Problem 102 - 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

To find the number of silver atoms in a $1.000\, \mathrm{cm}$ cube, first calculate the mass of the cube ($10.5\, \mathrm{g}$) and the number of moles ($0.09735\, \mathrm{mol}$). Multiply by Avogadro's number to obtain the total number of atoms ($\approx 5.86 \times 10^{22}\, \mathrm{atoms}$). Then, find the volume of the silver atoms ($0.740\, \mathrm{cm}^3$) and the volume of a single atom ($\approx 1.26 \times 10^{-23}\, \mathrm{cm}^3/\mathrm{atom}$). Finally, use the volume of a sphere formula to calculate the radius of a silver atom, which is approximately $1.44\, \mathrm{Å}$.

Step 1: Calculate the volume and mass of the cube

First, we need to determine the volume of the silver cube. Since it's a cube with an edge length of \(1.000\mathrm{~cm}\), the volume is given by: $$ V_\text{cube} = a^3 = (1.000\mathrm{~cm})^3 = 1.000\mathrm{~cm}^3, $$ where \(a\) is the length of one edge. Next, we'll find the mass of the cube. Given the density of silver \(\rho = 10.5\mathrm{~g/cm^3}\), we have: $$ m_\text{cube} = \rho \times V_\text{cube} = (10.5\mathrm{~g/cm^3})(1.000\mathrm{~cm}^3) = 10.5\mathrm{~g}. $$

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

To find the number of silver atoms in the cube, we'll need the molar mass of silver (\(M_\text{Ag} = 107.87\mathrm{~g/mol}\)). Then we can calculate the number of moles using the mass of the cube and the molar mass: $$ n_\text{cube} = \frac{m_\text{cube}}{M_\text{Ag}} = \frac{10.5\mathrm{~g}}{107.87\mathrm{~g/mol}} \approx 0.09735\mathrm{~mol}. $$ Finally, we can find the number of silver atoms, \(N_\text{cube}\), in the cube using Avogadro's number (\(N_\text{A} = 6.022\times10^{23}\mathrm{~atoms/mol}\)): $$ N_\text{cube} = n_\text{cube} \times N_\text{A} \approx 0.09735\mathrm{~mol}\times 6.022\times10^{23}\mathrm{{atoms/mol}} \approx 5.86\times10^{22} \mathrm{~atoms}. $$

Step 3: Calculate the volume of a single silver atom

Since only \(74\%\) of the volume of the cube is filled with silver atoms, the total volume occupied by the silver atoms is: $$ V_\text{atoms} = 0.74 \times V_\text{cube} = 0.74\times 1.000\mathrm{~cm}^3 = 0.740\mathrm{~cm}^3. $$ To find the volume of a single silver atom, we divide \(V_\text{atoms}\) by the total number of silver atoms: $$ V_\text{single\ atom} = \frac{V_\text{atoms}}{N_\text{cube}}\approx \frac{0.740\mathrm{~cm}^3}{5.86\times10^{22}\mathrm{~atoms}} \approx 1.26\times10^{-23}\mathrm{~cm}^3\mathrm{/atom}. $$

Step 4: Calculate the radius of a silver atom

Since a silver atom can be modeled as a sphere, we can use the formula for the volume of a sphere to calculate the radius of a silver atom. The volume of a sphere is given by: $$ V_\text{sphere} = \frac{4}{3}\pi r^3, $$ where \(r\) is the radius of the sphere. We can now use this formula to solve for the radius of the silver atom: $$ r = \left(\frac{3V_\text{single\ atom}}{4\pi}\right)^{1/3} \approx \left(\frac{3(1.26\times10^{-23}\mathrm{~cm^3})}{4\pi}\right)^{1/3} \approx 1.44\times10^{-8}\mathrm{~cm}. $$ Finally, we'll convert this radius from centimeters to angstroms, knowing that \(1\mathrm{~cm} = 10^8\mathrm{~Å}\): $$ r = 1.44\times10^{-8}\mathrm{~cm} \times \frac{10^8\mathrm{~Å}}{1\mathrm{~cm}} \approx 1.44\mathrm{~Å}. $$ In conclusion, the radius of a silver atom is approximately \(1.44\,\mathrm{Å}\).

Key Concepts

Density of Materials

Density is a fundamental property of materials that relates mass to volume. It's defined as the mass per unit volume and is usually expressed in units such as grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³). For instance, in the given exercise, the density of silver is stated as 10.5 g/cm³. This means that every cubic centimeter of silver has a mass of 10.5 grams.

To find the mass of an object, you can multiply its volume by its density, as demonstrated in the problem where the mass of a cube with a volume of 1.000 cm³ was calculated. Knowing the density of materials helps in understanding how much mass is present in a given volume, which is critical for numerous applications including calculating how many atoms are present in a given mass.

Molar Mass and Moles

The concept of the mole is essential for linking the macroscopic amounts of substances we see and measure to the microscopic level of atoms and molecules. The molar mass of an element is the mass of one mole of its atoms. This is usually expressed in grams per mole (g/mol). For silver in the problem, its molar mass is 107.87 g/mol.

To calculate the number of moles of a substance, you divide the mass of the substance by its molar mass. For example, given the mass of the silver cube, we can find the number of moles by dividing 10.5 g (mass of the cube) by 107.87 g/mol (molar mass of silver), resulting in approximately 0.09735 moles.

This conversion from mass to moles is crucial in chemistry because it allows us to calculate the number of atoms or molecules in a sample using Avogadro's number.

Avogadro's Number

Avogadro's number is a fundamental constant that tells us the number of atoms or molecules in one mole of a substance. It is approximately 6.022 x 10²³ particles per mole. This large number helps bridge the gap between the microscopic world of atoms and the macroscopic world that we experience.

For instance, to determine how many silver atoms are in the cube, we multiply the number of moles (0.09735 mol) by Avogadro's number, resulting in around 5.86 x 10²² atoms.

Understanding Avogadro's number allows chemists to translate their experimental results into the atomic scale, making it a cornerstone of chemical calculations.

Volume of a Sphere

A sphere is a perfectly symmetrical three-dimensional shape and its volume is given by the formula \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius of the sphere. In the context of atoms, which are often assumed to be spherical, this formula is used to calculate the space that an atom occupies.

In the exercise example, we determine the volume of a single silver atom based on its contribution to the cube's volume. Once this volume is found, the radius of the atom is derived using the formula for the volume of a sphere, allowing us to convert this radius into practical units such as angstroms.

Understanding the volume of a sphere is vital for estimating how individual particles fit together in a solid and how much space they occupy.