Problem 40 - Chemistry The Central Science In Si Units

Question

An element crystallizes in a face-centered cubic lattice. The edge of the unit cell is \(0.408 \mathrm{nm}\), and the density of the crystal is \(10.49 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the atomic weight of the element and identify the element.

Step-by-Step Solution

Answer

The atomic weight is approximately 107.88 g/mol; the element is Silver (Ag).

Step 1: Understand the Face-Centered Cubic (FCC) Structure

In FCC structures, there are 4 atoms per unit cell. This is due to 1/8 of an atom at each of the 8 corners and 1/2 of an atom on each of the 6 faces, totaling 4 whole atoms.

Step 2: Convert Unit Cell Dimension

We are given the edge length of the unit cell as 0.408 nm. Convert this length into cm: \[ 0.408 \text{ nm} = 0.408 \times 10^{-7} \text{ cm} \]

Step 3: Calculate Volume of Unit Cell

Calculate the volume of the unit cell using the edge length (in cm): \[ V = a^3 = (0.408 \times 10^{-7} \text{ cm})^3 \approx 6.785 \times 10^{-23} \text{ cm}^3 \]

Step 4: Understand Density and Mass Relationship

Density formula: \[ \rho = \frac{m}{V} \] With given density \( \rho = 10.49 \text{ g/cm}^3 \), multiply \( \rho \) by volume \( V \) to find mass of atoms in the unit cell: \[ m = \rho \times V = 10.49 \text{ g/cm}^3 \times 6.785 \times 10^{-23} \text{ cm}^3 \approx 7.111 \times 10^{-22} \text{ g} \]

Step 5: Calculate Atomic Mass

With 4 atoms per unit cell, the mass of a single atom \(m_a\) is \[ m_a = \frac{m}{4} = \frac{7.111 \times 10^{-22} \text{ g}}{4} \approx 1.778 \times 10^{-22} \text{ g} \]

Step 6: Calculate Molar Mass

Avogadro's number \(N_A = 6.022 \times 10^{23} \text{ atoms/mol}\). Calculate atomic weight as follows: \[ \text{Atomic Weight} = m_a \times N_A = 1.778 \times 10^{-22} \text{ g} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 107.88 \text{ g/mol} \]

Step 7: Identify the Element

The calculated atomic weight is approximately 107.88 g/mol, which corresponds to the atomic weight of Silver (Ag).

Key Concepts

Crystal Density

Understanding crystal density is crucial in determining the properties of crystalline materials. Crystal density refers to how much mass is compacted within a given volume of the crystal structure. In a practical scenario, it helps identify elements or compounds based on their structural arrangement. The formula to calculate density (\( \rho \)) is \( \rho = \frac{m}{V} \), where \( m \) is the mass, and \( V \) is the volume of the unit cell. For a face-centered cubic lattice, knowing the density involves calculating the volume and counting the number of atoms in the unit cell.

• Crystals with a high density typically have tightly packed atoms.
• Density is usually expressed in grams per cubic centimeter (\(\text{g/cm}^3\)).

Bearing in mind the density helps to further calculate other properties like atomic weight and verify the element's identity.

Atomic Weight Calculation

Atomic weight calculation is essential in the context of crystalline lattices, primarily when identifying an unknown element. After determining the mass of the entire unit cell, divide it by the number of atoms in the unit cell for the mass of a single atom. In a face-centered cubic lattice, there are 4 atoms per unit cell. Once you know the mass of one atom, you can extrapolate to find the molar mass using Avogadro's number.
As per the step-by-step solution, find the mass of one atom by dividing the total mass of atoms in the unit cell. These calculated values are cross-referenced with known atomic weights from the periodic table, thus helping identify the element.

• The atomic weight is often given in grams per mole (\(\text{g/mol}\)).
• Atomic weight is a weighted average of all isotopes of an element.
• Understanding atomic weight assists in further studies of chemical behaviors and reactions of the element.

Unit Cell Volume

For crystals with a face-centered cubic structure, calculating the unit cell volume is pivotal. The unit cell is the smallest repeating entity in the crystal, and its volume tells us about the space occupied by atoms. To find it, cube the edge length of the cell. In this exercise, the given edge length is in nanometers, but converting it to centimeters makes calculations consistent with density measurements.

• The unit cell volume for an FCC structure is \( V = a^3 \), where \( a \) is the edge length.
• Ensuring consistent units, especially converting nanometers to centimeters, avoids errors in calculations.
• This volume helps calculate density and further determines properties depending on how compact the structure is.

The thorough understanding of unit cell volume is instrumental in material science and engineering for predicting how materials will behave.

Avogadro's Number

Avogadro's number is a fundamental constant used to bridge atomic scale calculations to macroscopic scales. It represents the number of atoms or molecules in one mole of a substance, precisely \( 6.022 \times 10^{23} \) entities per mole.
This number is essential in converting atomic mass (which is often minuscule) to a more usable molar mass which can be compared to standard atomic weights. In the context of the face-centered cubic lattice, after calculating the mass of a single atom, you multiply by Avogadro's number to get the molar mass, which helps in identifying the element itself.

• Avogadro's number allows chemists to count particles by weighing them.
• Used in chemical equations, reactions, and stoichiometry.
• Helps in connecting physical observations (mass) with chemical properties (molecular interactions).

Without Avogadro's number, converting between microscopic and macroscopic scales in chemistry would be challenging.