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**Problem #1:** Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, sodium has a density of 0.968 g/cm^{3} and a unit cell side length (a) of 4.29 Å

a. How many sodium atoms are in 1 cm^{3}?

b. How many unit cells are in 1 cm^{3}?

c. How many sodium atoms are there per unit cell?

**Solution:**

1) Calculate the average mass of one atom of Na:

22.99 g mol¯^{1}÷ 6.022 x 10^{23}atoms mol¯^{1}= 3.82 x 10¯^{23}g/atom

2) Determine atoms in 1 cm^{3}:

0.968 g / 3.82 x 10¯^{23}g/atom = 2.54 x 10^{22}atoms in 1 cm^{3}

3) Determine volume of the unit cell:

(4.29 x 10¯^{8}cm)^{3}= 7.89 x 10¯^{23}cm^{3}

4) Determine number of unit cells in 1 cm^{3}:

1 cm^{3}/ 7.89 x 10¯^{23}cm^{3}= 1.27 x 10^{22}unit cells

5) Determine atoms per unit cell:

2.54 x 10^{22}atoms / 1.27 x 10^{22}unit cells = 2 atoms per unit cell

**Problem #2:** Metallic iron crystallizes in a type of cubic unit cell. The unit cell edge length is 287 pm. The density of iron is 7.87 g/cm3. How many iron atoms are there within one unit cell?

**Solution:**

1) Calculate the average mass of one atom of Fe:

55.845 g mol¯^{1}÷ 6.022 x 10^{23}atoms mol¯^{1}= 9.2735 x 10¯^{23}g/atom

2) Determine atoms in 1 cm^{3}:

7.87 g / 9.2735 x 10¯^{23}g/atom = 8.4866 x 10^{22}atoms in 1 cm^{3}

3) Determine volume of the unit cell:

287 pm x (1 cm / 10

^{10}pm) = 2.87 x 10¯^{8}cm(2.87 x 10¯

^{8}cm)^{3}= 2.364 x 10¯^{23}cm^{3}

4) Determine number of unit cells in 1 cm^{3}:

1 cm^{3}/ 2.364 x 10¯^{23}cm^{3}= 4.23 x 10^{22}unit cells

5) Determine atoms per unit cell:

8.4866 x 10^{22}atoms / 4.23 x 10^{22}unit cells = 2 atoms per unit cell

**Problem #3:** (a) You are given a cube of silver metal that measures 1.015 cm on each edge. The density of silver is 10.49 g/cm^{3}. How many atoms are in this cube?

**Solution to a:**

(1.015 cm)^{3}x (10.49 g / cm^{3}) x (1 mole Ag / 107.9 g) x (6.023 x 10^{23}atoms / 1 mole) = 6.12 x 10^{22}atoms Ag

(b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that 74% of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom.

**Solution to b:**

(1.015 cm)^{3}x 0.74 = 0.774 cm^{3}filled by Ag atoms

0.774 cm^{3}/ (6.12 x 10^{22}atoms) = 1.26 x 10^{-23}cm^{3}/ Ag atom

(c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

**Solution to c:**

V = (4/3) π r^{3}

r^{3}= V / 1.33 π = (1.26 x 10^{-23}cm^{3}) / (1.33)(3.14) = 3.03 x 10^{-24}cm^{3}r = 1.4 x 10

^{-8}cm = 144 pm

**Problem #4:** Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, gold has a density of 19.32 g/cm^{3} and a unit cell side length of 4.08 Å. (1 Å = 1 x 10^{-8} cm.)

(a) How many gold atoms are in exactly 1 cm^{3}?

(b) How many unit cells are in exactly 1 cm^{3}?

(c) How many gold atoms are there per unit cell?

(d) The atoms/unit cell suggests that gold packs as a (i) simple, (ii) body-centered or (iii) face-centered unit cell.

