Problem 123 - Chemistry: An Atoms-Focused Approach

Question

Under the appropriate reaction conditions, small cubes of molybdenum, \(4.8 \mathrm{nm}\) on a side, can be deposited on carbon surfaces. These "nanocubes" are made of bcc arrays of Mo atoms. a. If the edge of each nanocube corresponds to 15 unit cell lengths, what is the effective radius of a molybdenum atom in these structures? b. What is the density of each molybdenum nanocube? c. How many Mo atoms are in each nanocube?

Step-by-Step Solution

Answer

Answer: The effective radius of a molybdenum atom in these structures is approximately 0.139 nm. The density of each molybdenum nanocube is roughly 9.73 x 10^6 g/m^3, and there are 6750 Mo atoms in each nanocube.

Step 1: Determine the edge length of a unit cell

In a BCC array, the edge length of a unit cell is equal to \(4\mathrm{r}/\sqrt{3}\), where r is the effective radius of an atom. Given that the edge length of each nanocube corresponds to 15 unit cell lengths, we can find the length of a single unit cell. The edge length of a nanocube is \(4.8\,\text{nm}\), which corresponds to \(15\) unit cell lengths. Therefore, the edge length of a unit cell, denoted by \(a\), is: \(a = (4.8\,\text{nm})/15\) \(a = 0.32\,\text{nm}\)

Step 2: Determine the effective radius of molybdenum atom

Now, we can determine the effective radius of a molybdenum atom, which is included in a BCC unit cell: Using the formula \(a = 4r/\sqrt{3}\), we can solve for \(r\): \(r = \frac{a\sqrt{3}}{4} = \frac{0.32\,\text{nm} \cdot \sqrt{3}}{4}\) \(r \approx 0.139\,\text{nm}\)

Step 3: Calculate the volume of a molybdenum nanocube and a BCC unit cell

Before finding the density of each molybdenum nanocube, we need to find the volume of a nanocube and a BCC unit cell. Volume of a molybdenum nanocube: \(V_{cube} = (4.8\,\text{nm})^3 = 110.592\,\text{nm}^3\) Volume of a BCC unit cell: \(V_{unitcell} = a^3 = (0.32\,\text{nm})^3 = 0.032768\,\text{nm}^3\)

Step 4: Calculate the mass of a molybdenum atom

We know the molar mass of molybdenum, which is \(95.94\text{g/mol}\). We can find the mass of one atom as follows: \(m_{ atom} = \frac{M_{molybdenum}}{N_A}\), where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23}/\text{mol}\)) \(m_{atom} = \frac{95.94\,\text{g/mol}}{6.022\times 10^{23}/\text{mol}}\) \(m_{atom} \approx 1.595\times10^{-22}\,\text{g}\)

Step 5: Calculate the density of each molybdenum nanocube

To find the density of each molybdenum nanocube, we need to determine the mass of needed unit cells and the volume of one nanocube. There are two molybdenum atoms in each BCC unit cell, so the mass of needed unit cells (only considering the molybdenum given that it constitutes the whole BCC structure) is given by: \(15^3\cdot 2 \cdot m_{atom} = 6750 \cdot 1.595\times 10^{-22}\,\text{g} \approx 1.076\times 10^{-18}\,\text{g}\) Density of a molybdenum nanocube is given by the following formula: \(\rho_{nanocube} = \frac{m_{nanocube}}{V_{cube}}\) \(\rho_{nanocube} = \frac{1.076 \times 10^{-18}\,\text{g}}{110.592\,\text{nm}^3}\) \(\rho_{nanocube} \approx 9.73 \times 10^6\,\text{g/m}^3\)

Step 6: Calculate the number of Mo atoms in each nanocube

As there are two atoms per BCC unit cell and there are \(15^3\) unit cells in one nanocube: Total number of Mo atoms = \(2 \cdot 15^3\) Total number of Mo atoms = \(2 \cdot 3375 = 6750\) In conclusion: a. The effective radius of a molybdenum atom in these structures is \(\approx 0.139\,\text{nm}\) b. The density of each molybdenum nanocube is roughly \(9.73 \times 10^6\,\text{g/m}^3\) c. There are \(6750\) Mo atoms in each nanocube.

Key Concepts

Body-Centered Cubic (BCC)

When understanding crystalline structures, the body-centered cubic (BCC) is a common type. In a BCC unit cell, atoms are positioned at each corner of a cube, and there's one atom at the center of the cube. This configuration is quite efficient in packing, although not as dense as face-centered cubic structures.

Each BCC unit cell effectively contains two atoms. One full atom resides at the center, and each corner atom counts as \(1/8\) of an atom (since a cube has 8 corners). Thus, when you count all corner atoms, they combine to form one whole atom.

This arrangement is significant because it affects properties like density and packing efficiency. Molybdenum often forms such BCC structures, which are stable and have equal articulation in three-dimensional space.

Density Calculation

Density is a fundamental property that relates the mass of a substance to its volume. When calculating the density of a molybdenum nanocube, it's essential first to determine its mass and volume.

• The mass of the nanocube can be found by considering the number of atoms and the mass of a single atom.
• The volume is based on the cube's edge length since it's a three-dimensional object.

To calculate density, use the formula: \( {ho} = \frac{mass}{volume}\). This formula gives you an idea of how much matter is packed into a given space, which is crucial for many applications, including creating materials with specific weight or strength characteristics.

For the molybdenum nanocube, we follow the step-by-step calculations to find its density, which is found to be approximately \(9.73 \times 10^6 \text{g/m}^3\). This high density is typical of metals and results from their closely packed atomic structures.

Atomic Radius

The atomic radius is a measure of the size of an atom, specifically the distance from the center of the atom's nucleus to the outermost electrons. In crystalline metals like molybdenum, the effective atomic radius can slightly differ due to the tight packing of atoms within the structure.

In the context of BCC structures, the atomic radius is determined by the geometry of the unit cell. For a BCC lattice, the relationship between the atomic radius \(r\) and the edge length \(a\) is given by \(a = \frac{4r}{\sqrt{3}}\).

This formula arises from the geometric considerations of spheres touching along the diagonal of a cube. Knowing the edge length of the unit cell allows precise calculation of the atomic radius, which, for molybdenum, is approximately \0.139 \, \text{nm}\ in such nanocubes. This radius helps in understanding how atoms will interact and bond in the material.

Unit Cell

The unit cell represents the smallest repeating segment that can be geometrically used to build the whole of a crystal lattice. In the case of a body-centered cubic (BCC) lattice, this cell contains two complete atoms in its minimal configuration.

Determining the unit cell size is crucial for calculating other properties of the crystal, such as volume and density. It links the macroscopic properties of the substance to its microscopic structure.

• The unit cell's edge length can be determined when we know its relationship to the nanocube's edge, as shown in the exercise.
• This relationship and calculations help in determining the effective atomic size and overall molecular structure of the material.

In this exercise, the edge length of the unit cell in relation to the molybdenum nanocube was used to understand various properties such as the atomic radius and density of the material.

Understanding the unit cell and its implications can provide insights into the mechanical and physical properties of materials, such as strength, flexibility, and conductivity.