(b) X-ray diffraction studies of buckminsterfullerene show that it has a face-centered cubic lattice of C 60 molecules. The length of an edge of the unit cell is 14.2 Å. Calculate the density of buckminsterfullerene.

Hey everyone, we're told when lent crystallizes, it adopts a face centered cubic crystal structure. The edge length of the unit cell is 495 PICO meters, calculate the radius in PICO meters and density and grams per cubic centimeters of a lead adam. So for a face centered cubic cell, our edge length is equal to two times the square root of two times our radius with edge length being denoted by a. We were told that our edge length was 495 m. So this is going to be equal to two times the square root of two times our radius. Now, when we divide both sides by two times the square root of two, we get a radius of 175 PICO meters and this is going to be one of our answers. Now let's go ahead and calculate for the density, we know that density is equal to our mass divided by our volume and this is going to be in grams per cubic centimeters. Now we know for the volume of our unit cell. This is equal to our edge length to the cubic power. But since this needs to be in centimeters, let's first go ahead and take our edge length And changes into cm. So taking our edge length of 495 km, we're going to use our dimensional analysis. Now we know that per one PICO meter, we have 10 to the negative 12 m and we know that her one centimeter, we have 10 to the negative two m. Now when we calculate this out and cancel out our units, we end up with an edge length of 4.95 times 10 to the negative eight centimeters. Now that we've converted this into cm. Let's go ahead and determine our volume. Taking that value of 4.95 times 10 to the negative eight cm, we can go ahead and cube this. This will get us a volume of 1.2129 times 10 to the negative 22 cubic centimeters. Now, let's go ahead and determine our mass in order to get our density. So to calculate our mass, We know that the molar mass of lead is 207.2 g per one mole. And we can easily find this in our periodic table. Now we know that per one mole we have 6. times 10 to the 23rd atoms, which is avocados number. And since we do have a face centered cubic cell, this is going to be four atoms per one unit cell. So when we calculate this out and cancel out all of our units, We end up with a value of one 0.3763 times 10 to the negative 21st grams per unit cell. Now that we've determined our mass, we can go ahead and determine our density which is mass divided by volume. So taking the value of 1.3763 times 10 to the negative 21st grams and dividing it by the volume that we calculated, which was 1.2129 times 10 to the negative cubic centimeters. And when we calculate this out, we end up with a density of 11.3 g per cubic centimeter, which is going to be our other final answer. Now, I hope that made sense and let us know if you have any questions.

Ch.12 - Solids and Modern Materials

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Brown 14th Edition

Ch.12 - Solids and Modern Materials

Problem 122b

Chapter 12, Problem 122b

(b) X-ray diffraction studies of buckminsterfullerene show that it has a face-centered cubic lattice of C₆₀ molecules. The length of an edge of the unit cell is 14.2 Å. Calculate the density of buckminsterfullerene.

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Understand the structure: Buckminsterfullerene (C60) forms a face-centered cubic (FCC) lattice. In an FCC lattice, there are 4 molecules per unit cell.

Calculate the molar mass of C60: The molar mass of carbon (C) is approximately 12.01 g/mol. Therefore, the molar mass of C60 is 60 times the molar mass of carbon, which is 60 * 12.01 g/mol.

Convert the unit cell edge length from Ångströms to centimeters: 1 Å = 1 x 10^-8 cm, so 14.2 Å = 14.2 x 10^-8 cm.

Calculate the volume of the unit cell: Use the formula for the volume of a cube, V = a^3, where 'a' is the edge length in cm.

Calculate the density: Use the formula for density, ρ = mass/volume. First, find the mass of 4 C60 molecules using the molar mass and Avogadro's number, then divide by the volume of the unit cell.

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Textbook Question

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation 1l2, the angle at which the radiation is diffracted 1u2, and the distance between planes of atoms in the crystal that cause the diffraction (d) is given by nl = 2d sin u. X rays from a copper X-ray tube that have a wavelength of 1.54 Å are diffracted at an angle of 14.22 degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming n = 1 (first-order diffraction).

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Textbook Question

Germanium has the same structure as silicon, but the unit cell size is different because Ge and Si atoms are not the same size. If you were to repeat the experiment described in the previous problem but replace the Si crystal with a Ge crystal, would you expect the X rays to be diffracted at a larger or smaller angle u?

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Textbook Question

(a) The density of diamond is 3.5 g/cm³, and that of graphite is 2.3 g/cm³. Based on the structure of buckminsterfullerene, what would you expect its density to be relative to these other forms of carbon?

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Open Question

When you shine light of band gap energy or higher on a semiconductor and promote electrons from the valence band to the conduction band, do you expect the conductivity of the semiconductor to (a) remain unchanged, (b) increase, or (c) decrease?

Textbook Question

The karat scale used to describe gold alloys is based on mass percentages. (a) If an alloy is formed that is 50 mol% silver and 50 mol% gold, what is the karat number of the alloy? Use Figure 12.18 to estimate the color of this alloy.

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Textbook Question

The karat scale used to describe gold alloys is based on mass percentages. (b) If an alloy is formed that is 50 mol% copper and 50 mol% gold, what is the karat number of the alloy? What is the color of this alloy?

721

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(b) X-ray diffraction studies of buckminsterfullerene show that it has a face-centered cubic lattice of C60 molecules. The length of an edge of the unit cell is 14.2 Å. Calculate the density of buckminsterfullerene.

Verified Solution

(b) X-ray diffraction studies of buckminsterfullerene show that it has a face-centered cubic lattice of C₆₀ molecules. The length of an edge of the unit cell is 14.2 Å. Calculate the density of buckminsterfullerene.