**Solution:**

1) Part a:

19.32 g / 197.0 g/mol = 0.098071 mol0.098071 mol times 6.022 x 10

^{23}atoms/mol = 5.9058 x 10^{22}atoms

2) Part b:

4.08 Å = 4.08 x 10^{-8}cm1 cm divided by 4.08 x 10

^{-8}cm = 24509804 (this is how many 4.08 Å segments in 1 cm)24509804 cubed = 1.47238 x 10

^{22}unit cells

3) Part c:

5.9058 x 10^{22}atoms / 1.47238 x 10^{22}unit cells = 4 atom/unit cell

4) Part d:

face-centered

**Problem #5:** A metal nitride has a nitrogen atom at each corner and a metal atom at each edge. Which is the empirical formula for this nitride?

a. Ba_{3}N_{2}

b. Na_{3}N

c. AlN

d. Ti_{3}N_{4}

**Solution:**

A cube has eight corners and an atom at a corner is in eight different cubes; therefore 1/8 of an atom at each corner of a given cube. So:

1/8 times 8 = 1 total nitrogen atom in each cube

A cube has 12 edges and each edge is in 4 different cubes, so there is 1/4 of an atom in each individual cube. So:

1/4 times 12 = 3 total metal atoms in each cube

The only choice to fit the above criteria is answer choice b, Na_{3}N.

**Problem #6:** Calcium fluoride crystallizes with a cubic lattice. The unit cell has an edge of 546.26 pm and has a density of 3.180 g/cm^{3}. How many formula units must there be per unit cell?

**Solution:**

1) Convert pm to cm:

546.26 pm times (1 cm / 10^{10}pm) = 5.4626 x 10^{-8}cm

2) Determine volume of unit cell:

(5.4626 x 10^{-8}cm)^{3}= 1.63 x 10^{-22}cm^{3}

3) Determine mass of CaF_{2} in unit cell:

3.180 g/cm^{3}times 1.63 x 10^{-22}cm^{3}= 5.1835 x 10^{-22}g

4) Determine mass of one formula unit of CaF_{2}:

formula weight of CaF_{2}= 78.074 g/mol78.074 g/mol divided by 6.022 x 10

^{23}formula units / mole = 1.2965 x 10^{-22}g

5) Determine number of formula units in one unit cell:

5.1835 x 10^{-22}g divided by 1.2965 x 10^{-22}g = 3.998There are 4 formula units of CaF

_{2}per unit cell

**Problem #7:** Tungsten has an atomic radius of 137 pm and crystallizes in a cubic unit cell having an edge length d = 316 pm. What type of cubic unit cell does tungsten crystallize in?

**Solution:**

Let us assume the cell is face-centered. If this is the case, then this relationship should hold true:d

^{2}+ d^{2}= (4r)^{2}where r = the radius.

Please see a small discussion of this in problem #1 here.

Using 316 pm for d and 548 pm for 4r, we have this:

316

^{2}+ 316^{2}?=? 548^{2}We find 199712 for the left and 300304 for the right, so the idea that tungsten is fcc fails.

For body-centered, please see problem #2 here for this equation:

d

^{2}+ (d√2)^{2}= (4r)^{2}3d

^{2}= (4r)^{2}3(316

^{2}) ?=? 548^{2}299568 = 300304

Due to the fact that these numbers are roughly equivalent, we can conclude that tungsten is being body-centered cubic.

**Problem #8:** What is the formula of the compound that crystallizes with Ba^{2+} ions occupying one-half of the cubic holes in a simple cubic arrangement of fluoride ions?

**Solution:**

We must consider two cubic unit cells, one with the barium ion and one without (this gives us one-half of the cubic holes filled).Cell 1: 8 F atoms at the 8 vertices. Since each vertex is in a total of 8 cells, we have 1 F atom in the unit cell. The cubic hole in the middle of the cell is empty.

Cell 2: 8 F atoms at the 8 vertices. Since each vertex is in a total of 8 cells, we have 1 F atom in the unit cell. The cubic hole in the middle of the cell has a barium in it.

Total for the two cells: one Ba and two F

Formula: BaF

_{2}

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