Problem 32 - Chemistry: The Central Science

Question

Sodium metal (atomic weight \(22.99 \mathrm{~g} / \mathrm{mol}\) ) adopts a bodycentered cubic structure with a density of \(0.97 \mathrm{~g} / \mathrm{cm}^{3}\). (a) Use this information and Avogadro's number \(\left(N_{A}=6.022 \times 10^{23} / \mathrm{mol}\right)\) to estimate the atomic radius of sodium. (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

In summary, the atomic radius of sodium is approximately \(1.86 \times 10^{-8} \mathrm{~cm}\). If sodium had a cubic close-packed structure, its density would be \(0.596 \mathrm{~g} / \mathrm{cm}^3\), and it would float on water due to having a lower density than water (approximately \(1 \mathrm{~g} / \mathrm{cm}^3\)).

Step 1: (a) Calculate the atomic volume of one mole of sodium atoms

We can begin by finding the atomic volume of one mole of sodium atoms. To do this, we'll use the density formula: Density = mass / volume We know the density and atomic mass of sodium, so we can rearrange the formula to find the volume: Volume = mass / density Volume of one mole of sodium atoms = (molar mass of sodium) / (density) Volume of one mole = \((22.99 \mathrm{~g/mol}) / (0.97 \mathrm{~g/cm^3}) = 23.70 \mathrm{~cm^3/mol}\)

Step 2: (a) Calculate the cell volume of body-centered cubic structure

A body-centered cubic (BCC) structure has two atoms per unit cell. To find the cell volume, we need to divide the atomic volume of one mole of sodium atoms by Avogadro's number and then multiply by the number of atoms per unit cell: Cell volume = \((23.70 \mathrm{~cm^3/mol}) / (6.022 \times 10^{23} / \mathrm{mol}) × 2 \mathrm{atoms/cell}\) Cell volume = \(7.85 \times 10^{-23} \mathrm{~cm}^3 / \mathrm{cell}\)

Step 3: (a) Calculate the side length of the unit cell

The cell volume of a cubic structure is given by the side length cubed: Cell volume = side length\(^3\) To find the side length, we can take the cube root of the cell volume: Side length = \(\sqrt[3]{7.85 \times 10^{-23} \mathrm{ ~cm}^3 /( \mathrm{cell} )}\) Side length = \(4.28 \times 10^{-8} \mathrm{~cm}\)

Step 4: (a) Calculate the atomic radius of sodium

In a BCC structure, the atoms touch along the body diagonal. The body diagonal can be expressed as: Body diagonal = \(\sqrt{3}\) × side length Since the body diagonal consists of 1 atomic radius from the atom at the corner and 2 atomic radii from the atom in the center: Atomic radius = \( (4.28 \times 10^{-8} \mathrm{~cm}) / (4\sqrt{2}) = 1.86 \times 10^{-8} \mathrm{~cm}\) So, the atomic radius of sodium is approximately \(1.86 \times 10^{-8} \mathrm{~cm}\).

Step 5: (b) Estimate the density of sodium with a cubic close-packed structure

Cubic close-packed (CCP) structures have four atoms per unit cell. We will first find the cell volume for a CCP structure by multiplying the atomic volume of one mole of sodium atoms divided by Avogadro's number by the number of atoms per unit cell: Cell volume (CCP) = \((23.70 \mathrm{~cm^3/mol}) / (6.022 \times 10^{23} / \mathrm{mol}) × 4 \mathrm{atoms/cell}\) Cell volume (CCP) = \(1.57 \times 10^{-22} \mathrm{~cm}^3 / \mathrm{cell}\)

Step 6: (b) Calculate the side length of the unit cell in the CCP structure

To find the side length of the unit cell in the CCP structure, we can take the cube root of the cell volume (CCP): Side length (CCP) = \(\sqrt[3]{1.57 \times 10^{-22} \mathrm{~cm}^3 / (\mathrm{cell})}\) Side length (CCP) = \(2.52 \times 10^{-8} \mathrm{~cm}\)

