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Chapter 12: Problem 59

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

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

Step by step solution

01

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.
02

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.
03

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 = \((5.223 × 10^{-8} \mathrm{cm})^3\) = \(1.424 × 10^{-22}\) \(cm^3\).
04

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/\(cm^3\).

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