Problem 56
Question
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.
Step-by-Step Solution
Verified Answer
The length of an edge of the PbSe unit cell is approximately 4.10 cm.
1Step 1: Determine the formula mass of PbSe
First, determine the individual molar mass of each element according to the periodic table.
Molar Mass of Pb = 207.2 g/mol
Molar Mass of Se = 78.97 g/mol
Then, calculate the formula mass of PbSe:
Formula Mass = Molar Mass of Pb + Molar Mass of Se
Formula Mass = 207.2 g/mol + 78.97 g/mol = 286.17 g/mol
2Step 2: Determine the number of atoms in the rock salt unit cell
The rock salt structure is characterized by a face-centered cubic (fcc) lattice. In this structure, each lattice point has one other atom associated with it, so there are two atoms per unit cell.
3Step 3: Calculate the volume of the PbSe unit cell
Next, we need to calculate the volume of the unit cell using the provided density:
Density = Mass/Volume
\(8.27 \mathrm{~g/cm^{3}} = (286.17 \mathrm{~g/mol})/(Volume)\)
To find the mass of our PbSe unit cell, we'll determine the number of moles present in one unit cell. Since there are two atoms of PbSe in one unit cell, there are two moles of PbSe:
Mass = (2 moles)(286.17 g/mol) = 572.34 g
Now, we can solve for the volume of the unit cell:
Volume = Mass/Density
Volume = (572.34 g)/\(8.27 \mathrm{~g/cm^{3}}\)
Volume = 69.15 cm³
4Step 4: Calculate the edge length of the PbSe unit cell
We know that the unit cell of PbSe in rock salt structure is cubic. Therefore the relationship between the volume and edge length (a) of the cube is:
Volume = \(a^3\)
Now, we can solve for the edge length:
\(a^3\) = 69.15 cm³
a = \(69.15^{1/3}\)
a = 4.10 cm
The length of an edge of the PbSe unit cell is approximately 4.10 cm.
Key Concepts
Unit Cell CalculationMolar MassDensity of Minerals
Unit Cell Calculation
The concept of a unit cell is central in understanding the crystalline structure of minerals like Clausthalite, which adopts the rock salt structure. A unit cell is the smallest repeating unit in the crystal lattice and provides insight into the mineral's internal arrangement. In the rock salt structure, which looks like a face-centered cubic (fcc) lattice, each unit cell contains two complete formula units. This is because each corner atom is shared among eight unit cells, while face atoms are shared between two, but entirely internal atoms are exclusive to the unit cell.
The calculation involves determining the volume of the unit cell using the crystal's density and the formula mass. Here, to solve for the edge length of a cubic unit cell, we use the equation for volume, where Volume = \(a^3\). Given the calculated volume and solving for \(a\), we find the edge length of the unit cell as \(a = 69.15^{1/3} = 4.10\) cm.
The calculation involves determining the volume of the unit cell using the crystal's density and the formula mass. Here, to solve for the edge length of a cubic unit cell, we use the equation for volume, where Volume = \(a^3\). Given the calculated volume and solving for \(a\), we find the edge length of the unit cell as \(a = 69.15^{1/3} = 4.10\) cm.
Molar Mass
Molar mass is a key component when examining minerals and performing calculations for structures like lead selenide (PbSe), which Clausthalite is made of. Molar mass is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol).
To calculate the molar mass of a compound, you need the atomic masses of its constituent elements, which are found on the periodic table.
\(207.2\, \text{g/mol} + 78.97\, \text{g/mol} = 286.17\, \text{g/mol}\).
This value is crucial for determining the mass used in density and unit cell volume calculations.
To calculate the molar mass of a compound, you need the atomic masses of its constituent elements, which are found on the periodic table.
- For lead (Pb), the atomic mass is 207.2 g/mol.
- For selenium (Se), it is 78.97 g/mol.
\(207.2\, \text{g/mol} + 78.97\, \text{g/mol} = 286.17\, \text{g/mol}\).
This value is crucial for determining the mass used in density and unit cell volume calculations.
Density of Minerals
Density is an important property in studying minerals. It helps relate mass to the volume of the mineral structure and aid in calculating structural dimensions, like the edge of a unit cell.
Density is expressed in grams per cubic centimeter (g/cm³) and defined by the formula: \(\text{Density} = \text{Mass} / \text{Volume}\). For minerals adopting the rock salt structure like PbSe, knowing the density allows us to calculate the volume of the unit cell.
In the given problem, the density of PbSe is 8.27 g/cm³. By using the relationships from formula mass and having found the mass of two moles (from two atoms in the unit cell) of PbSe, we can solve the volume equation for the unit cell.
This calculated volume then enables determining the edge length when considering the cubic nature of the unit cell.
Density is expressed in grams per cubic centimeter (g/cm³) and defined by the formula: \(\text{Density} = \text{Mass} / \text{Volume}\). For minerals adopting the rock salt structure like PbSe, knowing the density allows us to calculate the volume of the unit cell.
In the given problem, the density of PbSe is 8.27 g/cm³. By using the relationships from formula mass and having found the mass of two moles (from two atoms in the unit cell) of PbSe, we can solve the volume equation for the unit cell.
This calculated volume then enables determining the edge length when considering the cubic nature of the unit cell.
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