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**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) (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:

$\sqrt[3]{\mathrm{5.8793545\; x\; 10\xaf23cm3}}$ = 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 edge 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}/ 8r = d /$\sqrt{8}$ <--- often left like this

r = d / (2$\sqrt{2}$) <--- an alternate formulation

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.

The above discusses how to determine r in terms of d in a face-centered unit cell. You may be asked to do the opposite, that is, to determine d in terms of r for a fcc cell. I'll repeat:

r = d /$\sqrt{8}$

followed by multiplying both sides by $\sqrt{8}$:

d = r$\sqrt{8}$

**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) (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:

$\sqrt[3]{\mathrm{4.376514\; x\; 10\xaf23cm3}}$ = 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$\sqrt{2}$) <--- one of two alternate formulations

1.97 x 10¯

^{8}cm = d / (2$\sqrt{2}$)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 / 6.022 x 10^{23}atoms/mol = 6.6556 x 10¯^{23}g/atom(6.6556 x 10¯

^{23}g/atom) (4 atoms) = 2.66224 x 10¯^{22}g

5) Determine density:

2.66224 x 10¯^{22}g / 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) (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}/ 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) (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 /$\sqrt{8}$ <--- one of two alternate formulations

r = 5.59 x 10¯

^{8}cm /$\sqrt{8}$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}) (4) = 1.29 x 10¯^{22}cm^{3}

2) Calculate volume of cell not filled with Kr:

1.7468 x 10¯^{22}cm^{3}− 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}/ 1.7468 x 10¯^{22}cm^{3}= 0.261526.15%

**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}) (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) (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$\sqrt{2}$) <--- one of two alternate formulations

1.97 x 10¯

^{8}cm = d / (2$\sqrt{2}$)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}) (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) (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 / $\sqrt{8}$ <--- one of two alternate formulations

2.15 x 10¯

^{8}cm = d / $\sqrt{8}$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}) (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^{½}L

(c) 2L

(d) L / 2^{½}

(e) None of the above answers are valid.

**Solution:**

Let 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

^{½}= dAnswer choice (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}) (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 / 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 Dr W.

**Problem #11:** Platinum has a density of 21.45 g/cm^{3} and a unit cell side length 'd' of 3.93 Ångstroms. What is the atomic radius of platinum? (1 Å = 10¯^{8} cm.)

**Solution:**

1) We need to determine if the unit cell is fcc or bcc.

Volume of unit cell:(3.93 x 10¯^{8}cm)^{3}= 6.0698 x 10¯^{23}cm^{3}Determine the mass of Pt in the unit cell:

(21.45 g/cm^{3}) (6.0698 x 10¯^{23}cm^{3}) = 1.302 x 10¯^{21}gHow many atoms is that?

(1.302 x 10¯^{21}g / 195.078 g/mol) * 6.022 x 10^{23}mol¯^{1}= 4The unit cell for Pt is fcc.

2) Use the Pythogrean Theorem to calculate the length of the hypotenuse which we know to be four times the radii of one Pt atom (see Problem #1 for a discussion).

We know this:d^{2}+ d^{2}= (4r)^{2}<--- where d is the edge length and r is the radius of the atom.Therefore:

r = d / (2$\sqrt{2}$) <--- one of two alternate formulationsr = (3.93 x 10¯

^{8}cm) / (2$\sqrt{2}$)r = 1.39 x 10¯

^{8}cm

3) Note that picometers is the preferred unit for atomic radii (with Ångstroms being the preferred unit of older vintage (for example, when the ChemTeam was in school).

139 pm = 1.39 x 10¯^{8}cm = 1.39 Å

**Problem #12:** The unit cell of platinum has a length of 392.0 pm along each side. Use this length (and the fact that Pt has a face-centered unit cell) to calculate the density of platinum metal in kg/m^{3} (Hint: you will need the atomic mass of platinum and Avogadro's number).

