Problem 150 - Chemistry and Chemical Reactivity

Question

Estimating the radius of a lead atom. (a) You are given a cube of lead that is \(1.000 \mathrm{cm}\) on each side. The density of lead is \(11.35 \mathrm{g} / \mathrm{cm}^{3} .\) How many atoms of lead are in the sample? (b) Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the available space. As an approximation, assume that \(60 \%\) of the space of the cube is filled with spherical lead atoms. Calculate the volume of one lead atom from this information. From the calculated volume (V) and the formula \((4 / 3) \pi r^{3}\) for the volume of a sphere, estimate the radius ( \(r\) ) of a lead atom.

Step-by-Step Solution

Answer

The radius of a lead atom is approximately \(2.62 \times 10^{-8}\) cm.

Step 1: Calculate Volume of Lead Cube

The volume of the cube is given by the formula for the volume of a cube: \[ V_{ ext{cube}} = ext{side}^3 = (1.000 ext{ cm})^3 = 1.000 ext{ cm}^3 \]

Step 2: Determine the Mass of Lead

Using the density formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), solve for mass:\[ \text{Mass} = \text{Density} \times \text{Volume} = 11.35 \text{ g/cm}^3 \times 1.000 \text{ cm}^3 = 11.35 \text{ g} \]

Step 3: Calculate Number of Lead Atoms

Using the molar mass of lead \( 207.2 \text{ g/mol} \) and Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mol}\), determine the number of atoms:\[ \text{Moles of lead} = \frac{11.35 \text{ g}}{207.2 \text{ g/mol}} = 0.05476 \text{ mol} \]\[ \text{Atoms of lead} = 0.05476 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 3.30 \times 10^{22} \text{ atoms} \]

Step 4: Calculate Volume Occupied by Spherical Atoms

Calculate the sphere-occupied volume, assuming 60% of the cube is filled with atoms:\[ V_{ ext{spheres}} = 0.60 \times 1.000 \text{ cm}^3 = 0.600 \text{ cm}^3 \]

Step 5: Determine Volume of a Single Lead Atom

The volume of one lead atom is:\[ V_{ ext{atom}} = \frac{V_{ ext{spheres}}}{\text{Number of atoms}} = \frac{0.600 \text{ cm}^3}{3.30 \times 10^{22} \text{ atoms}} = 1.818 \times 10^{-23} \text{ cm}^3 \]

Step 6: Calculate the Radius of a Lead Atom

Use the formula for the volume of a sphere, \(V = \frac{4}{3} \pi r^3\), to solve for the radius: \[ 1.818 \times 10^{-23} \text{ cm}^3 = \frac{4}{3} \pi r^3 \]Solve for \(r\):\[ r^3 = \frac{1.818 \times 10^{-23} \text{ cm}^3}{\frac{4}{3} \pi} \]\[ r^3 = 1.363 \times 10^{-23} \text{ cm}^3 \]\[ r = (1.363 \times 10^{-23} \text{ cm}^3)^{1/3} \]\[ r \approx 2.62 \times 10^{-8} \text{ cm} \]

Key Concepts

lead atom

A lead atom is the fundamental building block of the element lead, represented by the symbol Pb in the periodic table. To understand its properties, such as the atomic radius, we explore its behavior in a solid form. In this exercise, we dealt with a cube of solid lead to determine the characteristics of a single lead atom. At the atomic level, each lead atom is considerably small, with its dimensions often expressed in terms of Angstroms or nanometers. Being a heavy metal, lead has a relatively high atomic number of 82, which means it has a dense nucleus and a substantial number of electrons clouding its outer layers. Understanding the size of a lead atom starts with recognizing its atomic mass and how it contributes to the density and mass of a macroscopic piece of lead, from which we can infer its volume and size through mathematical relationships.

spherical atoms

Atoms, including those of lead, are often approximated as spheres. This assumption greatly simplifies calculations in physics and chemistry, even though real atoms may not be perfect spheres due to the complex shapes of electron orbitals. However, the spherical model allows us to use geometry to predict the volume of atoms. In our exercise, knowing that lead atoms are considered to be approximately spherical, we could deduce their spatial distribution within a solid block of lead. By assuming that only 60% of the space in the cubic lead sample is filled by these spherical atoms, we account for empty spaces or 'packing inefficiencies,' which naturally occur in real-world solids. This packing efficiency helps us calculate the number of spheres fitting into the total volume and, consequently, determine the average size of each sphere or atom.

volume calculation

Calculating the volume of an object, like a lead atom treated as a sphere, is pivotal in figuring out other properties such as the radius of the atom. The volume formula for a sphere, \( V = \frac{4}{3} \pi r^3 \), relates a sphere's volume directly to its radius. In our context, given the calculated aggregate volume occupied by the lead atoms in the cubic centimeter of lead, we divided this volume by the total number of atoms to find the volume of a single atom. This figure was then used within the spherical volume formula to solve for the radius. Such calculations link the microscopic scale (single atoms) to the macroscopic measurements (visible volumes), and are essential in fields like material science where understanding atomic-level behavior is crucial for designing and improving materials.

density and mass relationship

The relationship between density, mass, and volume is fundamental in understanding materials. For any material, density is defined as mass per unit volume and is a key parameter in identifying substances. In our exercise, the density of lead, given as 11.35 g/cm³, allowed us to calculate the mass of a lead cube by simply multiplying density by volume. Knowing the mass is vital for further calculations, such as determining the number of moles of elements present, using the substance's molar mass. Here, using the atomic mass of lead, we converted the macroscopic mass to a mole and then to the microscopic level number of atoms, courtesy of Avogadro's number. Thus, by understanding density and mass, we connect observed macroscopic properties to the atomic and molecular scale properties, which are crucial in both scientific inquiry and practical applications.