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.

Short answer

The number of silver atoms in the cube is approximately \(5.86 × 10^{22}\) atoms. The volume of a single silver atom is around \(1.26 × 10^{-23} \,\mathrm{cm}^3\). The radius of a silver atom is approximately 1.44 angstroms.

Step-by-step solution

Calculate the mass of the cube

First, we need to calculate the mass of the silver cube using its density and volume. We are given the density of silver (10.5 g/cm³) and the side length of the cube (1.000 cm). The volume of the cube is given by the formula: Volume = side³ So, the volume of the cube = \(1.000^3 \mathrm{cm}^3 = 1.000 \mathrm{cm}^3\). Now, using the formula: Mass = Volume × Density, We can calculate the mass of silver in the cube: Mass = \(1.000 \mathrm{cm}^3 × 10.5 \frac{\mathrm{g}}{\mathrm{cm}^3} = 10.5 \mathrm{g}\)

Calculate the number of silver atoms in the cube

Now we have the mass of silver. We can use the molar mass of silver to convert this to the number of silver atoms. Silver's molar mass is approximately \(107.87 \frac{\mathrm{g}}{\mathrm{mol}}\). First, we calculate the number of moles of silver in the cube: Moles of silver = \(\frac{\text{Mass of silver}}{\text{Molar mass of silver}}=\dfrac{10.5\,\mathrm{g}}{107.87\,\mathrm{g}/\mathrm{mol}}=0.0973\,\mathrm{mol}\) Now, using Avogadro's number (\(6.022 × 10^{23} \text{atoms/mol}\)), we can find the number of silver atoms in the cube: Number of silver atoms = Moles of silver × Avogadro's number = \(0.0973\,\mathrm{mol} × 6.022 × 10^{23} \text{atoms/mol} ≈ 5.86 × 10^{22} \text{atoms}\)

Calculate the total volume of silver atoms

We are given that only 74% of the cube's volume is occupied by silver atoms. Knowing the volume of the cube and this percentage, we can find the total volume occupied by silver atoms: Total volume of silver atoms = % of volume filled × Volume of the cube = \(0.74 × 1.000 \mathrm{cm}^3 = 0.740 \mathrm{cm}^3\)

Determine the volume of a single silver atom

We now have both the total volume occupied by silver atoms and the total number of silver atoms in that volume. We can use this to find the volume of a single silver atom. Volume of a single silver atom = \(\frac{\text{Total volume of silver atoms}}{\text{Number of silver atoms}} = \dfrac{0.740 \,\mathrm{cm}^3}{5.86 × 10^{22} \,\text{atoms}} ≈ 1.26 × 10^{-23} \,\mathrm{cm}^3\)

Calculate the radius of a silver atom

We will now use the formula for the volume of a sphere to calculate the radius of a silver atom. The formula for the volume of a sphere is: \(V = \frac{4}{3}\pi r^3\) Where V is the volume and r is the radius. We can rearrange the formula to solve for the radius and then insert the volume of a single silver atom: \(r = \sqrt[3]{\dfrac{3V}{4\pi}} = \sqrt[3]{\dfrac{3 × 1.26 × 10^{-23}\, \mathrm{cm}^3}{4\pi}} ≈ 1.44 × 10^{-8}\, \mathrm{cm}\) Finally, we can convert the radius from centimeters to angstroms by using the conversion factor \(1\, \mathrm{cm} = 10^{8}\, \mathrm{angstrom}\). Radius of a silver atom = \(1.44 × 10^{-8}\, \mathrm{cm} × 10^{8}\, \mathrm{angstrom/cm} ≈ 1.44 \mathrm{angstrom}\) So, the radius of a silver atom is approximately 1.44 angstroms.

Key concepts

Density

Density is a fundamental property of matter that is used in chemistry to relate the mass of a substance to its volume. It's expressed as mass per unit volume, typically in grams per cubic centimeter (g/cm³). The concept of density helps you understand how "packed" or concentrated a substance is. For silver, we know the density is 10.5 g/cm³.

To find the mass of an object knowing its density, you can use the formula:

• Density = Mass/Volume

Rearranging gives us:

• Mass = Density × Volume

This shows how you can use density to convert volume (a measure of size) into mass (a measure of amount).

In the context of the original exercise, we calculated the mass of a small, one cubic centimeter block of silver. Knowing both its density (10.5 g/cm³) and its volume (1.000 cm³), we were able to determine its mass (10.5 g). This relationship becomes crucial when we want to link physical space or dimensions in chemistry to more abstract measures like moles and atoms.

Avogadro's Number

Avogadro's Number is a key constant in chemistry, representing the number of atoms, ions, or molecules in one mole of substance. It is approximately equal to \(6.022 \times 10^{23}\) entities per mole. This number allows chemists to count particles by weighing them.

To find the number of atoms in a sample, such as our cube of silver, we first need to know the mass and convert that mass into moles. This involves dividing the mass by the molar mass of the substance.

• For silver: Molar mass ≈ 107.87 g/mol

Once we had the moles of silver (here, about 0.0973 mol), multiplying by Avogadro's Number gave us the total number of atoms. This relationship allows conversion between macroscopic (grams) and microscopic (atoms) scales in chemistry.

Atomic Radius

The atomic radius is the distance from the center of an atom's nucleus to its outermost electron shell. However, because atoms are not solid spheres and often "fuzzy", the radius is not fixed and varies slightly based on context, such as bonding states.

In the exercise, after calculating the volume occupied by silver atoms in the cube, the next task relied on the formula for the volume of a sphere, since atoms were approximated as spheres. The formula is:

• Volume \(V = \frac{4}{3}\pi r^3\)

Where \(r\) is the atomic radius. By rearranging this formula, we found \(r\), the radius, for a silver atom. Importantly, radius measurements are often converted into angstroms because these are more convenient for describing atomic-scale distances. One angstrom equals \(10^{-10}\) meters or \(10^{-8}\) centimeters.

Volume Calculation

Volume calculations are essential in understanding the physical properties of substances. For our exercise, we dealt with two main volumes: that of the cube itself and the space the atoms occupy. Calculations of both related directly to understanding how matter fills a certain space.

First, we calculated the cube's volume using the straightforward formula: 

• Volume = side³ = \(1.000^3\) cm³ = 1.000 cm³

Next, to find out how much of this volume was "filled" by the atoms, we used the packing efficiency of silver atoms in solid form, known to be 74%. Thus, the volume actually filled by atoms is:

• 0.74 × Cube's Volume = 0.740 cm³

Then, to find the volume of a single silver atom, we divided this total occupied volume by the number of atoms, allowing us to describe each atom individually. This breakdown clearly illustrates volume allocation and provides a basis for further calculations, like determining atomic radius.