Problem 107 - 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

In summary: (a) There are \(5.86 \times 10^{22}\) silver atoms in the cube. (b) The volume of a single silver atom is \(1.26 \times 10^{-23}\) cm³. (c) The radius of a silver atom is 1.78 angstroms.

Step 1: Determine the volume of the cube

The volume (V) of a cube of edge length (s) can be calculated by the formula: \[V = s^3\] Given edge length s = 1.000 cm, the volume (V) is: \[V = (1.000)^3 = 1.000\,\text{cm}^3\]

Step 2: Calculate the mass of the cube using density

The mass (m) of an object is related to its density (ρ) and volume (V) by the formula: \[m = ρ \cdot V\] Using the given density of silver, ρ = 10.5 g/cm³, we can find the mass of the silver cube: \[m = (10.5\,\text{g/cm}^3)(1.000\,\text{cm}^3) = 10.5\,\text{g}\] #b. Calculate the number of atoms in the cube#

Step 3: Determine the molar mass of Silver (Ag)

From the periodic table, the molar mass of silver is: \(M_{Ag} = 107.87 \frac{\text{g}}{\text{mol}}\)

Step 4: Find the number of moles in the cube

The number of moles (n) is calculated using the mass (m) and molar mass (M) through the formula: \[n = \frac{m}{M}\] Using the mass of the silver cube, we can calculate the number of moles: \[n = \frac{10.5\,\text{g}}{107.87 \frac{\text{g}}{\text{mol}}} = 0.0973\,\text{mol}\]

Step 5: Calculate the number of atoms in the cube

To find the number of atoms (N), we multiply the number of moles (n) by Avogadro's number (\(6.022 \times 10^{23}\)): \[N = n \times N_A\] \[N = 0.0973\,\text{mol} \times 6.022 \times 10^{23}\,\frac{\text{atoms}}{\text{mol}} = 5.86 \times 10^{22}\,\text{atoms}\] #c. Calculate the volume of a single Silver atom#

Step 6: Determine the volume occupied by silver atoms

Using the given information that 74% of the volume of the solid is filled with silver atoms, we can find the volume occupied by silver atoms (V_occupied): \[V_\text{occupied} = 0.74 \times 1.000\,\text{cm}^3 = 0.740\,\text{cm}^3\]

Step 7: Calculate the volume of a single silver atom

Given the number of atoms in the cube, we can divide the occupied volume by the number of atoms to find the volume of a single silver atom (V_atom): \[V_\text{atom} = \frac{V_\text{occupied}}{N}\] \[V_\text{atom} = \frac{0.740\,\text{cm}^3}{5.86 \times 10^{22}\,\text{atoms}} = 1.26 \times 10^{-23}\,\text{cm}^3\,\text{per atom}\] #d. Calculate the radius of a Silver atom#

Step 8: Use the formula for the volume of a sphere

The volume (V) of a sphere is given by the following formula, involving the sphere's radius (r): \[V = \frac{4}{3}\pi r^3\] Now, we have the volume of an individual silver atom, V_atom. We can isolate r and solve for it: \[r = \sqrt[3]{\frac{3V_\text{atom}}{4\pi}}\]

Step 9: Calculate the radius in centimeters

Using the volume of a single silver atom, we calculate the radius in centimeters: \[r = \sqrt[3]{\frac{3 \times 1.26 \times 10^{-23}\,\text{cm}^3}{4\pi}} = 1.78 \times 10^{-8}\,\text{cm}\]

Step 10: Convert the radius to angstroms

To convert the radius from centimeters to angstroms, we multiply by the conversion factor (\(1 \,\text{cm} = 10^8 \, \text{Å}\)): \[r = 1.78 \times 10^{-8}\,\text{cm} \times \frac{10^8\,\text{Å}}{1\,\text{cm}} = 1.78\,\text{Å}\] In summary: (a) There are \(5.86 \times 10^{22}\) silver atoms in the cube. (b) The volume of a single silver atom is \(1.26 \times 10^{-23}\) cm³. (c) The radius of a silver atom is 1.78 angstroms.

Key Concepts

Density of Silver

Density is a crucial property that relates mass to volume. For silver, the given density is typically expressed as \( 10.5 \text{ g/cm}^3 \). This means that for every one cubic centimeter of silver, it weighs 10.5 grams. Knowing the density allows us to calculate the mass of any silver object if we know its volume. In our case, we have a cube with a volume of \( 1.000 \text{ cm}^3 \), so the cube's mass can be calculated by multiplying its volume by the density:

• Density formula: \( \rho = \frac{m}{V} \)
• Mass calculation: \( m = 10.5 \text{ g/cm}^3 \times 1.000 \text{ cm}^3 = 10.5 \text{ g} \)

With this simple multiplication, we establish that the mass of our silver cube is 10.5 grams.

Molar Mass of Silver

The molar mass of a substance tells us how much one mole of that substance weighs. For silver (Ag), the molar mass is \( 107.87 \text{ g/mol} \). This information is directly retrieved from the periodic table.
When we know the mass of our sample, we can calculate how many moles it constitutes. This is done by dividing the sample's mass by the molar mass. In our exercise, the silver cube weighs 10.5 grams. To find the number of moles of silver, use the following steps:

• Formula: \( n = \frac{m}{M} \)
• Calculation: \( n = \frac{10.5 \text{ g}}{107.87 \text{ g/mol}} = 0.0973 \text{ mol} \)

Understanding the concept of molar mass lets us transition from mass to moles, which is critical for further calculations involving Avogadro's number.

Avogadro's Number

Avogadro's number \( ( 6.022 \times 10^{23} \text{ atoms/mol} ) \) is a fundamental constant used to convert moles of a substance into the number of atoms or molecules present. It represents the number of atoms contained in one mole of a substance, providing a bridge between the macroscopic and atomic worlds.
We calculated 0.0973 moles of silver in our cube. To find the number of atoms, we need to multiply this number by Avogadro's number.

• Formula: \( N = n \times N_A \)
• Atom calculation: \( N = 0.0973 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 5.86 \times 10^{22} \text{ atoms} \)

Thus, there are \( 5.86 \times 10^{22} \) silver atoms in our cube. This calculation is essential for understanding the sheer scale of atomic structures.

Volume and Radius Calculations

Volume and radius calculations are necessary to understand the physical dimensions of silver atoms and their arrangement in a metal. Given that the atoms occupy 74% of the total cube volume due to packing, we first calculate the actual volume occupied by the atoms:

• Volume occupied: \( V_\text{occupied} = 0.74 \times 1.000 \text{ cm}^3 = 0.740 \text{ cm}^3 \)
• Volume of a single atom: \( V_\text{atom} = \frac{0.740 \text{ cm}^3}{5.86 \times 10^{22} \text{ atoms}} = 1.26 \times 10^{-23} \text{ cm}^3 \)

To find the radius of a silver atom, we use the volume formula for spheres \( V = \frac{4}{3}\pi r^3 \), solving for the radius \( r \):

• Radius formula: \( r = \sqrt[3]{\frac{3V_\text{atom}}{4\pi}} \)
• Calculation: \( r = \sqrt[3]{\frac{3 \times 1.26 \times 10^{-23} \text{ cm}^3}{4\pi}} = 1.78 \times 10^{-8} \text{ cm} \)

Voila, converting this radius to Angstroms using the conversion \( 1 \text{ cm} = 10^8 \text{ Å} \), gives \( 1.78 \text{ Å} \). These calculations help visualize the size of atoms in a tangible way.