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.

Short answer

The radius of a lead atom is approximately \(3.5 \times 10^{-8} \, \text{cm}\).

Step-by-step solution

Calculate the Mass of Lead

Given the cube's dimensions and density, calculate its mass. Volume of the cube = \(1.000 \, \text{cm}^3\). Density of lead = \(11.35 \, \text{g/cm}^3\). Mass = Density \( \times \) Volume = \(11.35 \, \text{g/cm}^3 \times 1.000 \, \text{cm}^3 = 11.35 \, \text{g}\).

Determine Atoms in the Cube

First, find the number of moles of lead in the cube.Using the molar mass of lead \( = 207.2 \, \text{g/mol}\), calculate:\[ \text{Moles of lead} = \frac{11.35 \, \text{g}}{207.2 \, \text{g/mol}} \approx 0.0548 \, \text{mol} \] Using Avogadro's number \(6.022 \times 10^{23} \) atoms/mol, calculate the number of atoms:\[ \text{Number of atoms} = 0.0548 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 3.30 \times 10^{22} \, \text{atoms} \]

Calculate Space Occupied by Atoms

Assume that only 60% of the cube's volume is occupied by atoms.Calculate actual occupied volume: \(0.60 \times 1.000 \, \text{cm}^3 = 0.600 \, \text{cm}^3\).

Find Volume of One Atom

Use the occupied volume to find the volume of a single lead atom:\[ \text{Volume of one lead atom} = \frac{0.600 \, \text{cm}^3}{3.30 \times 10^{22} \, \text{atoms}} \approx 1.82 \times 10^{-23} \, \text{cm}^3 \]

Estimate the Radius of a Lead Atom

Use the formula for the volume of a sphere \( V = \frac{4}{3} \pi r^3 \)Set \( V = 1.82 \times 10^{-23} \, \text{cm}^3 \)\[ r^3 = \frac{V \times 3}{4 \pi} = \frac{1.82 \times 10^{-23} \, \text{cm}^3 \times 3}{4 \pi} \approx 4.34 \times 10^{-24} \, \text{cm}^3 \] \[ r = \left(4.34 \times 10^{-24} \right)^{1/3} \approx 3.5 \times 10^{-8} \, \text{cm} \]

Conclusion

The estimated radius of a lead atom is approximately \(3.5 \times 10^{-8} \, \text{cm}\).

Key concepts

Density of Lead

The term "density" is crucial for understanding the properties of materials. In the case of lead, its density is given as 11.35 g/cm³. This means that every cubic centimeter of lead weighs 11.35 grams. Density is calculated as mass divided by volume. Therefore, it plays an essential role in determining how much mass a certain volume of a material has.

Understanding density helps clarify why lead is relatively heavy compared to other metals. Due to its high density, lead has numerous applications in industries such as construction and radiation shielding.

To calculate density, use the formula:

• Density = Mass/Volume

This relationship allows us to easily find one of the variables if the others are known, making it straightforward to calculate the volume or mass of lead when its density is a given factor. High-density materials like lead can help students understand the impact of material choice on design and functionality.

Molar Mass of Lead

Molar mass is another key term when dealing with chemical elements. For lead, its molar mass is approximately 207.2 g/mol. This implies that one mole of lead, which is about Avogadro's number of atoms, weighs 207.2 grams.

Molar mass serves as a bridge between the mass of a substance and the amount of substance present in moles, making stoichiometric calculations possible. This is particularly useful in chemistry and physics.

Understanding Moles and Molar Mass

A mole is a unit used to express amounts of a chemical substance. Its significance lies in its capacity to connect macroscopic quantities that we can measure with microscopic quantities that we cannot directly observe. With Avogadro's number, you can determine how many atoms or molecules are in a mole.

When working with elements like lead in reactions or processes, knowing the molar mass allows you to translate the data you have into useful information about the structure at an atomic level. It's essential for converting between grams and moles, using the formula:

• Moles of substance = Mass of substance / Molar mass

Avogadro's Number

Avogadro's number is a quintessential concept in chemistry, represented with the value: \(6.022 \times 10^{23}\). This number signifies the number of atoms or molecules contained in one mole of any substance. It's like a bridge connecting the microscopic world of atoms to the macroscopic world we can see and measure.

Why Use Avogadro's Number?

Because atoms are incredibly small and practically innumerable in even tiny amounts of matter, Avogadro's number offers a usable quantity to relate atomic-scale events to observable scales. We can use it to determine how many atoms or molecules are present in a given mass of a substance.

In practical applications, this number aids in comprehending chemical reactions quantitatively. It tells us exactly how many atoms are participating in a reaction and helps to quantify and predict the outcomes accurately.

Use the following relationship:

• Number of Particles = Moles × Avogadro's Number

Volume of a Sphere

The volume of a sphere is a fundamental concept in geometry and is particularly useful in estimating the size of spherical atoms, like the lead atoms in this exercise. The formula for calculating the volume of a sphere is:

• Volume \( V = \frac{4}{3} \pi r^3 \)

where \( r \) is the radius of the sphere. This formula helps in estimating the radius when the volume is known, as demonstrated in the original problem. Understanding how to manipulate this formula allows us to solve for the radius when the volume is a given variable.

Application in Atomic Radius Estimation

When atoms are assumed to be spherical, as in this case with lead, using the sphere volume formula is a convenient way to estimate atomic radii. For example, you can plug in the known volume of one atom to solve for the radiusUsing:

• \( r = \left( \frac{3V}{4\pi} \right)^{1/3} \)

This approach is crucial in fields like materials science and nanotechnology, where understanding and manipulating atomic scales are fundamentally important.