Step 7: (b) Calculate the density of the cubic close-packed sodium

Since we know the side length of the unit cell, we can easily calculate the cell volume: Cell volume (CCP) = (side length (CCP))\(^3\) Cell volume (CCP) = \((2.52 \times 10^{-8}\mathrm{~cm})^3\) = \(1.60 \times 10^{-23} \mathrm{~cm}^3/cell\) Since there are four atoms per unit cell in a CCP structure, we can calculate the atomic volume by dividing the cell volume by the number of atoms per unit cell and then multiply by Avogadro's number: Atomic volume (CCP) = \((1.60 \times 10^{-23} \mathrm{~cm}^3/cell) (4 \mathrm{atoms/cell}) (6.02 \times 10^{23} / \mathrm{mol})\) = \( 38.56\ (\mathrm{ cm}^3/\mathrm{mol})\) Now, we can calculate the density of the cubic close-packed sodium: Density (CCP) = molar mass of sodium / atomic volume (CCP) Density (CCP) = \((22.99 \mathrm{~g} / \mathrm{mol}) / (38.56 \mathrm{~cm}^3 / \mathrm{mol}) = 0.596 \mathrm{~g} / \mathrm{cm}^3\)

Step 8: (b) Determine if CCP sodium would float on water

The density of water is approximately \(1 \mathrm{~g} / \mathrm{cm}^3\). For sodium to float on water, its density must be less than the density of water. Since the density of cubic close-packed sodium is \(0.596 \mathrm{~g} / \mathrm{cm}^3\), which is less than the density of water, it would float on water.

Key Concepts

Density and Molar Mass Relationship

Understanding the relationship between density and molar mass is vital when dealing with material properties. Density is defined as mass per unit volume and depends on the amount of substance (measured in moles) and how much space it occupies. The molar mass, on the other hand, is the mass of one mole of a substance.

When you have a pure element or compound, the density can provide insights into its molecular or atomic structure. For example, in a body-centered cubic or a cubic close-packed structure, you can determine the volume occupied by the atoms using the molar mass and density. By rearranging the density formula, \( \text{Volume} = \frac{\text{molar mass}}{\text{density}} \), and using Avogadro's number to relate moles to the number of atoms, you can deduce atomic volumes and, subsequently, atomic radii.

In practical exercises like the one we have for sodium, the calculation starts with the molar mass and density of the element to find the volume that one mole would occupy. Knowing this volume, you can delve deeper into the lattice structure to make further estimations about atomic positioning and spacing.

Body-Centered Cubic Structure

A body-centered cubic (BCC) structure is one of the ways atoms can be arranged in a crystal lattice. In this structure, each cube-shaped unit cell contains one atom at each of its eight corners and one atom at the center, making a total of two atoms per unit cell. The corner atoms are shared among eight unit cells, which is why only one-eighth of each corner atom belongs to a unit cell, leading to the two-atom count after accounting for the one at the center.

To understand the BCC structure, it's required to grasp the concept of the unit cell. The BCC’s unit cell's side length can be calculated from the cell volume, and then the atomic radius can be estimated as the atoms are assumed to touch along the body diagonal.

In our sodium example, calculations of the atomic radius take into account the length of the cube's side and the geometry of the cube to understand the spacing of the atoms within the BCC lattice. This mathematical approach is essential to determine the atomic radius accurately, which in turn can be used to understand various physical properties of the metal.

Cubic Close-Packed Structure

Another common structure in crystalline materials is the cubic close-packed (CCP) structure, also known as face-centered cubic (FCC). This lattice structure is characterized by atoms located at each of the corners and the centers of all the cube faces. Each unit cell in a CCP structure effectively contains four atoms— one-eighth of an atom at each of the eight corners and one-half of an atom on each of the six faces.

The CCP structure is known for its high packing efficiency and is found in many metals. Calculating properties like the atomic radius within a CCP structure requires understanding the cell volume and how atoms are arranged within that volume. The steps taken in the example show that once you know the atomic volume from the molar volume and Avogadro's number, determining the cell volume for a CCP structure and, subsequently, its density is straightforward.

The exercise demonstrates that if sodium adopted a CCP structure, its density would decrease compared to its BCC structure. This fact supports the concept that different structures of the same substance can have different physical properties, such as density, which was crucial in determining whether sodium would float on water — a question grounded in real-world applications of material science.