Problem 68 - Chemistry: The Central Science

Question

A particular form of cinnabar (HgS) adopts the zinc blende structure, Figure \(11.42(b)\). The length of the unit cell side is \(5.852 \AA\). (a) Calculate the density of \(\mathrm{HgS}\) in this form. (b) The mineral tiemmanite (HgSe) also forms a solid phase with the zinc blende structure. The length of the unit cell side in this mineral is \(6.085 \AA\). What accounts for the larger unit cell length in tiemmanite? (c) Which of the two substances has the higher density? How do you account for the difference in densities?

Step-by-Step Solution

Answer

In summary, cinnabar (HgS) has a density of 7.70 g/cm³, while tiemannite (HgSe) has a higher density of 8.24 g/cm³. The larger unit cell length in tiemannite is due to the larger covalent radius of Se compared to S. The difference in densities can be attributed to the higher molar mass of HgSe compared to HgS and a comparatively smaller increase in unit cell volume in HgSe as a result of the larger covalent radius of Se compared to S.

Step 1: Determine the number of formula units in a unit cell

In the zinc blende structure, there are 4 formula units per unit cell.

Step 2: Calculate the molar mass of HgS

We'll find the molar mass of HgS by adding the molar masses of Hg and S: Hg: \(200.59 g/mol\) S: \(32.07 g/mol\) Molar mass of HgS = Hg + S = \(200.59 + 32.07 = 232.66 g/mol\)

Step 3: Calculate the volume of HgS unit cell

The length of the unit cell side is given as \(5.852 \AA\). We can calculate the volume of the unit cell by cubing the side length: Volume = \((5.852 \AA)^3 = 200.0 \AA^3\) Convert the volume to cubic centimeters: \(1 \AA = 10^{-8} cm\) Volume = \(200.0 \AA^3 × (10^{-8} cm/\AA)^3 = 2.00 × 10^{-22} cm^3\)

Step 4: Determine the density of HgS

Now, we'll calculate the density of HgS using the following formula: Density = \(\frac{mass}{volume}\) The mass can be determined by multiplying the number of formula units (4) by the molar mass of HgS and dividing by Avogadro's number (6.022 × 10^23): Mass = \(\frac{4 \times 232.66 g/mol}{6.022 × 10^{23} formula units/mol} = 1.54 × 10^{-22} g\) Now calculating the density: Density of HgS = \(\frac{1.54 × 10^{-22} g}{2.00 × 10^{-22} cm^3} = 7.70 g/cm^3\) ## Part (b) Reason for larger unit cell length in Tiemannite ## The larger unit cell length in tiemannite (HgSe) compared to cinnabar (HgS) can be attributed to the larger covalent radius of Se compared to S, resulting in a slightly longer bond distance between Hg and Se atoms in the unit cell of tiemannite. ## Part (c) Comparison of densities ## We now need to follow the same process as in part (a) to find the density of HgSe, and then compare it with the density of HgS.

Step 5: Calculate the molar mass of HgSe

Molar mass of HgSe = Hg + Se = \(200.59 + 78.96 = 279.55 g/mol\)

Step 6: Calculate the volume of HgSe unit cell

Volume of HgSe unit cell = \((6.085 \AA)^3 = 225.7 \AA^3 = 2.26 × 10^{-22} cm^3\)

Step 7: Determine the density of HgSe

Mass of HgSe = \(\frac{4 \times 279.55 g/mol}{6.022 × 10^{23} formula units/mol} = 1.86 × 10^{-22} g\) Density of HgSe = \(\frac{1.86 × 10^{-22} g}{2.26 × 10^{-22} cm^3} = 8.24 g/cm^3\) Now, we can compare the densities: Density of HgS = 7.70 g/cm³ Density of HgSe = 8.24 g/cm³ Tiemannite (HgSe) has a higher density than cinnabar (HgS). The difference in densities can be attributed to the higher molar mass of HgSe compared to HgS and a comparatively smaller increase in unit cell volume in HgSe as a result of the larger covalent radius of Se compared to S.

Key Concepts

Unit Cell Volume

Understanding unit cell volume is essential in materials science, particularly when we are calculating properties such as density. The unit cell volume represents the three-dimensional space occupied by the smallest repeating unit in a crystal structure. In calculations, it's often necessary to cube the length of the unit cell edge to find this volume. For instance, in the exercise involving cinnabar with a zinc blende structure, the side length of the unit cell is given as 5.852 angstroms. To find the volume, we cube this value:

Unit cell volume = Side length³ = (5.852 Å)³ = 200.0 ų

Since calculations commonly require volume in cubic centimeters (cm³), we must convert from angstroms (1 Å = 1 x 10^-8 cm) to cm³. The result is invaluable for subsequent density calculations.

Molar Mass

The molar mass of a compound is another vital concept that plays a significant role in density calculations. Molar mass refers to the mass of one mole of a substance, measured in grams per mole (g/mol). To calculate it, we sum up the molar masses of all individual atoms in the compound. For instance, when solving our density problem for cinnabar (HgS), we combine the molar mass of mercury (Hg, 200.59 g/mol) with that of sulfur (S, 32.07 g/mol), giving us:

Molar mass of HgS = 200.59 g/mol + 32.07 g/mol = 232.66 g/mol

Determining the molar mass allows us to convert between the number of particles in a substance and its mass, which we'll need for finding the mass present in one unit cell.

Avogadro's Number

Avogadro's number, a fundamental constant, is the bridge between the atomic scale and the macroscopic world. Represented by 6.022 x 10²³, it defines the number of atoms, ions, or molecules in one mole of a substance. In the context of density calculations for crystalline materials, we use Avogadro's number to convert from moles to formula units. For example, given that there are four formula units per unit cell of cinnabar, we need to determine the mass of one unit cell by multiplying the four formula units by the molar mass of HgS and then dividing by Avogadro's number:

Mass = (4 x 232.66 g/mol) / (6.022 x 10²³ formula units/mol)

This yields the mass of one unit cell, which we'll need when we calculate the density.

Covalent Radius

The covalent radius measures the size of an atom that forms part of a single covalent bond—in other words, half the distance between two atoms bonded together. In crystal structures like the zinc blende structure, this radius affects the distances between atoms and hence the overall size of the unit cell. A larger covalent radius will often result in a larger unit cell, as observed in tiemannite (HgSe) compared to cinnabar (HgS). This is because selenium (Se) has a larger covalent radius than sulfur (S), leading to a larger length of the unit cell side in tiemannite:

Length of HgSe unit cell: 6.085 Å
Length of HgS unit cell: 5.852 Å

This illustrates how atomic size differences in similarly structured crystals can influence their macroscopic physical properties.

Zinc Blende Structure

The zinc blende structure, a common motif for various compounds, is critical in understanding the problem at hand. It is a type of cubic crystal structure where each atom is tetrahedrally coordinated with its neighbors. When calculating the density of materials like cinnabar and tiemannite that crystallize in this structure, one key piece of information is the number of formula units present within a unit cell. For the zinc blende structure, this number is four. Knowledge of the structure allows us to better comprehend why materials behave differently, such as why tiemannite has a larger unit cell and higher density in comparison to cinnabar. Studying material structures and their implications on physical properties is fundamental in materials science and engineering.