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Chapter 11: Problem 75

Lithium hydride (LiH) has the sodium chloride structure, and the length of the edge of the unit cell is \(4.086 \times 10^{-8} \mathrm{~cm} .\) Calculate the density of this solid.

Short Answer

Expert verified
0.774 g/cm^3

Step by step solution

01

Identify the Type of Unit Cell

Lithium hydride (LiH) has the sodium chloride structure. The sodium chloride structure is characterized by a face-centered cubic (FCC) unit cell.
02

Determine the Number of Atoms per Unit Cell

In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell. This is because there are 8 atoms at the corners (contributing 1/8 atom each) and 6 atoms on the faces (contributing 1/2 atom each), totaling 4 full atoms.
03

Calculate the Molar Mass

The molar mass of lithium hydride is calculated as the sum of the atomic masses of lithium (Li) and hydrogen (H). - Atomic mass of Li = 6.94 g/mol. - Atomic mass of H = 1.01 g/mol. - Therefore, molar mass of LiH = 6.94 + 1.01 = 7.95 g/mol.
04

Find the Volume of the Unit Cell

The edge length of the unit cell is given as \(4.086 \times 10^{-8} \text{ cm}\). The volume of a cubic unit cell \(V\) is calculated as the cube of the edge length:\[ V = (4.086 \times 10^{-8} \, \text{cm})^3 = 6.819 \times 10^{-23} \, \text{cm}^3 \]
05

Calculate the Density

Density \( \rho \) is given by the formula\[ \rho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \]1. The mass of one mole (Avogadro's number \(6.022 \times 10^{23}\)) of LiH is 7.95 g.2. Therefore, the mass of one unit cell (4 atoms) is \[\frac{7.95}{6.022 \times 10^{23}} \times 4 = 5.278 \times 10^{-23} \text{ g}\]3. The density: \[\rho = \frac{5.278 \times 10^{-23} \text{ g}}{6.819 \times 10^{-23} \text{ cm}^3} = 0.774 \text{ g/cm}^3\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Face-Centered Cubic Structure
A face-centered cubic (FCC) structure is a common arrangement of atoms within a crystalline solid, like lithium hydride. In this type of crystal structure, the unit cell, which is the smallest repeating unit within the crystal lattice, is a cube. This cube has atoms positioned at each of its corners and at the center of each of its faces.
  • This arrangement results in each cell containing a total of 4 atoms.
  • The corner atoms are shared among eight neighboring unit cells, so each corner atom contributes 1/8 of an atom per unit cell.
  • There are also six face-centered atoms, each contributing 1/2 atom to the unit cell, shared between two adjacent cells.
A face-centered cubic unit cell is a highly efficient structure in packing, meaning it utilizes space efficiently and offers stability to the compound.
Molar Mass
Molar mass is an important property when calculating the density of a compound. It is defined as the mass of one mole of a substance (in grams per mole). To find the molar mass of lithium hydride (LiH), you need to add the atomic masses of its constituent elements:
  • The atomic mass of lithium (Li) is 6.94 g/mol.
  • The atomic mass of hydrogen (H) is 1.01 g/mol.
Combine these values to find the molar mass of LiH:\[\text{Molar mass of LiH} = 6.94 + 1.01 = 7.95 \text{ g/mol}\]Knowing the molar mass helps in determining the mass of atoms in a specific number of unit cells, which is crucial for computing the density of the material.
Unit Cell Volume
The unit cell volume is essential for calculating density as it gives you the space one unit cell occupies. For lithium hydride, the unit cell is a cube because it follows the cubic crystal system.To find the volume of the unit cell, cube the edge length, which for LiH is given as:\[4.086 \times 10^{-8} \text{ cm}\]Hence, the volume is computed by:\[V = (4.086 \times 10^{-8} \, \text{cm})^3 = 6.819 \times 10^{-23} \, \text{cm}^3\]This tiny volume is typical and reflects the scale of atomic dimensions in crystal structures, which is pivotal when determining density.
Density Formula
The density of a material measures its mass per unit volume. For solids like lithium hydride, we need to find the mass of the atoms within a single unit cell and divide it by the volume of that cell.The density \( \rho \) is calculated using the formula:\[ ho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}}\]Steps to calculate this for LiH:
  • Calculate the mass of the unit cell by dividing the molar mass by Avogadro鈥檚 number \(6.022 \times 10^{23}\), and then multiply by 4 (since there are 4 formula units of LiH in one FCC unit cell).
  • Once you have the mass, in this case, \(5.278 \times 10^{-23} \text{ g}\), divide by the unit cell volume, \(6.819 \times 10^{-23} \, \text{cm}^3\), to determine density.
Thus, the density of LiH is:\[0.774 \, \text{g/cm}^3\]This result helps in understanding how tightly packed the atoms are within the crystal structure.

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