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?

Short answer

The density of cinnabar (HgS) is 7.721 g/cm³, and the density of tiemmanite (HgSe) is 8.217 g/cm³. The unit cell side length in tiemmanite is larger due to the larger selenium atom compared to sulfur, but the higher atomic mass of Se in HgSe compared to HgS contributes to a higher density for HgSe despite having a larger unit cell.

Step-by-step solution

Calculate the density of cinnabar (HgS)

The density of a compound is given by the formula: Density = \( \frac{mass}{volume} \) To find the mass, we need to calculate the molar mass of HgS: Molar mass of HgS = Molar mass of Hg + Molar mass of S = 200.59 g/mol + 32.07 g/mol = 232.66 g/mol We know that the unit cell side length (a) is 5.852 Å; let's convert that to meters: a = 5.852 Å * \(1 \times 10^{-10}\) m/Å = 5.852 × 10⁻¹⁰ m Volume of the unit cell is \(a^3\): Volume = \((5.852 \times 10^{-10})^3\) m³ = 2.00419 × 10⁻²⁹ m³ Now, we need to calculate how many HgS molecules are in the unit cell. The zinc blende structure has two formula units per unit cell. Therefore, the mass of the unit cell is: Mass = 2 ( number of molecules in unit cell) × Molar mass of HgS / Avogadro's number Mass = 2 × 232.66 g/mol / 6.022 × 10²³ mol⁻¹ = 7.725 × 10⁻²² g Now, we calculate the density of HgS: Density = \( \frac{mass}{volume} \) = \( \frac{7.725 \times 10^{-22} g}{2.00419 \times 10^{-29} m^3} \) = 7.721 g/cm³

Discuss the larger unit cell length in tiemmanite (HgSe)

The unit cell side length in tiemannite (HgSe) is 6.085 Å, which is larger than the unit cell side length in cinnabar (HgS). This is because the selenium (Se) atom is larger than the sulfur (S) atom. As a result, the unit cell length has to be greater for HgSe than for HgS.

Compare densities and discuss differences

We have already calculated the density of HgS, now let's calculate the density of HgSe. To find the mass, we need to calculate the molar mass of HgSe: Molar mass of HgSe = Molar mass of Hg + Molar mass of Se = 200.59 g/mol + 78.96 g/mol = 279.55 g/mol We know that the unit cell side length (a) is 6.085 Å; let's convert that to meters: a = 6.085 Å * \(1 \times 10^{-10}\) m/Å = 6.085 × 10⁻¹⁰ m Volume of the unit cell is \(a^3\): Volume = \((6.085 \times 10^{-10})^3\) m³ = 2.25888 × 10⁻²⁹ m³ Now, we calculate how many HgSe molecules are in the unit cell. The zinc blende structure has two formula units per unit cell. Therefore, the mass of the unit cell is: Mass = 2 ( number of molecules in unit cell) × Molar mass of HgSe / Avogadro's number Mass = 2 × 279.55 g/mol / 6.022 × 10²³ mol⁻¹ = 9.286 × 10⁻²² g Now, we calculate the density of HgSe: Density = \( \frac{mass}{volume} \) = \( \frac{9.286 \times 10^{-22} g}{2.25888 \times 10^{-29} m^3} \) = 8.217 g/cm³ The density of HgSe (8.217 g/cm³) is higher than the density of HgS (7.721 g/cm³). This difference in densities is due to the larger mass of HgSe molecules compared to HgS molecules because of the difference in the mass of S and Se atoms. The higher atomic mass of Se contributes to the higher overall molar mass of HgSe and, therefore, results in a higher density despite having a larger unit cell.

Key concepts

Unit Cell Geometry

Understanding the geometry of the unit cell is crucial when it comes to exploring crystalline structures like zinc blende. The zinc blende structure has a cubic unit cell, meaning all its edges are of equal length, forming a perfect cube. In this type of lattice, the atoms are positioned at each corner of the cube, with an additional atom at the center of the face on each cube side, characteristic of a face-centered cubic (fcc) arrangement.

Now, when we talk about unit cells, we are referring to the smallest volume that, when repeated across space, builds up the entire crystal structure of a material. In the exercise, the side length (or edge length) of the unit cell is a critical factor in determining various properties of the crystal, including its density. For instance, in the given example, cinnabar (HgS) has a unit cell edge length of 5.852 Å, and so we calculate its volume by raising this length to the third power (cubing it), because for a cube, the volume is the edge length cubed.

Molar Mass Determination

Molar mass serves as a bridge between the mass of a substance and the number of atoms or molecules present. The molar mass is essentially the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To determine this in practice, you sum the atomic masses of all the atoms within a molecule. These atomic masses can be found on the periodic table.

For compounds like HgS and HgSe in the exercise, we add the molar masses of mercury (Hg) and sulfur (S) or selenium (Se), respectively. This provides us with the total molar mass of HgS and HgSe, which is an essential step for calculating the mass of a mole of these substances. Once we have the molar mass, we can figure out the mass of one unit cell by considering how many formula units are in one unit cell of the compound and then using Avogadro's number to find out how many grams this corresponds to.

Avogadro's Number Application

Avogadro's number (\(6.022 \times 10^{23}\)) is a fundamental constant in chemistry that represents the number of entities (usually atoms or molecules) in one mole of any substance. It bridges the microscopic world of atoms and molecules with the macroscopic world we measure in the lab. In the context of our density calculation, we use Avogadro's number to convert between the number of unit cells and the mass of atoms present.

By dividing the molar mass of the substance by Avogadro's number, we obtain the mass of a single formula unit. In the case of the zinc blende structure, with two atoms per unit cell, this calculation helps us to find the total mass present in one unit cell. With that and the volume of the unit cell, we can then calculate the density of the substance. It is a remarkable and powerful number that allows us to scale from individual atoms all the way up to macroscopic amounts of material.