Problem 34 - Chemistry The Central Science

Question

Sodium metal (atomic weight 22.99 \(\mathrm{g} / \mathrm{mol}\) ) adopts a body- centered cubic structure with a density of 0.97 \(\mathrm{g} / \mathrm{cm}^{3}\) . (a) Use this information and Avogadro's number \(\left(N_{\mathrm{A}}=6.022 \times 10^{23} / \mathrm{mol}\right)\) to estimate the atomic radius of sodium. \((\mathbf{b})\) If sodium didn't react so vigorously, it could float on water. Use the answer from part (a) to estimate the density of Na if its structure were that of a cubic close packed metal. Would it still float on water?

Step-by-Step Solution

Answer

The atomic radius of sodium in a body-centered cubic structure is \(1.86 \times 10^{-8}\, \text{cm}\). In the cubic close-packed structure, the density of sodium would be lower than the density of water (1 g/cm³), which means that it would indeed float on water, provided it didn't react vigorously.

Step 1: Calculate the volume of a unit cell

The volume of a unit cell in a body-centered cubic structure is given by: \[ V_{\text {unit cell}}=a^{3} \] Where 'a' is the edge length of the unit cell. In a body-centered cubic structure, there are two atoms per unit cell. So, the volume of a unit cell can be written as: \[ V_{\text {unit cell}} = \frac{2 \times m_{\text {Na}}}{\rho_{\text {Na}}} \] Where \(\rho_{Na}\) is the density of sodium, and \(m_{Na}\) is the mass of a single sodium atom.

Step 2: Calculate the mass and edge length of the unit cell

The mass of a single sodium atom can be found using its atomic weight and Avogadro's number: \[ m_{\text {Na}} = \frac{\text {Atomic weight of Na}}{N_{\text {A}}} \] Substitute the given values: \[ m_{\text {Na}} = \frac{22.99 \text { g/mol}}{6.022 \times 10^{23} \, \text {atoms/mol}} \] \[ m_{\text {Na}} = 3.82 \times 10^{-23}\,\text{g} \] Now, replace the mass of sodium atom in the volume formula: \(V_{\text {unit cell}} = \frac{2 \times m_{\text {Na}}}{\rho_{\text {Na}}}\) and solve for the edge length: \[ a^{3} = \frac{2 \times 3.82 \times 10^{-23}\, \text{g}}{0.97 \,\text{g/cm}^{3}} \] \[ a^{3} = 7.88 \times 10^{-23}\, \text{cm}^{3} \] \[ a = 4.29 \times 10^{-8}\, \text{cm} \]

Step 3: Calculate the atomic radius of sodium

For a body-centered cubic structure, the face diagonal length is given by: \[ a\sqrt{3} = 4r \] where r is the atomic radius. Rearranging the equation to solve for the atomic radius of sodium: \[ r = \frac{a\sqrt{3}}{4} \] Now, substitute the value of 'a': \[ r = \frac{(4.29 \times 10^{-8}\,\text{cm})\sqrt{3}}{4} \] \[ r = 1.86 \times 10^{-8}\, \text{cm} \] Now, we address the second part of the question.

Step 4: Calculate the volume and mass of a unit cell in a cubic close-packed structure

In a cubic close-packed structure, there are 4 atoms per unit cell, so the volume of a unit cell can be written as: \[ V_{\text { unit cell }}=\frac{4 \times m_{\text { Na }}}{\rho_{\text { new }}} \] The edge length of the unit cell can be given by: \[ a_{\text { new }}=\frac{4r}{\sqrt{2}} \] Now, solve for the volume \(V_{\text {unit cell new}}=a_{\text {new}}^{3}\).

Step 5: Calculate the new density of sodium in the cubic close-packed structure

To find the new density, solve for \(\rho_{new}\): \[ \rho_{\text {new}} = \frac{4 \times m_{\text {Na}}}{V_{\text {unit cell new}}} \] Substitute the values calculated earlier in the equation and calculate the new density. Finally, compare the results with the density of water (1 g/cm³) to determine if sodium would float on water in the new structure.

Key Concepts

Body-Centered Cubic Structure

Understanding the body-centered cubic (BCC) structure is vital when studying materials like sodium that adopt this arrangement. The BCC structure is one of the simplest crystalline arrangements found in metals where atoms are positioned at the eight corners of a cube and one atom at the center of the cube. This central atom is equally distanced from all corner atoms, creating a distinct geometric pattern.

Each corner atom is shared by eight adjacent unit cells, so within a single unit cell, a corner atom contributes only 1/8th of its volume. Since there are eight corners, this contribution accounts for one complete atom. Adding the central atom, a total of two complete atoms are present in a single BCC unit cell. The atomic radius estimation for elements with a BCC structure can be determined by the arrangement and spacing of the atoms.

Avogadro's Number

Avogadro's number, denoted as \(N_A\), is a fundamental constant in chemistry and physics that represents the number of constituent particles, typically atoms or molecules, in one mole of a substance. Its value is approximately \(6.022 \times 10^{23}\) particles per mol. This constant allows chemists and physicists to convert between the mass of a substance and the number of atoms or molecules it contains.

When dealing with substances at the atomic or molecular scale, knowing Avogadro's number is essential for calculating the mass of a single atom or molecule. The step-by-step solution utilizes Avogadro's number to determine the molar mass of sodium, which is crucial for further calculations related to atomic radius and density.

Density Calculations

Density calculations are commonly used in material science and engineering to ascertain how much mass is contained within a unit volume of a material. The density \(\rho\) of a substance is mathematically expressed as \(\rho = \frac{m}{V}\), where 'm' is the mass and 'V' is the volume. For crystalline structures like those of metals, understanding density is particularly important as it relates to their packing arrangements and atomic radii.

Determining the density of a crystalline arrangement involves knowing the mass of the atoms within a unit cell and its volume. As laid out in the step-by-step solution, the mass of a single sodium atom is calculated, which is then used to estimate the volume of its BCC unit cell. Using this volume, along with the known mass of the atoms in the cell, one can determine the density of sodium.

Cubic Close-Packed Structure

The cubic close-packed (CCP) structure is another way atoms can organize themselves in a crystalline solid. This arrangement is highly efficient in terms of space utilization, as 74% of the space in a unit cell is filled—the highest possible for spherical particles. In the CCP structure, atoms are located at each of the corners and the center of each face of the cube, which amounts to four atoms per unit cell.

Transitioning from BCC to CCP affects the density of the material. In the provided solution, we estimate the new density of sodium assuming that it changed to a CCP crystalline structure. This shift results in different packing efficiency and thus a different density compared to its BCC arrangement. By comparing the new density to the density of water, we can infer whether sodium would float or sink if its structure were analogous to that of a CCP metal.