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Chapter 12: Problem 61

A particular form of cinnabar (HgS) adopts the zinc blende structure. The length of the unit cell edge is 5.852 A. (a) Calculate the density of HgS in this form. (b) The mineral tiemannite (HgSe) also forms a solid phase with the zinc blende structure. The length of the unit cell edge in this mineral is 6.085 A. 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

Expert verified
The density of cinnabar (HgS) is 7.72 g/cm³. Tiemannite (HgSe) has a larger unit cell length due to the larger size of the selenium atom compared to sulfur in HgS. HgSe has a higher density of 8.23 g/cm³, which can be attributed to the higher atomic weight of selenium compared to sulfur, resulting in a greater mass for the unit cell.

Step by step solution

01

Calculate the molecular weight of HgS and HgSe

First, let's calculate the molecular weight of the substances HgS (cinnabar) and HgSe (tiemannite). This will be useful for calculating their densities. For HgS: Molecular weight (HgS) = Atomic weight of Hg + Atomic weight of S = 200.59 g/mol (Hg) + 32.07 g/mol (S) = 232.66 g/mol (HgS) For HgSe: Molecular weight (HgSe) = Atomic weight of Hg + Atomic weight of Se = 200.59 g/mol (Hg) + 78.96 g/mol (Se) = 279.55 g/mol (HgSe)
02

Calculate moles per unit cell

In the zinc blende structure, there are 4 formula units of the compound per unit cell. Thus, the moles per unit cell for both HgS and HgSe are: Moles per unit cell = 4 × (1 formula unit / Avogadro's number) = 4 × (1 / (6.022 x 10^23)) = 6.644 x 10^(-24) moles
03

Calculate the volume of the unit cell

Now, we will calculate the volume of the unit cell for both compounds, using the unit cell length provided in the exercise. For HgS: a (unit cell edge) = 5.852 Å Convert Å to cm: 1 Å = 10^(-8) cm 5.852 Å = 5.852 x 10^(-8) cm Volume (HgS) = (5.852 × 10^(-8) cm)^3 = 2.002 x 10^(-22) cm^3 For HgSe: a (unit cell edge) = 6.085 Å 6.085 Å = 6.085 x 10^(-8) cm Volume (HgSe) = (6.085 × 10^(-8) cm)^3 = 2.257 x 10^(-22) cm^3
04

Calculate the densities of HgS and HgSe

Now, we will calculate the densities of both compounds using the molecular weight, moles per unit cell, and volume of the unit cell. Density = (mass of unit cell) / (volume of unit cell) Mass of unit cell = moles per unit cell × molecular weight For HgS: mass (HgS) = (6.644 x 10^(-24) moles) × (232.66 g/mol) = 1.545 x 10^(-21) g Density (HgS) = (1.545 x 10^(-21) g) / (2.002 x 10^(-22) cm^3) = 7.72 g/cm^3 For HgSe: mass (HgSe) = (6.644 x 10^(-24) moles) × (279.55 g/mol) = 1.857 x 10^(-21) g Density (HgSe) = (1.857 x 10^(-21) g) / (2.257 x 10^(-22) cm^3) = 8.23 g/cm^3
05

Analyzing the results

a) The density of HgS is calculated to be 7.72 g/cm^3. b) Tiemannite has a larger unit cell length than cinnabar because the selenium atom in HgSe is larger than the sulfur atom in HgS. This causes an increase in the unit cell edge length to accommodate the larger atom in the zinc blende structure. c) The density of HgSe was calculated to be 8.23 g/cm^3, which is higher than the density of HgS. The difference in density can be accounted for by the fact that selenium has a higher atomic weight than sulfur (78.96 g/mol for Se versus 32.07 g/mol for S). As a result, the unit cell of HgSe has more mass, leading to a higher density.

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Most popular questions from this chapter

Clausthalite is a mineral composed of lead selenide (PbSe). The mineral adopts the rock salt structure. The density of PbSe at \(25^{\circ} \mathrm{C}\) is 8.27 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the length of an edge of the PbSe unit cell.

You are given a white substance that melts at \(100^{\circ} \mathrm{C}\) . The substance is soluble in water. Neither the solid nor the solution is a conductor of electricity. Which type of solid (molecular, metallic, covalent- network, or ionic) might this substance be?

Which of these statements about alloys and intermetallic compounds is false? (a) Bronze is an example of an alloy. (b) "Alloy" is just another word for "a chemical compound of fixed composition that is made of two or more metals." (c) Intermetallics are compounds of two or more metals that have a definite composition and are not considered alloys. (d) If you mix two metals together and, at the atomic level, they separate into two or more different compositional phases, you have created a heterogeneous alloy.(e) Alloys can be formed even if the atoms that comprise them are rather different in size.

Covalent bonding occurs in both molecular and covalent network solids. Which of the following statements best explains why these two kinds of solids differ so greatly in their hardness and melting points? $$ \begin{array}{l}{\text { (a) The molecules in molecular solids have stronger covalent bonding than covalent-network solids do. }} \\ {\text { (b) The molecules in molecular solids are held together by weak intermolecular interactions. }}\end{array} $$ $$ \begin{array}{l}{\text { (c) The atoms in covalent-network solids are more polarizable than those in molecular solids. }} \\ {\text { (d) Molecular solids are denser than covalent-network solids. }}\end{array} $$

For each of the following pairs of semiconductors, which one will have the larger band gap: (a) CdS or CdTe, (b) GaN or InP, ( c) GaAs or InAs?
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