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

(a) There are approximately \(5.85 \times 10^{22}\) silver atoms in the cube. (b) The volume of a single silver atom is approximately \(1.26 \times 10^{-23} \mathrm{ ~cm}^{3} \mathrm{~/atom}\). (c) The radius of a silver atom is approximately \(144\,\mathrm{angstroms}\).

Step-by-step solution

Find the mass of the cube

First, we find the mass of the cube by using the density formula. The density of silver is given as \(10.5\) \(\mathrm{g} / \mathrm{cm}^{3}\), and the edge of the cube is \(1.000 \mathrm{~cm}\). Let's find the volume \(\mathrm{cm}^{3}\) and then its mass. Volume = edge length x edge length x edge length = \(1.0 \times 1.0 \times 1.0 \mathrm{~cm}^{3} = 1.0 \mathrm{~cm}^{3}\) Now, we can calculate the mass of the cube. Mass = Volume × Density = \(1.0 \mathrm{~cm}^{3} \times 10.5 \mathrm{g}/\mathrm{cm}^{3} = 10.5 \mathrm{ ~g}\)

Calculate the number of silver atoms in the cube

To find the number of atoms in the cube, we will use the molar mass of silver (\(107.87 \mathrm{~g/mol}\)) to calculate the number of moles. Number of moles = Mass / Molar mass = \(10.5 \mathrm{~g} / 107.87 \mathrm{~g/mol} \approx 0.0973 \mathrm{ ~mol}\) Now, we can find the number of atoms using Avogadro's number, \(N_A = 6.022 \times 10^{23} \mathrm{~atoms/mol}\). Number of atoms = Number of moles × Avogadro's number = \(0.0973 \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{~atoms/mol} \approx 5.85 \times 10^{22} \mathrm{ ~atoms}\)

Calculate the volume of one silver atom

Given that silver atoms occupy \(74 \%\) of the total volume, we can calculate the volume occupied by the atoms. Volume occupied by atoms = Total volume × Packing efficiency = \(1.0 \mathrm{~cm}^{3} \times 0.74 \approx 0.74 \mathrm{ ~cm}^{3}\) Next, we can find the volume of a single silver atom by dividing the total volume occupied by atoms by the number of atoms. Volume of one silver atom = Volume occupied by atoms / Number of atoms = \(0.74 \mathrm{ ~cm}^{3} / (5.85 \times 10^{22} \mathrm{ ~atoms}) \approx 1.26 \times 10^{-23} \mathrm{ ~cm}^{3} \mathrm{~/atom}\)

Calculate the radius of a silver atom

To find the radius of a silver atom, we will use the volume of a sphere formula: \(V = \frac{4}{3} \pi r^3\), where \(V\) is the volume and \(r\) is the radius of the sphere (the silver atom). First, rearrange the formula to solve for the radius: \(r = \sqrt[3]{\frac{3V}{4\pi}}\) Now, we can plug in the volume of one silver atom: \(r \approx \sqrt[3]{\dfrac{3 \times 1.26 \times 10^{-23} \mathrm{ ~cm}^{3}}{4\pi}} \approx 1.44 \times 10^{-8} \mathrm{ ~cm}\) Finally, we'll convert the radius from centimeters to angstroms: \(r \approx 1.44 \times 10^{-8} \mathrm{ ~cm} \times \frac{10^{10} \mathrm{~angstroms}}{1 \mathrm{~cm}} = 1.44 \times 10^{2} \mathrm{ ~angstroms}\) So, the radius of a silver atom is approximately \(144 \mathrm{ ~angstroms}\).

Key concepts

Density Calculation

Understanding density calculation is critical in various scientific fields. In general terms, density is the measure of how much mass is contained within a specific volume. It's a simple yet powerful concept that helps us quantify the concentration of matter in an object. If you have a solid block of a material, like silver in our example, its density can be calculated using the formula:
\( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).
In the provided exercise, you measure the mass of a silver cube by multiplying its given density with the volume, which is conveniently a 1-cubic centimeter cube. This makes the mass calculation straightforward:
\( \text{Mass} = 1.0 \, \text{cm}^{3} \times 10.5 \, \text{g/cm}^{3} = 10.5 \, \text{g} \).
This foundational step is essential before proceeding to more complex parts of the problem, such as determining the number of atoms within the cube.

Avogadro's Number

Diving deeper into our exploration, we encounter Avogadro's number, a cornerstone of chemistry and physics. Avogadro's number, denoted by \(N_A\), is approximately \(6.022 \times 10^{23}\) and represents the number of units (often atoms or molecules) in one mole of a substance. It bridges the macroscopic and atomic scales by allowing scientists to work with amounts of substance in practical, measurable terms.
Using the mass of the silver cube calculated in the first step and the molar mass of silver (\(107.87 \, \text{g/mol}\)), we employ Avogadro's number to transition from macroscopic mass to microscopic quantification of atoms:
\( \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \) and then \( \text{Number of atoms} = \text{Number of moles} \times N_A \).
This yields the staggering number of atoms contained in a seemingly tiny cube of silver, exemplifying how Avogadro's number links tangible measurements to the world of the unseen.

Atomic Volume

Another exciting concept illustrated by this exercise is the volume of a single atom, often referred to as atomic volume. In solids, atoms are packed together in specific arrangements, and the atomic volume can be found by considering the proportion of space these atoms actually occupy, known as packing efficiency.
In the case of silver, the atoms fill about \(74\%\) of the cube's volume. To calculate the volume of an individual silver atom, we divided the cube's volume occupied by atoms (which comes from the product of the total volume and the packing efficiency) by the number of silver atoms obtained using Avogadro's number:
\( \text{Volume of one silver atom} = \frac{\text{Volume occupied by atoms}}{\text{Number of atoms}} \).
This calculation grants us insight into the atomic scale—understanding the size and space each atom inhabits is vital for grasping properties of materials and phenomena in physics and chemistry.