**Solution:**

1) Calculate the volume of the unit cell in meters cubed:

392.0 pm times (1 m / 10^{12}pm) = 392.0 x 10¯^{12}m = 3.920 x 10¯^{10}m(3.920 x 10¯

^{10}m)^{3}= 6.0236288 x 10¯^{29}m^{3}

2) Calculate the mass of Pt in the unit cell in kg:

195.078 g/mol / 6.022 x 10^{23}mol¯^{1}= 3.239422 x 10¯^{22}g(3.239422 x 10¯

^{22}g) (4) = 1.2957688 x 10¯^{21}g(1.2957688 x 10¯

^{21}g) (1 kg / 1000 g) = 1.2957688 x 10¯^{24}kg

3) Calculate the density:

1.2957688 x 10¯^{24}kg / 6.0236288 x 10¯^{29}m^{3}= 21511 kg / m^{3}The book value is 21450 kg / m

^{3}.

Note the use of the SI-approved unit for density as opposed to the more commonly-used unit of g/cm^{3}.

**Problem #13:** A metal crystallizes in a face-centered cubic structure and has a density of 11.9 g cm¯^{3}. If the radius of the metal atom is 138 pm, what is the most probable identity of the metal.

**Solution:**

1) Determine the atomic radius in cm:

(138 pm) (100 cm / 10^{12}pm) = 138 x 10¯^{10}cm = 1.38 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 / $\sqrt{8}$ <--- one of two alternate formulations

1.38 x 10¯

^{8}cm = d / $\sqrt{8}$d = 3.90323 x 10¯

^{8}cm

2) Determine the volume of the unit cell:

(3.90323 x 10¯^{8}cm)^{3}= 5.94665 x 10¯^{23}cm^{3}

3) Determine the mass of the metal inside the unit cell:

(11.9 g cm¯^{3}) (5.94665 x 10¯^{23}cm^{3}) = 7.0765 x 10¯^{22}g

3) The above mass is that of 4 atoms (based on our knowledge that the unit cell is fcc). Scale the mass to that of Avogadro Number of atoms:

7.0765 x 10¯^{22}g is to 4 atoms as x is to 6.022 x 10^{23}atoms/molex = 106.5 g/mol

The metal is palladium.

**Problem #14:** Nickel oxide (NiO) crystallizes in the NaCl type of crystal structure. The length of the unit cell of NiO is 4.20 Å. Calculate the density of NiO.

**Solution:**

1) A brief discussion . . .

. . . of the NaCl structure is found here. Ignore the question and the last half of the answer.The key point is that in the NaCl unit cell, there are 4 Na

^{+}and 4 Cl¯. You can think of it as a face-centered unit cell of chloride ions has been interpenetrated with a face-centered unit cell of sodium ions.Following the above, a unit cell of NiO will contain 4 Ni

^{2+}and 4 O^{2}¯

2) Convert Å to cm:

1 Å = 10¯^{8}cm (Ångströn is an old unit but still remains in limited use today)4.20 Å = 4.20 x 10¯

^{8}cm

3) The volume of one NiO unit cell is this:

(4.20 x 10¯^{8}cm)^{3}= 7.4088 x 10¯^{23}cm^{3}

4) The weight of four NiO in the unit cell:

(4) (74.692 g/mol / 6.022 x 10^{23}mol¯^{1}) = 4.96128 x 10¯^{22}g

5) Determine the density:

4.96128 x 10¯^{22}g / 7.4088 x 10¯^{23}cm^{3}= 6.70 g/cm^{3}This compares to the book value of 6.67 g/cm

^{3}

**Problem #15:** NiO adopts the face-centered-cubic arrangement. Given that the density of NiO is 6.67 g/cm^{3}, calculate the length of the edge of its unit cell (in pm).

