Problem 106 - Chemistry The Central Science

Question

(a) You are given a cube of silver metal that measures \(1.000 \mathrm{~cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

Step-by-Step Solution

Answer

The number of silver atoms in the cube is approximately \(5.87 \times 10^{22}\) atoms. The volume of a single silver atom is approximately \(1.26 \times 10^{-23} \mathrm{~cm}^3\). The radius of a silver atom is approximately 1.63 Å.

Step 1: Calculate the mass of the silver cube

To calculate the mass of the silver cube, we can multiply its volume (since it's a cube with each side measuring 1.000 cm, the volume is equal to \(1.000^3 = 1.000 \mathrm{~cm}^3\)) by its density: \[ \text{mass} = \text{volume} \times \text{density} = 1.000 \mathrm{~cm}^3 \times 10.5 \mathrm{~g} / \mathrm{cm}^3 = 10.5 \mathrm{~g}. \]

Step 2: Calculate the number of silver atoms in the cube

To calculate the number of silver atoms in the cube, first we need the molar mass of silver - which is 107.87 g/mol. Then, we can convert the mass of the silver cube, calculated in Step 1, into moles of silver, and finally calculate the number of atoms using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol): \[ \text{moles of silver} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.5 \mathrm{~g}}{107.87 \mathrm{~g/mol}}. \] Now, multiplying the moles of silver by Avogadro's number to obtain the number of atoms: \[ \text{number of atoms} = \text{moles of silver} \times \text{Avogadro's number} = \frac{10.5 \mathrm{~g}}{107.87 \mathrm{~g/mol}} \times 6.022 \times 10^{23} \mathrm{~atoms/mol} \approx 5.87 \times 10^{22} \mathrm{~atoms}. \]

Step 3: Calculate the volume of a single silver atom

We are given that 74% of the volume of the cube is filled with silver atoms, so we can calculate the total volume occupied by these atoms: \[ \text{volume of atoms} = \text{total volume of cube} \times \frac{74}{100} = 1.000 \mathrm{~cm}^3 \times 0.74 = 0.740 \mathrm{~cm}^3. \] Now, divide the volume occupied by silver atoms by the number of silver atoms to get the volume of a single silver atom: \[ \text{volume of a single atom} = \frac{\text{volume of atoms}}{\text{number of atoms}} = \frac{0.740 \mathrm{~cm}^3}{5.87 \times 10^{22} \mathrm{~atoms}} \approx 1.26 \times 10^{-23} \mathrm{~cm}^3. \]

Step 4: Calculate the radius of a silver atom

We know that the volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), where \(V\) is the volume and \(r\) is the radius. Using the volume of a single silver atom obtained from Step 3, we can solve for the radius. First, rearrange the formula for the volume of a sphere to find the radius: \[ r = \sqrt[3]{\frac{3V}{4\pi}}. \] Now,Plug in the volume of a single silver atom calculated in step 3: \[ r = \sqrt[3]{\frac{3(1.26 \times 10^{-23} \mathrm{~cm}^3)}{4\pi}} \approx 1.63 \times 10^{-8} \mathrm{~cm}. \] Finally, convert radius from cm to angstroms (1 cm = 1 × 10^8 angstroms): \[ r = 1.63 \times 10^{-8} \mathrm{~cm} \times \frac{1 \times 10^{8} \mathrm{~\AA}}{1 \mathrm{~cm}} \approx 1.63 \mathrm{~\AA}. \] So, the radius of a silver atom is approximately 1.63 Å.

Key Concepts

Density of Silver

Density is a key property that relates the mass of a material to its volume. For silver, the density is given as \(10.5 \, \text{g/cm}^3\). This means that each cubic centimeter of silver weighs 10.5 grams.
This concept helps us understand how tightly packed the silver atoms are in a given space. By knowing the density:

• You can calculate the mass of a sample if you know its volume.
• It indicates how heavy the material is for a particular space.

In the given problem, we used the density to find the mass of a cube of silver with sides measuring 1 cm each. The mass of this cube is calculated as:
\[1.000 \, \text{cm}^3 \times 10.5 \, \text{g/cm}^3 = 10.5 \, \text{g}\]Understanding the density is crucial when dealing with materials in physics and chemistry, as it often leads to other important calculations, like determining the number of atoms present.

Atomic Structure

The atomic structure of a material defines its basic physical properties. Silver atoms are spherical in nature, which means they do not fill all available space when packed together.
In crystalline metals like silver, atoms usually arrange themselves in specific patterns known as packing structures. This packing results in a fraction of the volume being occupied, often not reaching 100% because of the gaps between spheres.

• For silver, the atoms fill approximately 74% of the available volume.
• The empty space is due to the spherical shape.

This packing efficiency affects calculations for atomic volume and mass. To understand how this affects a silver cube, we calculate that only a portion of the cube is filled with atoms.
Knowing the structure and arrangement helps in understanding how different substances behave, form, and interact at the atomic level.

Avogadro's Number

Avogadro's number is a critical concept in chemistry and physics. It represents the number of atoms or molecules in one mole of a substance, valued at \(6.022 \times 10^{23}\).
This number allows us to convert between the mass of a substance and the number of atoms it contains. Here's how it works in practice:

• First, convert the mass of the silver cube to moles using the molar mass of silver, which is 107.87 g/mol.
• Then, use Avogadro's number to find the total number of atoms.

The calculation steps are:
\[\text{moles of silver} = \frac{10.5 \, \text{g}}{107.87 \, \text{g/mol}}\]
\[\text{number of atoms} = \text{moles of silver} \times 6.022 \times 10^{23}\]
This results in approximately \(5.87 \times 10^{22}\) silver atoms in the cube. Mastering this concept is essential for understanding molecular scales in chemistry.

Volume and Radius Calculations

Calculating the volume and radius of atoms involves understanding the geometry of spheres. Each atom of silver is roughly spherical, filling only a part of the cube's volume due to its shape.

• The volume of a single atom is calculated by dividing the total occupied volume by the number of atoms.
• Then, using the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), we can solve for the radius \(r\).

Given that the calculated volume for one silver atom is \(1.26 \times 10^{-23} \, \text{cm}^3\), we rearrange the sphere formula to find:
\[r = \sqrt[3]{\frac{3V}{4\pi}}\]
By inserting the calculated volume, the radius is found to be approximately \(1.63 \times 10^{-8} \, \text{cm}\). Converting to angstroms (where 1 cm = 1 × 10^8 Å), the radius is about 1.63 Å. Understanding these calculations highlights the spatial arrangement and size of atoms, foundational for many scientific fields.