For the three cubic unit cells that you built (simple cubic, body-centered cubic, and face-centered cubic), you were asked to determine 1.) the coordination number for the unit cell, 2.) the number of atoms per unit cell, and 3.) the fraction of unit cell volume occupied by atoms. Below are the calculations necessary to obtain the answer to question #3 for each of the three unit cells.

To find the fraction of volume of the simple cubic unit cell occupied by atoms, we need to find the volume of one atom (there is one atom within this unit cell), then divide that volume by the volume of the unit cell. We can use the volumes of the spheres (even though these are obviously larger than real atoms!) because we are taking a ratio, which will be the same on both the atomic and the model scale.

So, first the volume of an atom. We can measure the diameter of one of the colorless spheres (2.5 cm), and use the resulting radius (1.25 cm) to calculate the volume of a sphere...

Next, let's find the volume of the unit cell. If we look at one face of the simple cubic unit cell (remembering that each atom is shared by eight unit cells), we can see that the length of one side of this cubic unit cell is equal to twice the atomic radius (r).

We can, therefore, determine the volume of the unit cell by cubing d...

V_{unit cell} = d³ = (2r)³ = 8r³ = 15.6 cm³

Now, let's divide the volume of one atom by the volume of the unit cell...

V_atom/V_{unit cell} = 8.18 cm³/15.6 cm³ = 0.524

This tells us that roughly half of the volume of a simple cubic unit cell is occupied by atoms. The rest is just space.

The procedure for the bcc unit cell is the same, except for two changes: 1.) there are two atoms in one bcc unit cell, and 2.) the procedure for finding the volume of the unit cell is different and quite a bit more complicated.

First, the easy part...we can simply multiply the volume of one sphere (which we calculated for the simple cubic unit cell) by two.

V_atoms = 2 x 8.18 cm³ = 16.4 cm³

Next, the hard part...we need to find the volume of one bcc unit cell. The length of one side of the unit cell is not a simple multiple of the atomic radius (as in the simple cubic unit cell). But if we look at the bcc unit cell closely, we can see that the length of the "body diagonal" of the cell (from one corner of the cube to the other) is equal to four times the atomic radius (r). We can draw a right triangle the sides of which are this "body diagonal" (4r), one side of the unit cell (d) and diagonal line across the "bottom" of the unit cell.

The length of the remaining side of this triangle is the length of the diagonal line through the bottom of the unit cell, the length of which we can determine by using the Pythagorean Theorem on another triangle...

The Pythagorean Theorem states that the square of the length of the hypotenuse of a right triangle (c) is equal to the sum of the squares of the lengths of the other two sides (a and b)...

c² = a² + b²

Thus, we can relate the length of the hypotenuse of the second triangle to the lengths of the sides of the unit cell....

d² + d² = 2d²

...and the length of the remaining side of the first triangle is √2d².

Now, we use the Pythagorean Theorem once again to relate the length of one side of the unit cell to the atomic radius (d to r).

(4r)² = d² + (√2d²)²
(4r)² = d² + 2d² = 3d²
16r² = 3d²
d² = 16/3 r²
d = √(16/3 r²)

If r = 1.25, d = 2.89 cm, and the volume of the bcc unit cell can be obtained by cubing this value...

V_{unit cell} = d³ = (2.89)³ = 24.0 cm³

Therefore, the fraction of the volume of the unit cell occupied by atoms is...

V_atoms/V_{unit cell} = 16.4 cm³/24.0 cm³ = 0.683

There are four atoms within the face-centered cubic unit cell, so we can obtain V_atoms by multiplying the sphere volume by four...

V_atoms = 4 x 8.18 cm³ = 32.7 cm³

The calculation of unit cell volume for the fcc unit cell is a little easier than for the bcc unit cell...we can express the length of one side of the unit cell in terms of the atomic radius (r) rather easily. Looking at one "face" of the fcc unit cell shows us that the length of the diagonal of the face is equal to 4r.

Using the Pythagorean Theorem...

(4r)² = d² + d² = 2d²
16r² = 2d²
d = √8r² = r√8

A sphere radius of 1.25 cm gives us a value for d equal to 3.54 cm...cubing this value to obtain unit cell volume gives us...

V_{unit cell} = d³ = (3.54)³ = 44.2 cm³

The fraction of unit cell volume occupied by atoms, therefore, is...

V_atoms/V_{unit cell} = 32.7 cm³/44.2 cm³ = 0.740

If we compare the values obtained for the three unit cells, we see that "packing efficiency" increases in the order simple cubic < body-centered cubic < face-centered cubic.

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This problem asks us to determine the distance in the NaCl unit cell between the center of a Na^+! ion to the center of the nearest Cl^-1 ion, using the density of NaCl (2.17 g/cm³). In order to do this, we must first look at the NaCl unit cell...

Notice that we can express the length of one side of the unit cell (d) in terms of the distance between a Na⁺¹ ion and a Cl^-1 ion (x). Therefore, if we can find out what d is for NaCl, we can find x. In turn, if we know the volume of one unit cell, we can take the cubed root of that value to obtain d. So...how can we get the volume of the unit cell? Well, we know the density of NaCl (2.17 g/cm³), and that can be applied to a solid no matter how small of a sample you have...even if your "sample" is one unit cell. If we have the density, we need the mass of a unit cell to find the volume of that unit cell.

So, what's the mass of one unit cell of NaCl? Remember that there are four Na⁺¹ and four Cl^-1 ions per unit cell. Let's use those atomic masses to start out with, then use Avogadro's number to determine the actual mass of four Na⁺¹ and four Cl^-1 ions.

4 x 35.45 g/mol = 141.8 g Cl/mol of unit cells
4 x 23 g/mol = 92 g Na/mol of unit cells

141.8 + 92 = 234 g/mol of unit cells

If one unit cell has a mass of 3.88 x 10^-22 g, and the density of NaCl is 2.17 g/cm³, then we can find the volume of one unit cell as follows...

density = mass/volume
volume = mass/density
volume = 3.88 x 10^-22 g/2.17 g/cm³ = 1.79 x 10^-22 cm³

Take the cubed root of 1.79 x 10^-22 cm³, and we get 5.63 x 10^-8 cm...this number is d, the length of one side of the unit cell. Divide d by two to get x, which is the distance between a Na⁺¹ and a Cl^-1 in the unit cell.

5.63 x 10^-8 cm/2 = 2.81 x 10^-8 cm