Problem 120 - Chemistry: An Atoms-Focused Approach

Question

An element crystallizes in the cubic closest-packed structure. The length of an edge of the unit cell is \(408 \mathrm{pm}\). The density of the element is \(19.27 \mathrm{g} / \mathrm{cm}^{3} .\) Identify the element.

Step-by-Step Solution

Answer

Answer: Gold (Au)

Step 1: Calculate the unit cell volume

The length of an edge of the unit cell is given as \(408 \mathrm{pm}\). First, we need to convert this length to cm, so we can work with the given density unit (\(\mathrm{g} / \mathrm{cm}^{3}\)). \(1 \mathrm{pm} = 10^{-10} \mathrm{cm}\), so \(408 \mathrm{pm} = 408 \times 10^{-10} \mathrm{cm}\). Now, calculate the unit cell volume by cubing the edge length: \(V = (4.08\times10^{-8})^3 =4.288\times10^{-23} \mathrm{cm}^{3}\).

Step 2: Determine the number of atoms in the cubic closest-packed unit cell

A cubic closest-packed (ccp) structure has a coordination number of 12, which means that each lattice point (atom) has 12 nearest neighbors. In a ccp unit cell, there are 8 corner atoms with \(\frac{1}{8}\) of their volume inside the unit cell, and 6 face-centered atoms with \(\frac{1}{2}\) of their volume inside the unit cell. Therefore, the total number of atoms in the unit cell is: \(n = 8\left(\frac{1}{8}\right) + 6\left(\frac{1}{2}\right) = 4\).

Step 3: Calculate the mass of the unit cell

Using the given density of the element, \(\rho = 19.27 \mathrm{g} / \mathrm{cm}^{3}\), we can calculate the mass (m) of the unit cell by multiplying the density by the unit cell volume: \(m = \rho \times V = 19.27 \times 4.288\times10^{-23} = 8.259\times10^{-22} \mathrm{g}\).

Step 4: Calculate the atomic mass of the element

Divide the mass of the unit cell by the number of atoms in the unit cell to find the atomic mass: \(M = \frac{m}{n} = \frac{8.259\times10^{-22}}{4} = 2.065\times10^{-22} \mathrm{g}\). To convert this value to atomic mass units (u), divide by the conversion factor: \(1 \mathrm{u} = 1.66054\times10^{-24} \mathrm{g}\). So, the atomic mass of the element is \(M = \frac{2.065\times10^{-22} \mathrm{g}} {1.66054\times10^{-24} \mathrm{g/u}} =197.11 ~ \mathrm{u}\).

Step 5: Identify the element

The atomic mass we calculated, \(197.11 ~ \mathrm{u}\), is close to the atomic mass of gold (Au) \(196.97 ~ \mathrm{u}\). Therefore, the element crystallizing in the cubic closest-packed structure is gold (Au).

Key Concepts

Unit Cell Volume

The unit cell volume is an essential parameter in crystallography as it helps determine the packing and properties of a crystalline structure. For a cubic system, the unit cell is a perfect cube, and its volume is found by cubing the edge length. In the given exercise, the element has a cubic closest-packed structure with an edge length of 408 picometers (pm). First, we convert this measurement from pm to centimeters (cm), as scientific calculations usually require standard units.

• Conversion: 1 pm = \(10^{-10}\) cm, so \(408 \text{ pm} = 408 \times 10^{-10} \text{ cm}\).
• Volume Calculation: Using the formula \(V = a^3\), where \(a\) is the edge length, we obtain the volume as \(V = (4.08\times10^{-8})^3 = 4.288\times10^{-23} \text{ cm}^3\).

Thus, knowing the unit cell volume assists in further calculations like determining the mass and density of the material. Understanding unit cell volume is vital in identifying elements based on crystalline structure.

Density of Element

The density of an element is a key property that describes how much mass is contained within a specific volume. It is usually expressed in grams per cubic centimeter (g/cm³). In crystallography, density helps identify unknown materials based on their structural parameters. For the given element with a cubic closest-packed structure, we use the calculated unit cell volume and given density to find out more about the element.

• The element's density is given as \(19.27 \text{ g/cm}^3\).
• Using the direct relationship between density (\(\rho\)), mass (\(m\)), and volume (\(V\)), we calculate mass as \(m = \rho \times V = 19.27 \times 4.288\times 10^{-23} = 8.259\times 10^{-22} \text{ g}\).

Understanding how density is utilized in these calculations provides insights into the material's characteristics and helps in identifying it.

Atomic Mass Calculation

Atomic mass calculations play a critical role in identifying elements, especially when analyzing crystalline structures. In our exercise, after obtaining the mass of the unit cell, we calculate the atomic mass by dividing this mass by the number of atoms within the unit cell.

• The cubic closest-packed structure has 4 atoms per unit cell.
• Using the formula \(M = \frac{m}{n}\), where \(m\) is the mass of the unit cell and \(n\) is the number of atoms, we find \(M = \frac{8.259 \times 10^{-22}}{4} = 2.065 \times 10^{-22} \text{ g}\).
• Convert grams to atomic mass units (u): \(1 \text{ u} = 1.66054 \times 10^{-24} \text{ g}\), thus \(M = \frac{2.065 \times 10^{-22} \text{ g}}{1.66054 \times 10^{-24} \text{ g/u}} = 197.11 \text{ u}\).

This method allows us to narrow down the possible elements by comparing the computed atomic mass to known values, leading us to conclude the identity of the element.

Crystallography

Crystallography is the study of crystal structures and their properties, and it applies to the arrangement of atoms in solids. In this context, understanding cubic closest-packed (ccp) structures is vital since they are one of the most efficient ways to pack spheres in three-dimensional space. This type of lattice has a coordination number of 12 and forms part of the group of face-centered cubic structures.

• Each sphere (atom) in a ccp structure is in contact with 12 others, showcasing efficient packing.
• The arrangement can be visualized as layered, where each atom's centers are above and below the gaps of adjacent layers.
• This type of structure is typical for metals, providing characteristics like high density and low reactivity.

By analyzing the atomic arrangements and densities, crystallography aids in uncovering the identity of materials. For this exercise, an understanding of ccp structures helps confirm the element's identity as gold (Au), aligning with its measured atomic mass.