Question

A metallic solid with atoms in a face-centered cubic unit cell with an edge length of 392 \(\mathrm{pm}\) has a density of 21.45 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the atomic mass and the atomic radius of the metal. Identify the metal.

Short answer

The metallic solid in the face-centered cubic unit cell has an atomic mass of 195.8 g/mol and an atomic radius of 137.9 pm, indicating that it is likely Platinum (Pt).

Step-by-step solution

Finding the number of atoms per unit cell

In an fcc structure, there are 8 corner atoms shared by 8 neighboring unit cells, and 6 face atoms shared by 2 neighboring unit cells. Therefore, the number of atoms per unit cell in an fcc structure can be calculated as: Number of atoms per unit cell = (8 corner atoms ÷ 8) + (6 face atoms ÷ 2) = 1 + 3 = 4

Finding the volume of the unit cell

We are given that the edge length of the unit cell is 392 pm. To find the volume, take the cube of the edge length and convert from pm³ to cm³: Volume of unit cell = (392 pm)³ * (1 cm / 10^10 pm)³ = 6.027 * 10^(-23) cm³

Calculating the molar mass

We are given the density (ρ) of the metal as 21.45 g/cm³. We can now calculate the molar mass by using the formula: ρ = (mass of unit cell) × (Avogadro's number) / (volume of unit cell × molar mass) Rearranging to solve for molar mass, we have: Molar mass = (mass of unit cell) × (Avogadro's number) / (ρ × volume of unit cell) The mass of the unit cell can be calculated as follows: mass of unit cell = (number of atoms per unit cell) × (atomic mass) Now we can substitute this formula back into the molar mass formula: Molar mass = ((number of atoms per unit cell) × (atomic mass)) × (Avogadro's number) / (ρ × volume of unit cell) Plugging in the values we know: Molar mass = (4 × atomic mass) × (6.022 * 10^23 atoms/mol) / (21.45 g/cm³ * 6.027 * 10^(-23) cm³) From this equation, we can calculate the atomic mass: Atomic mass = 195.8 g/mol

Calculating the atomic radius

We know the atomic radius (r) of a face-centered cubic unit cell can be determined using the edge length (a) and the following formula: 4r = √(2) × a Rearrange the formula to solve for r and plug in the known values: Atomic radius (r) = (edge length / √(2)) / 4 r = (392 pm / √2) / 4 = 137.9 pm

Identifying the metal

Based on the calculated atomic mass of 195.8 g/mol and atomic radius of 137.9 pm, we can identify the metal as Platinum (Pt), which has an atomic mass of approximately 195 g/mol. In conclusion, the metallic solid has an atomic mass of 195.8 g/mol and an atomic radius of 137.9 pm, indicating that it is likely Platinum.

Key concepts

Atomic Mass Calculation

The process of determining the atomic mass of a metal in a face-centered cubic (fcc) structure involves a few crucial steps. First, it's important to recognize that in an fcc structure, there are four atoms per unit cell. This comes from 8 corner atoms each contributing 1/8 of an atom and 6 face atoms each contributing half an atom.

To calculate the atomic mass, we use the formula for density (\( \rho \)), which is defined as mass per unit volume. In formula terms, \( \rho = \frac{(\text{mass of unit cell}) \times (\text{Avogadro's number})}{\text{volume of unit cell} \times \text{molar mass}} \). By rearranging the formula, we can solve for the molar mass of the atomic structure:

• Molar mass = \( \frac{(\text{number of atoms per unit cell}) \times \text{atomic mass}}{\text{volume of unit cell} \times \rho} \times \text{Avogadro's number} \)

Once we have all the necessary values such as density, volume of the unit cell, and Avogadro's number, we can plug them into the formula to find the atomic mass, which in this case is 195.8 \( \text{g/mol} \).

Atomic Radius Determination

Determining the atomic radius in an fcc unit cell requires specific formulas that take into account the unit cell’s geometry. The formula used to find the atomic radius (r) for a face-centered cubic structure is:

• \( 4r = \sqrt{2} \times a \)

Here, \( a \) is the edge length of the unit cell. This formula results from the arrangement of atoms such that atoms on diagonal lines form a triangle inside the cube. Solving for the atomic radius requires rearranging this formula and substituting the given edge length \( a = 392 \, \text{pm} \).

Thus, the calculation is:

• Atomic radius \( r = \frac{a}{\sqrt{2} \times 4} \)
• \( r = \frac{392 \, \text{pm}}{\sqrt{2} \times 4} \approx 137.9 \, \text{pm} \)

These calculations tell us how much space one atom occupies along the edge of the unit cell.

Metal Identification

Identifying a metal based on calculated properties like atomic mass and radius is fascinating. In this exercise, once the atomic mass was determined to be approximately 195.8 \( \text{g/mol} \), and the atomic radius was determined to be about 137.9 \( \text{pm} \), we looked for metals with close matching characteristics.

In this particular case, Platinum (Pt) has an atomic mass of roughly 195 \( \text{g/mol} \) and fits within the calculated parameters. It’s vital to compare the calculated properties against known data from a periodic table to make an accurate identification.

This process showcases how physical properties and mathematical calculations work together to uncover the identity of an unknown metal. Using these calculated parameters is a common method in material sciences to identify unknown samples.

Density Calculation

Density plays a central role in determining the atomic mass and other structural details of metals. It tells us how much mass is contained in a given volume — an important clue when examining crystalline structures like fcc.

The density \( \rho \) is given by:

• \( \rho = \frac{\text{mass}}{\text{volume}} \)

In the context of crystal lattices, we need to consider the mass of the unit cell as a product of the number of atoms and their individual masses. Using the calculated volume and the given density, we can isolate unknowns like atomic mass from this direct application of the density formula:

• Use the equation: \( \rho = \frac{(\text{number of atoms per unit cell}) \times (\text{atomic mass})}{\text{volume of unit cell}} \)

This simplifies the process of solving for particular properties. Understanding how to manipulate these equations allows us to extract detailed physical information from observed data and is crucial for practical applications in science and engineering.