**Solution:**

1) Calculate the average mass of one NiO formula unit:

74.692 g/mol / 6.022 x 10^{23}mol¯^{1}= 1.24032 x 10¯^{22}g

2) NiO has the NaCl structure, so 4 Ni and 4 O per unit cell. Determine the total mass of NiO in one unit cell:

(4) (1.24032 x 10¯^{22}g) = 4.96128 x 10¯^{22}g

3) Determine the volume of the unit cell:

4.96128 x 10¯^{22}g / 6.67 g/cm^{3}= 7.4382 x 10¯^{23}cm^{3}

4) Determine the edge length:

$\sqrt[3]{\mathrm{7.4382\; x\; 10\xaf23cm3}}$ = 4.2055 x 10¯^{8}cm

5) Convert from cm to pm:

4.2055 x 10¯^{8}cm times (10^{10}pm / 1 cm) = 420.55 pmA value of 420. pm seems reasonable as a final answer.

**Problem #16:** Aluminum crystallizes in a face-centered cubic unit cell and has an atomic radius of 143 pm. What is the density of aluminum?

**Solution:**

1) There are 4 atomic radii (r) along the face of an fcc cube. Thus, the edge (d) can be found:

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

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

^{2}= 8r^{2}d = r $\sqrt{8}$

2) Convert the atomic radius from pm to cm

(143 pm) (1 m/10^{12}pm) (10^{2}cm/1 m) = 1.43 x 10¯^{8}cmBy the way, note how the conversion went through the unit meters. Sometimes, you see this:

(143 pm) (10

^{2}cm/10^{12}pm) = 1.43 x 10¯^{8}cmOr this:

(143 pm) (1 cm/10

^{10}pm) = 1.43 x 10¯^{8}cm

3) Calculate the value of 'd:'

d = (1.43 x 10¯^{8}cm) ($\sqrt{8}$)d = 4.04465 x 10 x 10¯

^{8}cm

4) Calculate volume of unit cell:

(4.04465 x 10¯^{8}cm)^{3}= 6.6167 x 10¯^{23}cm^{3}

5) There are 4 atoms in each fcc unit cell. Calculate the mass of 4 atoms of aluminum:

(4 atoms) (26.98 g/mol / 6.022 x 10^{23}atoms/mol) = 1.7921 x 10¯^{22}g

6) Density of Al:

1.7921 x 10¯^{22}g / 6.6167 x 10¯^{23}cm^{3}= 2.71 g/cm^{3}(to three sig figs)

**Problem #17:** The density of an unknown metal is 8.94 g/cm^{3} and its atomic radius is 0.126 nm. It has a face-centered cubic lattice. Find the atomic weight of this metal.

**Solution:**

1) Face-centered means two things:

(a) four atoms in a unit cell

(b) the atoms touch along a face diagonal

2) With 'd' representing the edge of the cube and '4r' representing the 4 atomic radii across the face diagonal, we use the Pythagorean Theorem to determine the value of 'd' in nanometers;

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

^{2}= (4*0.126)^{2}2d

^{2}= 0.254016d

^{2}= 0.254016d = 0.35638 nm

This formulation could have been used as well:

d = r$\sqrt{8}$d = (0.126) ($\sqrt{8}$) = 0.35638 nm

3) Our nm value must be converted to cm. I will convert by going through meters:

(0.35638 nm) (1 m / 10^{9}nm) (100 cm / 1 m) = 3.5638 x 10¯^{8}cm

4) Determine the volume of the unit cell:

(3.5638 x 10¯^{8}cm)^{3}= 4.526265 x 10¯^{23}cm^{3}

5) Use the density to determine the mass of the metal inside the unit cell:

(8.94 g/cm^{3}) (4.526265 x 10¯^{23}cm^{3}) = 4.046481 x 10¯^{22}g

6) The mass just calculated is the mass of the 4 atoms in the unit cell. Scale that ratio to the one that uses Avogadro Number of atoms:

4.046481 x 10¯ ^{22}gx ––––––––––––––– = –––––––––––––––––––– 4 atoms 6.022 x 10 ^{23}atoms / molx = 60.92 g/mol

7) There is no element with an atomic weight of 60.92 g/mol. There is an isotope of copper (Cu-61) that weighs 60.93. It has a half-life of a bit more than 3 hors. There is also a Ni-61 isotope and it is stable, making up about 1.14% of naturally-obtained nickel. I didn't look at any other elements.

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