Question

The radius of an atom of gold $(\mathrm{Au})$ is about $1.35\text{AA}$. (a) Express this distance in nanometers $(\mathrm{nm})$ and in picometers $(\mathrm{pm})$. (b) How many gold atoms would have to be lined up to span $1.0\mathrm{mm}$ ? (c) If the atom is assumed to be a sphere, what is the volume in ${\mathrm{cm}}^3$ of a single $\mathrm{Au}$ atom?

Short answer

The radius of a gold atom is 1.35 Å, which is equivalent to 0.135 nm and 135 pm. To span a distance of 1.0 mm, approximately $7.41\times {10}^6$ gold atoms would be needed. Assuming the gold atom is a sphere, the volume of a single gold atom is approximately $1.03\times {10}^{-23}{\mathrm{cm}}^3$.

Step-by-step solution

Convert the given distance to nanometers and picometers

To express the given distance (1.35 Å) in nanometers (nm) and picometers (pm), we first need to find the conversion factors between the various units. 1 Å = 0.1 nm = 100 pm. Now, to convert the distance (1.35 Å) to nanometers and picometers, we can multiply by the respective conversion factors: To nanometers: 1.35 Å × 0.1 nm/Å To picometers: 1.35 Å × 100 pm/Å

Calculate the number of gold atoms required to span 1.0 mm

First, let's convert 1.0 mm to Å as this is the unit of measurement we have for the gold atom's radius: 1.0 mm = 10^7 Å Since we know the diameter of a gold atom (twice the radius), we can determine how many gold atoms would be needed to span 1.0 mm: Number of gold atoms = Total distance (Å) / Diameter of a gold atom (Å)

Determine the volume of a single gold atom

We can assume that the gold atom is a sphere, and use the formula for the volume of a sphere: $V=\frac{4}{3}\pi r^3$, where V is the volume, and r is the radius of the sphere. Using the given radius value in Å, first convert it to cm: 1.35 Å = 1.35 × 10^{-8} cm After converting the radius to cm, we can calculate the volume of a single gold atom by plugging the radius into the sphere volume formula. Remember to convert the volume to cubic centimeters (${\mathrm{cm}}^3$).

Key concepts

Unit Conversion

Unit conversion is the process of converting a measure of physical quantity from one unit to another. It's an essential skill, particularly in science and engineering, because different fields and countries use different measurement systems. To convert units, you need the appropriate conversion factor, which is a numerical value that allows you to multiply or divide to perform the conversion.

For example, in the exercise, the radius of a gold atom is given in angstroms (Å), a unit commonly used in atomic physics. However, it's often necessary to convert this to more familiar units like nanometers (nm) and picometers (pm), which are more commonly used in different branches of science. To make these conversions:

• 1 Å (angstrom) is equal to 0.1 nm (nanometers).
• 1 Å is also equal to 100 pm (picometers).

Therefore, to convert the atomic radius from angstroms to nanometers or picometers, we multiply the value by the conversion factors. For instance,$1.35\text{ Å}\times 0.1\frac{\text{nm}}{\text{Å}}=0.135\text{ nm}$, and $1.35\text{ Å}\times 100\frac{\text{pm}}{\text{Å}}=135\text{ pm}$. Such conversions are not just academic exercises; they are crucial for understanding the scale at which atomic-level phenomena occur.

Avogadro's Number

Avogadro's number, named after the scientist Amedeo Avogadro, is the number of constituent particles, often atoms or molecules, that are contained in one mole of a substance. One mole is defined as exactly $6.02214076\times {10}^{23}$ particles of the substance. This constant is fundamental in the realm of chemistry because it establishes a link between the macroscopic and microscopic worlds, allowing for the calculation of particles in a given amount of material.

In calculations involving Avogadro's number, it's often needed to determine how many atoms, ions, or molecules are in a certain sample of substance. For instance, in a related exercise, we might be asked to determine how many atoms of gold are present in a mole of gold. Using Avogadro's number, we can calculate that one mole of gold, weighing about 197 grams based on its atomic mass, contains $6.02214076\times {10}^{23}$ atoms. This concept aids in understanding the scale and amount of substances involved in chemical reactions and is a cornerstone of stoichiometry in chemistry.

Volume of a Sphere

The volume of a sphere is a measure of how much space it occupies, and it's crucial when dealing with spherical objects, such as atoms in this case. The formula to calculate the volume of a sphere is $V=\frac{4}{3}\pi r^3$, where $V$ represents the volume and $r$ is the radius of the sphere.

In the exercise, it’s assumed that a gold atom is spherical, which allows us to use this volume formula to determine how much space a single atom occupies. Given the atomic radius in angstroms, one needs to convert it to a more standard volume unit, such as cubic centimeters, before using the formula. After conversion, you plug in the radius thus converted to calculate the volume. This helps in visualizing the space taken by an atom, and when scaled up, it helps understand the structure and properties of materials at the atomic level. It's this kind of calculation that can feed into broader applications like understanding the porosity of materials or the efficiency of packing in crystal structures.