Question

Solid lithium has a body-centered cubic unit cell with the length of the edge of \(351 \mathrm{pm}\) at \(20{ }^{\circ} \mathrm{C}\). Calculate the density of lithium at this temperature.

Short answer

The density of lithium at \(20^{ ext{o}} ext{C}\) is calculated using its unit cell properties in a BCC structure.

Step-by-step solution

Understand the Structure

Lithium has a body-centered cubic (BCC) structure. This means each unit cell contains two lithium atoms: one at the center and parts of eight atoms at the corners shared between eight unit cells.

Edge Length of Unit Cell

The given edge length of the unit cell is \(351 \text{ pm}\). Convert it to centimeters for ease of density calculation. Since \(1 \text{ pm} = 10^{-12} \text{ m}\) and \(1 \text{ cm} = 10^{-2} \text{ m}\), the edge length is \(351 \times 10^{-10} \text{ cm}\).

Calculate Volume of the Unit Cell

The volume \(V\) of the cubic unit cell can be calculated using \(V = a^3\), where \(a\) is the edge length. So, \(V = (351 \times 10^{-10} \text{ cm})^3\).

Calculate Mass of Lithium Atoms in Unit Cell

The atomic mass of lithium is approximately \(6.94 \text{ g/mol}\). Use Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mol}\) to find the mass of one lithium atom: \(\frac{6.94}{6.022 \times 10^{23}} \text{ g/atom}\). Since there are two atoms per unit cell in BCC, the mass of the unit cell is \(2 \times \frac{6.94}{6.022 \times 10^{23}} \text{ g}\).

Calculate Density of Lithium

Density \(\rho\) is mass divided by volume. Using the mass of lithium atoms per unit cell and the volume of the unit cell, calculate the density as: \[ \rho = \frac{\text{mass of lithium atoms in unit cell}}{\text{volume of unit cell}}. \] Substitute the respective values from Steps 3 and 4 to find the density.

Key concepts

Body-centered cubic (BCC) structure

The body-centered cubic (BCC) structure is a common crystalline arrangement found in metals, including lithium. Understanding this structure is crucial for calculating properties like density. In a BCC structure, each unit cell contains two atoms:

• One atom is positioned at the very center of the cube.
• The other atoms are situated at the corners of the cube, shared among adjacent cells.

Although there are eight corners in a cube, each corner atom is shared among eight neighboring unit cells. Thus, each corner contributes only one-eighth of an atom to the central unit cell. Therefore, the total number of atoms per unit cell in a BCC structure is:

• Center atom = 1
• Corner atoms = 8 corners × \( \frac{1}{8} = 1 \)
• Total = 1 (center) + 1 (corners) = 2 atoms per unit cell.

Atomic mass of lithium

The atomic mass of lithium is a fundamental concept for density calculation. It refers to the mass of a single atom of lithium, usually expressed in atomic mass units (amu) or grams per mole (g/mol). For lithium, the atomic mass is approximately 6.94 g/mol. This measure helps convert bulk atomic data to individual atomic characteristics:

• It tells us the amount of lithium in grams when we consider one mole of lithium, which consists of \(6.022 \times 10^{23}\) atoms (Avogadro's number).
• The atomic mass allows us to determine how much a single lithium atom weighs by dividing the atomic mass by Avogadro's number.
These conversions are essential in calculating the mass of lithium atoms within the BCC unit cell and, subsequently, finding the density of lithium.

Avogadro's number

Avogadro's number is a constant that describes the number of constituent particles (usually atoms or molecules) in one mole of a given substance. The value of Avogadro's number is \(6.022 \times 10^{23}\) particles per mole. It plays a critical role in bridging the macroscopic world with the atomic scale:

• It allows chemists to count atoms by weighing macroscopic quantities of material.
• By knowing the atomic mass and Avogadro's number, the mass of a single atom can be determined.
• In the context of the BCC structure and lithium density calculation, Avogadro's number helps compute the mass of the lithium atoms contained in one unit cell.

By using Avogadro's number, we can transition from the number of atoms to macroscopic mass measurements effectively.

Unit cell volume calculation

Calculating the volume of a unit cell is an essential step in determining the density of a crystalline material. For lithium in a body-centered cubic (BCC) arrangement, the edge of the cube (notated as \(a\)) is crucial. The volume \(V\) of a cubic unit is found using the formula:\[ V = a^3 \]where \(a\) is the edge length of the unit cell. In the case of lithium, this edge length is given as 351 pm (picometers):

• To simplify calculations, convert picometers to centimeters: \(1\text{ pm} = 10^{-12}\text{ m} = 10^{-10}\text{ cm}\).
• Thus the edge length \(a = 351 \times 10^{-10}\text{ cm}\).
• The volume of the unit cell becomes \((351 \times 10^{-10})^3\) cm³.

Knowing this volume allows us to further calculate how densely packed the atoms are within the structure, which is critical for finding the substance's density. This concept links the spatial arrangement of atoms to tangible properties like mass and density.