Question

Mn crystallizes in the same type of cubic unit cell as Cu. Assuming that the radius of \(\mathrm{Mn}\) 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 approximately 7.87 g/cm³.

Step-by-step solution

Determine the structure type of the unit cell

Mn and Cu are crystallized in the same type of cubic unit cell. Copper crystallizes in a face-centered cubic (FCC) unit cell, where it has atoms in the corners of the cube and in the center of the faces. Mn will also have an FCC unit cell structure.

Calculate the edge length of the unit cells based on their radius

In an FCC unit cell, the relationship between the radius (r) and the edge length (a) of the unit cell is given by \(a = 2\sqrt{2}r\). Since the radius of Mn is 5.6% larger than the radius of Cu, we can express the edge length ratio between Mn and Cu as follows: \( \frac{a_{Mn}}{a_{Cu}} = \frac{2\sqrt{2}r_{Mn}}{2\sqrt{2}r_{Cu}} \) Since, \(r_{Mn} = (1 + 0.056)r_{Cu}\) \( \frac{a_{Mn}}{a_{Cu}} = \frac{(1 + 0.056)r_{Cu}}{r_{Cu}} \)

Calculate the volume ratio between the unit cells

The volume of a cube is given by \(V = a^3\). To find the relationship between the volumes of Mn and Cu unit cells, we can use the edge length ratio obtained in step 2: \( \frac{V_{Mn}}{V_{Cu}} = \left(\frac{a_{Mn}}{a_{Cu}}\right)^3\) Substituting the ratio of the edge lengths we found in step 2 \( \frac{V_{Mn}}{V_{Cu}} = \left(\frac{1.056 r_{Cu}}{r_{Cu}}\right)^3 = 1.056^3\)

Calculate the density ratio between Mn and Cu

The density of a substance is given by the mass per unit volume. In the case of the FCC unit cells of Mn and Cu, the mass of Mn atoms will be the same as the mass of Cu atoms because they are in the same atomic structure, so the density will only depend on the unit cell volume. We can then use the volume ratio calculated in step 3 to find the density ratio between Mn and Cu: \( \frac{\rho_{Mn}}{\rho_{Cu}} = \frac{V_{Cu}}{V_{Mn}} \) Substituting the volume ratio \( \frac{\rho_{Mn}}{\rho_{Cu}} = \frac{1}{1.056^3}\)

Calculate the density of Mn

Now we can use the density ratio between Mn and Cu to find the density of Mn: \( \rho_{Mn} = \rho_{Cu} \cdot \frac{1}{1.056^3} \) Given that the density of Cu is 8.96 g/cm³, we can calculate the density of Mn: \( \rho_{Mn} = 8.96 \cdot \frac{1}{1.056^3} \: \mathrm{g/cm^3}\) \( \rho_{Mn} \approx 7.87 \: \mathrm{g/cm^3}\) The density of Mn is approximately 7.87 g/cm³.

Key concepts

Face-Centered Cubic Unit Cell

In crystallography, a face-centered cubic (FCC) unit cell is one of the most efficient ways atoms can arrange themselves in space. Materials like copper (Cu) and manganese (Mn) crystallize in an FCC structure. In an FCC unit cell, every cube corner atom is shared by eight neighboring unit cells, while each face-centered atom is shared between two unit cells. This leads to a packing efficiency of about 74%, meaning that 74% of the space is occupied by atoms, and the rest is voids.

One key feature of the FCC lattice is that it contains four atoms per unit cell. This can be visualized as:

• 1/8th of an atom at each of the 8 corners (totaling 1 atom)
• 1/2 of an atom on each of all 6 faces (totaling 3 atoms)

Understanding this structure is crucial for computing properties like density since the arrangement defines how tightly atoms pack within a cell.

Atomic Radius

The atomic radius is a measure of the size of an atom's electron cloud. It is essential for determining how closely atoms can pack in a lattice. Within an FCC lattice, the relationship between the atomic radius (r) and the unit cell's edge length (a) can be described as \( a = 2\sqrt{2}r \).

In our exercise, manganese (Mn) has an atomic radius that is 5.6% larger than that of copper (Cu). This size increase translates directly into a longer edge length and affects the overall geometry of the unit cell. As a result, this new atomic size impacts calculations for volume and subsequently the density.

Volume Ratio

Understanding volume ratio is essential when discussing crystalline structures. The volume of a cube, which is the shape of our unit cells, is calculated using \( V = a^3 \), where \( a \) is the edge length.

When comparing two crystalline cells, like Mn and Cu, we use their edge length ratio to find their volume ratio. For instance, in the case of Mn and Cu, with Mn's edge length being 1.056 times that of Cu, the volume ratio would be calculated as \( \left(1.056\right)^3 \). This calculation reveals how much larger the manganese unit cell is compared to the copper unit cell.

Volume significantly influences density since density is mass per unit volume. Meaning, even a small adjustment in volume can lead to noticeable differences in density values.

Density Ratio

The density ratio reflects the relative density between two substances by comparing their mass to volume ratios. Density itself is defined as \( \rho = \frac{m}{V} \), where \( m \) represents mass and \( V \) is the volume.

For FCC unit cells of Mn and Cu, mass is consistent due to atomic similarities. The differing factor is volume, arising from differences in atomic size. With a confirmed volume ratio of 1.056^3, the density ratio between Mn and Cu becomes inverted— \( \frac{\rho_{Mn}}{\rho_{Cu}} = \frac{V_{Cu}}{V_{Mn}} \).

This inverted relationship arises because if the same mass occupies a larger volume, the overall density diminishes. Consequently, calculating back using the copper density of 8.96 g/cm³, we find Mn’s density is approximately 7.87 g/cm³. This illustrates how even small atomic size changes can significantly affect material properties like density.