Question

Alabandite is a mineral composed of manganese(II) sulfide (MnS). The mineral adopts the rock salt structure. The length of an edge of the MnS unit cell is 5.223 A at \(25^{\circ} \mathrm{C}\) . Determine the density of Mns in \(\mathrm{g} / \mathrm{cm}^{3} .\)

Short answer

The density of MnS in the rock salt structure is 4.06 g/\(cm^3\).

Step-by-step solution

Calculate the number of moles of MnS in one unit cell

Since there are 4 formula units of MnS per unit cell in a rock salt structure, we have: Number of moles of MnS per unit cell = 4 formula units × (1 mole/6.022 × 10^23 formula units) = \(6.650 × 10^{-24}\) moles.

Calculate the mass of MnS in one unit cell

To calculate the mass of MnS, we need to find the molar mass and then multiply by the number of moles in one unit cell. Molar mass of MnS = Molar mass of Mn + Molar mass of S = (54.94 + 32.07) g/mol = 87.01 g/mol Mass of MnS per unit cell = Number of moles of MnS × Molar mass of MnS = \(6.650 × 10^{-24}\) moles × 87.01 g/mol = \(5.784 × 10^{-22}\) g.

Calculate the volume of the unit cell

The given length of an edge of the MnS unit cell is 5.223 A (Angstroms). First, we need to convert it to cm: 1 A = \(1 × 10^{-8}\) cm Length of an edge in cm = 5.223 A × (\(1 × 10^{-8}\) cm/A) = \(5.223 × 10^{-8}\) cm Now, we can calculate the volume of the cubic unit cell: Volume of unit cell = (Edge length)³ = \((5.223 × 10^{-8} \mathrm{cm})^3\) = \(1.424 × 10^{-22}\) \(cm^3\).

Calculate the density of MnS

We can now calculate the density of MnS by dividing the mass of MnS per unit cell by the volume of the unit cell: Density of MnS = Mass of MnS / Volume of unit cell = \(\frac{5.784 × 10^{-22} \mathrm{g}}{1.424 × 10^{-22} cm^3}\) = 4.06 g/\(cm^3\). Thus, the density of MnS is 4.06 g/\(cm^3\).

Key concepts

Rock Salt Structure

The rock salt structure is a common type of ionic crystalline structure. It is named after the mineral rock salt, or sodium chloride (NaCl), which has the same arrangement of atoms. This structure is essential because it dictates how ions are packed in a crystal and, consequently, affects the material's properties.
In the rock salt structure, each ion (e.g., Mn in MnS) is surrounded by six ions of opposite charge (e.g., S in MnS) in an octahedral configuration. This means that the manganese ions are enclosed by sulfur ions, and vice versa. The cube's face centers and corners contain these ions, forming a tightly packed lattice.
Some key features of this structure include:

• Coordination number of 6 for each ion, indicating that each ion is in contact with six oppositely charged neighboring ions.
• Four formula units per unit cell, which means four pairs of Mn and S ions are present in one unit cell.
• Cubic symmetry, leading to uniform distribution of ions in the crystal lattice.

Understanding the rock salt structure helps explain why certain materials have particular densities and mechanical properties.

Molecular Formula

A molecular formula indicates the types and number of atoms in a molecule, helping to identify the compound and predict its behavior. For Alabandite, the molecular formula is MnS. This tells us that each molecule consists of one manganese (Mn) atom and one sulfur (S) atom.
Knowing the molecular formula allows us to calculate the molar mass, which is crucial in determining the properties of substances in unit cells. For example, the molar mass of MnS can be found by adding the atomic masses of manganese (54.94 g/mol) and sulfur (32.07 g/mol), resulting in 87.01 g/mol.
By multiplying the molar mass by the number of moles present in one unit cell (which is calculated using Avogadro's number), we can determine the mass of the compound within the cell. This mass is necessary for calculating the density of mineral structures, which is an essential property in geology and material science.

Unit Cell Volume

The volume of a unit cell is a fundamental aspect of crystal structure analysis. It represents the three-dimensional space occupied by the repeating unit of a crystalline solid. The unit cell's dimensions influence physical properties such as density and stability.
In a cubic structure like the rock salt structure, the unit cell's volume can be calculated by cubing the length of one edge. If we take the length of an edge as 5.223 Å and convert it to centimeters (1 Å = 1 × 10^-8 cm), the calculation proceeds as follows:
\(5.223 Å = 5.223 × 10^{-8} cm\)
Volume = \((5.223 × 10^{-8} cm)^3 = 1.424 × 10^{-22} cm^3\)

Understanding how to calculate the unit cell volume is critical for determining the density of a mineral. Density is derived by dividing the mass of the unit cell by its volume. Therefore, accurate measurement of the unit cell size directly impacts material density calculations in scientific fields like mineralogy and materials engineering.