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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}

**Problem #9:** The radius of gold is 144 pm, and the density is 19.32 g/cm^{3}. Does gold crystallize in a face-centered cubic structure or a body-centered cubic structure?

**Solution:**

1) I will assume the unit cell is face-centered cubic. I will use that assumption and the atomic radii to calculate the volume of the cell. From there, I will use the fact that there are 4 atoms of gold in the unit cell to determine the density. The final step will be to compare it to the 19.32 value. Here is one face of a face-centered cubic unit cell:

2) Across the face of the unit cell, there are 4 radii of gold, hence 576 pm. Using the Pythagorean Theorem, we determine the edge length of the unit cell:

d^{2}+ d^{2}= 576^{2}d = 407.2935 pm

3) Let us convert the pm to cm:

4407.2935 pm times (1 cm / 10^{10}pm) = 4.072935 x 10^{-8}cm

4) The volume of the unit cell:

(4.072935 x 10^{-8}cm)^{3}= 6.75651 x 10^{-23}cm^{3}

5) The mass of 4 gold atoms:

(196.96655 g/mol divided by 6.022 x 10^{23}atoms/mol) times 4 atoms = 1.308313 x 10^{-21}g

6) Let's see what density results:

1.308313 x 10^{-21}g / 6.75651 x 10^{-23}cm^{3}= 19.36 g/cm^{3}

We conclude that gold crystallizes fcc because we were able to reproduce the known density of gold.

Let's do the bcc calculation (which we know will give us the wrong answer). Here's an image showing what to do with the Pythagorean Theorem:

The rest of the calculation with minimal comment:

d^{2}+ (d√2)^{2}= 576^{2}d = 332.55 pm

332.55 pm = 3.3255 x 10

^{-8}cm(3.3255 x 10

^{-10}cm)^{3}= 3.6776 x 10^{-23}cm^{3}There are two atoms in a body-centered cubic.

(197 g/mol divided by 6.022 x 10

^{23}atoms/mol) times 2 atoms = 6.5427 x 10^{-22}g6.5427 x 10

^{-22}g / 3.6776 x 10^{-23}cm^3 = 17.79 g/cm^3Gold does not crystallize bcc because bcc does not reproduce the known density of gold.

**Problem #10:** Avogadro's number has been determined by about 20 different methods. In one approach, the spacing between ions in an ionic substance is determined by using X-ray diffraction. X-ray diffraction of sodium chloride have shown that the distance between adjacent Na^{+} and Cl¯ ions is 2.819 x 10^{-8} cm. The density of solid NaCl is 2.165 g/cm^{3}. By calculating the molar mass to four significant figures, you can determine Avogadro's number. What value do you obtain?

**Solution:**

1) Imagine a cube with 4 Na and 4 Cl at adjacent vertices. I'll call it the reference cube.

There's going to be a twist and it involves how many Na and Cl are in the cube. Think about it before the reveal in the last step.

2) Determine the volume of the cube:

(2.819 x 10^{-8}cm)^{3}= 2.2402 x 10^{-23}cm^{3}

3) Calculate the mass of NaCl inside the cube:

2.2402 x 10^{-23}cm^{3}times 2.165 g/cm^{3}= 4.85 x 10^{-23}g

4) The molar mass divided by the mass inside the cube equals Avogadro's Number. Here's where the twist comes into play.

Each Na and each Cl at a vertex of the reference cube is shared by a total of 8 cubes. (You may verify this on your own.) At any one vertex, there is 1/8 of a Na atom and 1/8 of a Cl atom inside the reference cube.With the reference cube having 4 vertices of Na and 4 vertices of Cl, this means there is a total of 1/2 of a Na atom and 1/2 of a Cl atom inside the reference cube.

The "molar mass" of (1/2)NaCl is half of 58.443.

29.2215 g/mol divided by 4.85 x 10

^{-23}g = 6.025 x 10^{23}mol^{-1}

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