculate the density:

8.96 x 10¯^{23}g / 3.6264691 x 10¯^{23}cm^{3}= 2.47 g/cm^{3}

**Problem #12:** The density of TlCl(s) is 7.00 g/cm^{3} 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 / 10^{12}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}cm^{3}

3) Calculate the mass of TlCl in one unit cell:

(7.00 g/cm^{3}) (5.7066625 x 10¯^{23}cm^{3}) = 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 10^{23}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.

**Bonus Problem:** 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.

**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?

(2) (9.63343408 Å^{3}) = 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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