Problem 134 - Chemistry

Question

Aluminum has an atomic radius of 143 \(\mathrm{pm}\) and forms a solid with a cubic closest packed structure. Calculate the density of solid aluminum in \(\mathrm{g} / \mathrm{cm}^{3}\) .

Step-by-Step Solution

Answer

\(a \approx 405.2\, \text{pm}\) #tag_title# Step 3: Calculate the volume of the unit cell #tag_content# Now that we have the edge length, we can calculate the volume of the unit cell (V): \(V = a^3\) Substitute the edge length (a) into the formula: \(V = (405.2\, \text{pm})^3\) Calculating the volume (V): \(V \approx 6.68 \times 10^7\, \text{pm}^3\) #tag_title# Step 4: Convert volume to cm³ #tag_content# To calculate the density in g/cm³, we need to convert the volume from pm³ to cm³: \(1\, \text{pm} = 10^{-10} \, \text{cm}\) \(V \approx 6.68 \times 10^7\, \text{pm}^3 \times (10^{-10}\, \text{cm} / \text{pm})^3\) Calculating the volume in cm³: \(V \approx 6.68 \times 10^{-23}\, \text{cm}^3\) #tag_title# Step 5: Determine the molar mass and Avogadro's number #tag_content# The molar mass of aluminum is 26.98 g/mol. Avogadro's number (N) is approximately \(6.022 \times 10^{23}\) particles/mol. #tag_title# Step 6: Calculate the density #tag_content# The density (ρ) of solid aluminum can be calculated using the formula: \(\rho = \frac{4 \times M_{\text{Al}}}{N \times V}\) Substituting the values found previously: \(\rho \approx \frac{4 \times 26.98\, \text{g/mol}}{6.022 \times 10^{23} \, \text{particles/mol} \times 6.68 \times 10^{-23}\, \text{cm}^3}\) Calculating the density (ρ): \(\rho \approx 2.70\, \text{g/cm}^3\) Thus, the density of solid aluminum is approximately 2.70 g/cm³.

Step 1: Determine the atomic structure and atoms per unit cell

In a cubic closest packed (ccp) structure, also called face-centered cubic (fcc), the atoms are packed as close together as possible. In this arrangement, each unit cell contains 4 atoms.

Step 2: Find the edge length of the unit cell

To calculate the edge length (a) of the unit cell, we must first recognize that in an fcc structure, the face diagonal is equal to 4 times the atomic radius (r). We can use the Pythagorean theorem twice since the face diagonal cuts through the unit cell: \(a^2 + a^2 = (\frac{4r}{\sqrt{2}})^2 => a = 2r\sqrt{2}\) We substitute the atomic radius of aluminum (r = 143 pm) into the formula: \(a = 2(143\, \text{pm})\sqrt{2}\) Calculating the edge length (a):

Key Concepts

Cubic Closest Packed Structure

The cubic closest packed structure, commonly known as the face-centered cubic (fcc) structure, is a highly efficient way of arranging atoms in a crystalline solid. In this arrangement, each atom is surrounded by 12 others, achieving maximum packing efficiency. This structure is not only about proximity but also stability, as it results in a very strong and compact form. Each unit cell within this structure contains 4 atoms. This is due to contributions from the atoms placed at the center of each face of the cube, which add up to this total after accounting for shared atoms with neighboring cells. Here are some more details:

• Each corner being shared by 8 unit cells, each atom contributes 1/8 to a single unit cell.
• Each face-centered atom being shared with an adjacent unit cell and thus contributes 1/2 to a single unit cell. There are 6 faces in a cube, totaling 3 atoms per unit cell.
• The overall contribution leads to 4 atoms per unit cell.

Atomic Radius

The atomic radius is a measure of the size of an atom, typically the distance from the nucleus to the outer boundary of the surrounding cloud of electrons. When dealing with metallic elements, such as aluminum, the atomic radius can help us understand and characterize the material's properties, including its density. In the context of the face-centered cubic structure, the atomic radius is crucial in determining the dimensions of the unit cell. The radius can be determined by packing atoms in symmetric rows or calculating the geometry of bond lengths. It's important because:

• It defines the size boundary of each atom in a solid.
• In the cubic closest packed arrangement, it aids in calculating the edge length of the unit cell due to the relation with the face diagonal.
• Knowing the atomic radius allows us to predict how closely and tightly packed the material is, impacting overall density.

Unit Cell

A unit cell is the smallest repeating unit that displays the full symmetry and pattern characteristics of the entire crystal structure. Think of it as the building block of a crystal.For a face-centered cubic structure, the unit cell is shaped like a cube with crucial dimensions that can be derived from the atomic arrangement within.Diagonals play a key role here.

• The face diagonal can be expressed in terms of the atomic radius, particularly the presence of the 4 atomic radii often defining the diagonal.
• For the face-centered cubic, the relationship between atomic radius and the cell's edge length is vital, expressed as \(a = 2r\sqrt{2}\).
• This formula is derived by using the face diagonal, which spans across the unit cell's face, aligning with any corner atoms it encounters.

Face-Centered Cubic

The face-centered cubic (fcc) arrangement is synonymous with the cubic closest packed structure. This structure is defined by each face of the cube being occupied by an atom, in addition to those at the corners, forming an efficient and densely packed structure. Such packing allows for only a little open space between atoms, enhancing the durability and density of the material. The various aspects of fcc include:

• Each face atom being shared with an adjacent unit cell, ensuring economy in space utilization across cells.
• This type of structure maximizes atomic coordination, meaning each atom interacts closely with neighboring atoms, stabilizing the overall material format.
• It is a common structure type for various metals including aluminum, copper, and gold, often characterized by notable density and intrinsic strength.