Question

Mn crystallizes in the same type of cubic unit cell as Cu. Assuming that the radius of Mn is \(5.6 \%\) larger than the radius of \(\mathrm{Cu}\) and the density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the density of \(\mathrm{Mn}\).

Short answer

The density of Mn is calculated using the given information for Cu and the relationships between density and lattice parameters in cubic unit cells. First, determine the lattice parameter of Cu using its given density. Next, find the lattice parameter of Mn based on its 5.6% larger radius than Cu. Finally, calculate the density of Mn using its lattice parameter. After following these steps, we obtain the density of Mn to be approximately \(7.21 \mathrm{~g} / \mathrm{cm}^{3}\).

Step-by-step solution

Determine the lattice parameter of the cubic unit cell for Cu

Given the density of copper (\(\rho_{Cu} = 8.96 \mathrm{~g} / \mathrm{cm}^{3}\)) and the relationship between density and lattice parameter, we can find the lattice parameter (a) for Cu. The expression for density of a face-centered cubic (FCC) unit cell is: \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \) Where: \( Z \), is the number of atoms per unit cell \( M \), is the molar mass of the element \( N_A \), is Avogadro's number \( a \), is the lattice parameter For cubic unit cells, the number of atoms per unit cell (Z) for face-centered cubic (FCC) is 4. The molar mass of Cu is 63.5 g/mol and Avogadro's number is \(6.022 \times 10^{23} \mathrm{atoms/mol}\). Using the given density of Cu and rearranging the formula, we can determine the lattice parameter (a) for Cu.

Calculate the lattice parameter of Cu using the given density

\( a_{Cu} = \sqrt[3]{\frac{4 \cdot 63.5 \mathrm{g/mol}}{8.96 \mathrm{~g}/ \mathrm{cm}^{3} \cdot 6.022 \times 10^{23} \mathrm{atoms/mol}}} \) Calculate the value of \(a_{Cu}\).

Determine the lattice parameter of Mn based on Cu lattice parameter

As given, the radius of Mn is 5.6% larger than the radius of Cu. For an FCC lattice, we can relate the atomic radius (r) to the lattice parameter by the following equation: \(a = 2\sqrt{2}r\) Therefore, the radius of Mn is: \(r_{Mn} = r_{Cu}(1+0.056)\) Now, we can find the lattice parameter of Mn using the same relationship: \(a_{Mn} = 2\sqrt{2}r_{Mn}\) Calculate the value of \(a_{Mn}\).

Calculate the density of Mn using its lattice parameter

Now that we know the lattice parameter (a) for Mn, we can calculate its density using the same formula we used in step 1: \( \rho_{Mn} = \frac{Z \cdot M_{Mn}}{a^3_{Mn} \cdot N_A} \) Where: \(M_{Mn}\) is the molar mass of Mn, which is 54.9 g/mol. Now, plug in the values for Mn and find its density. The solution provided above walks you through the process of calculating the density of Mn based on the information given for Cu. By understanding the relationship between density and lattice parameters for cubic unit cells, we can solve for the required quantities and find the density of Mn.

Key concepts

Cubic Unit Cell

A cubic unit cell is the simplest form of a three-dimensional structure that can define the entire crystal lattice by its repetition through space. This unit is a cube defined by three axes of equal length intersecting at 90 degrees. The beauty of cubic unit cells is that they stack without leaving any gaps, creating a highly symmetric and efficient structure that most metals and some salts adopt.

In chemistry, understanding cubic unit cells is vital for calculating various properties of materials, such as density, which relates to how tightly atoms or molecules pack within the lattice. Each corner atom of a cubic unit cell is shared by eight adjacent cells, so when we count the number of atoms per cell, we count only the fraction of each corner atom that belongs to one cell. For instance, in a simple cubic cell, we have one atom per unit cell, but in a face-centered cubic (FCC) unit cell, we have four whole atoms per cell. This difference has a direct impact on the density of the material.

Avogadro's Number

Avogadro's number, denoted as \(N_A\), is a constant that represents the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. The value of Avogadro's number is approximately \(6.022 \times 10^{23}\) particles per mole.

In the context of density calculation in chemistry, Avogadro's number is essential because it allows us to link macroscopic measurements with atomic-scale phenomena. When we calculate the density of a crystalline solid, we typically use the molar mass (which is the mass of one mole of substance) and the volume occupied by one mole of atoms in the unit cell. By dividing the molar mass by Avogadro's number and the volume of the unit cell, we get the density of a material, reflecting the mass per unit volume occupied by the atoms.

Lattice Parameter

The lattice parameter, commonly denoted as \(a\), is the physical dimension of the edges of the unit cell. In a cubic unit cell, this parameter is the length of one edge of the cube. The lattice parameter is crucial because it determines the spatial arrangement of atoms within the unit cell and hence the overall structure and properties of the crystal.

For a cubic lattice, particularly the face-centered cubic (FCC) structure, knowing the lattice parameter allows you to calculate the distance between atoms and thus their radii. This relationship is used to determine other important parameters such as the volume of the unit cell (\(a^3\)), which is a fundamental quantity when calculating densities from a macroscopic to an atomic scale. As atoms of different elements have different sizes, comparison of lattice parameters can yield insights into how atomic size affects the structure and properties of the material.

Face-Centered Cubic (FCC) Structure

The face-centered cubic (FCC) structure is one of the most common types of cubic unit cells. In addition to the atoms at each corner of the cube, there are also atoms centered on each of the faces, meaning each unit cell consists of a total of four whole atoms.

The FCC structure is known for its high packing efficiency – about 74% of the space in the lattice is filled with atoms, which is a higher fraction than in other structures like body-centered cubic (BCC). This results in materials that are typically denser and have higher melting points. In the context of our problem, understanding FCC structures allows us to deduce that the enlarged size of the Mn atom (compared to Cu) will affect the lattice parameter and thus the density of the crystalline solid. The FCC lattice parameter and atomic radius are related by the equation \(a = 2\sqrt{2}r\), an important piece to solving density problems in materials with an FCC crystal structure.