Problem 34 - Chemistry: The Central Science in SI Units, Global Edition

Question

Potassium metal (atomic weight \(39.10 \mathrm{~g} / \mathrm{mol}\) ) adopts a body-centered cubic structure with a density of \(0.856 \mathrm{~g} / \mathrm{cm}^{3}\). (a) Use this information and Avogadro's number \((6.022 \times\) \(10^{23}\) ) to estimate the atomic radius of potassium. (b) If potassium didn't react so vigorously, it could float on water. Use the answer from part (a) to estimate the density of \(\mathrm{K}\) 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 potassium in a body-centered cubic structure can be found by first calculating the volume of one potassium atom using atomic weight, density, and Avogadro's number, then relating the volume of a unit cell to the atomic radius. The result is: \(r = \frac{\sqrt{3} * 39.10}{8 * 0.856 * 6.022 * 10^{23}}\) Assuming cubic close-packed potassium, the density can be calculated using the same atomic radius and compared to the density of water (1 g/cm³). The density of cubic close-packed potassium is: Density = \(\frac{39.10 * √2}{(2r)^{3} * 6.022 * 10^{23}}\) If the calculated density is less than the density of water, potassium would float on water.

Step 1: Calculate the volume of one potassium atom

By using the formula for density, mass and volume, we can find the volume of one potassium atom. Density = Mass / Volume Given, Atomic weight = 39.10 g/mol Density = 0.856 g/cm³ Avogadro's number = 6.022 × 10^23 atoms/mol First, we need to find the mass of one potassium atom: Mass of one potassium atom = Atomic weight / Avogadro's number = \(\frac{39.10}{6.022 * 10^{23}}\) Next, we'll find the volume of one potassium atom: Volume = Mass / Density = \(\frac{39.10}{0.856 * 6.022 * 10^{23}}\)

Step 2: Calculate the volume of a unit cell of body-centered cubic potassium

The body-centered cubic structure has two atoms per unit cell. Therefore, the volume of a unit cell can be found by multiplying the volume of one potassium atom by the number of atoms in the unit cell. Volume of a unit cell = Volume of one potassium atom * Number of atoms per unit cell = \(\frac{39.10}{0.856 * 6.022 * 10^{23}}\) * 2

Step 3: Determine the atomic radius of potassium

For a body-centered cubic unit cell, the relationship between its volume (V) and the atomic radius (r) is as follows: V = a³ where a is the length of the side of the cubic unit cell a = 4r / √3 Now, substituting V (found in step 2) in the volume equation: \(\frac{39.10}{0.856 * 6.022 * 10^{23}}\) * 2 = (\(4 * r / √3\))³ Now, solve for r (atomic radius of potassium):

Step 4: Estimate the density of cubic close-packed potassium

Assuming that potassium has a cubic close-packed structure, the relationship between its volume V' and atomic radius r (the same atomic radius as in step 3) is as follows: V' = a'³ where a' is the length of the side of the cubic close-packed unit cell a' = 2 * √2 * r Density = Mass / Volume Mass = Density of cubic close-packed potassium * Volume Density of cubic close-packed potassium = Mass / V' Now, substitute V' (from the above equation) and calculate the density of cubic close-packed potassium.

Step 5: Compare the new density with the density of water

The density of water is about 1 g/cm³. Compare the density of cubic close-packed potassium (from step 4) with the density of water to determine if potassium would float or not. If the density of cubic close-packed potassium is less than the density of water, then potassium would float on water. Otherwise, it would sink.

Key Concepts

Body-Centered Cubic Structure

The body-centered cubic (BCC) structure is a type of crystal lattice found in many metals, including potassium. In this structure, one atom is located at each corner of the cube, and one atom is positioned at the center of the cube as well. This results in each unit cell containing two atoms in total. The BCC structure is important because it influences the metal's density and stability.

• Each corner atom is shared by eight different cubes, meaning that each cube effectively contributes 1/8 of an atom per corner position.
• The center atom is not shared with any other unit cell.

This arrangement of atoms affects how they pack together and how closely they are packed. The atomic radius in a BCC structure can be calculated using the formula involving the cube's side length (aaa):\[a = \frac{4r}{\sqrt{3}}\] Here, \(r\) is the atomic radius. This relationship is crucial for determining the atomic radius of an element like potassium, helping us understand its physical properties.

Avogadro's Number

Avogadro's number, which is approximately \(6.022 \times 10^{23}\), is a fundamental constant in chemistry representing the number of atoms or molecules in one mole of a substance. This large number helps bridge the gap between the macroscopic and atomic worlds. Understanding Avogadro's number is essential when dealing with calculations involving the atomic scale, such as:

• Determining the number of particles in a given amount of substance in moles.
• Converting between atomic and macroscopic quantities in stoichiometric calculations.

In the context of potassium and its crystalline structure, Avogadro's number allows us to calculate the mass of a single potassium atom. By dividing the atomic weight of potassium by Avogadro's number, we determine the mass, giving key insights into density and structural properties. This understanding enables the computation of volume and further density calculations crucial for different atomic arrangements.

Density of Potassium

Density is a measure of mass per unit volume, and in the case of potassium metal, it is given as \(0.856 \text{ g/cm}^3\) for the body-centered cubic structure. The density of potassium is a crucial property for understanding how potassium atoms are arranged in a lattice.To calculate the density, we use:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]With the atomic mass known, and using Avogadro's number to find the mass of a single atom, the formula helps derive the volume occupied by each atom in the structure.Understanding potassium's density helps predict how potassium might behave in different chemical environments and whether it would float on water under various crystalline structures. For example, if potassium adopted a cubic close-packed structure, its density would differ, possibly affecting whether it floats or sinks in water.

Cubic Close-Packed Structure

The cubic close-packed (CCP) structure, also known as face-centered cubic (FCC), is another common crystal lattice found in metals. Unlike body-centered cubic, CCP features atoms at each corner and the center of all faces of the cube. This arrangement leads to a higher packing efficiency.In CCP:

• Atoms touch each other along the face diagonals of the cube.
• Each unit cell effectively contains four atoms as each face-centered atom is shared by two unit cells, and each corner atom is shared by eight.

The relationship between the lattice parameter \(a'\) and atomic radius \(r\) in CCP is given by:\[a' = 2 \sqrt{2} \times r\]This higher packing efficiency often results in a different density. If potassium were to adopt a CCP structure instead of its normal BCC, calculations would show a different density value, which determines whether it could float on water. Understanding CCP is essential for predicting material properties and guiding their use in real-world applications.