Problem 19 - Chemistry The Central Science In Si Units

Question

he radius of an atom of tungsten (W) is about \(2.10 \AA\). (a) Express this distance in nanometers (nm) and in picometers (pm). (b) How many tungsten atoms would have to be lined up to create a wire of \(2.0 \mathrm{~mm} ?(\mathbf{c})\) If the atom is assumed to be a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single \(\mathrm{W}\) atom?

Step-by-Step Solution

Answer

The radius of a tungsten atom is 0.21 nm and 210 pm. Approximately 4.76 million atoms line up in 2 mm. The volume of one atom is about \(3.88 \times 10^{-29} \text{ m}^3\).

Step 1: Convert Angstroms to Nanometers

The conversion factor between angstroms and nanometers is 1 Å = 0.1 nm. Given that the radius of a tungsten atom is 2.10 Å, you can convert it to nanometers by multiplying by 0.1. \[ 2.10 \, \text{Å} \times 0.1 \, \frac{\text{nm}}{\text{Å}} = 0.21 \, \text{nm} \]

Step 2: Convert Angstroms to Picometers

The conversion factor between angstroms and picometers is 1 Å = 100 pm. Given that the radius of a tungsten atom is 2.10 Å, you can convert it to picometers by multiplying by 100. \[ 2.10 \, \text{Å} \times 100 \, \frac{\text{pm}}{\text{Å}} = 210 \, \text{pm} \]

Step 3: Calculate Number of Atoms for the Wire

First, convert the wire length from millimeters to angstroms, as 1 mm = 10^7 Å. For a 2.0 mm wire: \[ 2.0 \, \text{mm} \times 10^7 \, \frac{\text{Å}}{\text{mm}} = 2.0 \times 10^7 \, \text{Å} \] Divide this by the diameter of a single tungsten atom (which is twice the radius). \[ \frac{2.0 \times 10^7 \, \text{Å}}{2 \times 2.10 \, \text{Å}} \approx 4.76 \times 10^6 \] Thus, approximately 4.76 million tungsten atoms are needed.

Step 4: Calculate the Volume of a Sphere

The volume of a sphere is given by the formula \( V = \frac{4}{3} \pi r^3 \). First, convert the radius to meters. Given the radius in nanometers is 0.21 nm, and 1 nm = 10^{-9} m: \[ 0.21 \, \text{nm} = 0.21 \times 10^{-9} \, \text{m} = 2.1 \times 10^{-10} \, \text{m} \] Using this radius in the volume formula: \[ V = \frac{4}{3} \pi (2.1 \times 10^{-10} \, \text{m})^3 \approx 3.88 \times 10^{-29} \, \text{m}^3 \]

Key Concepts

Angstrom to Nanometer Conversion

The conversion from angstroms to nanometers is a fundamental unit conversion in chemistry and physics, allowing scientists to express measurements at an atomic scale in terms of slightly larger but more universally used units. An angstrom (Å) is defined as 0.1 nanometers (nm).
To convert from angstroms to nanometers, simply multiply the value in angstroms by the conversion factor 0.1.
For example, if the radius of a tungsten atom is given as 2.10 Å, converting it to nanometers involves:

• Multiplying the angstrom value by 0.1: \(2.10 \times 0.1 = 0.21\) nm.

This straightforward process simplifies visualizing incredibly small distances in more manageable terms.

Angstrom to Picometer Conversion

Similar to the conversion between angstroms and nanometers, changing units to picometers helps provide a different perspective of size at the atomic scale. A picometer (pm) is one-thousandth of a nanometer, which makes it extremely useful for atomic measurements. In essence, 1 Å is equivalent to 100 pm.
Given this relationship, converting an atomic radius from angstroms to picometers involves multiplying the angstrom measurement by 100.
To illustrate, for a tungsten atom with a radius of 2.10 Å, calculate:

• \(2.10 \times 100 = 210\) pm.

Expressing measurements in picometers often aids in calculations that require higher precision at such small scales.

Volume of a Sphere

In geometry, the volume of a sphere is determined using the formula: \(V = \frac{4}{3} \pi r^3\).
Where \(r\) is the radius of the sphere.
This formula captures how much space the sphere occupies.
To apply this to an atom like tungsten, you first need to convert the atomic radius into meters because meters are the SI units for measuring volume.
Given the radius in nanometers, 0.21 nm, convert to meters by knowing 1 nm equals \(10^{-9}\) m:

• \(0.21 \times 10^{-9} = 2.1 \times 10^{-10}\) meters.

Insert this value into the volume formula:

• \(V = \frac{4}{3} \pi (2.1 \times 10^{-10})^3 \approx 3.88 \times 10^{-29}\) m³.

This calculation demonstrates the incredibly diminutive volume of a single atom, emphasizing how densely packed they can be.

Unit Conversion in Chemistry

Unit conversion is a crucial skill in chemistry, enabling scientists to engage in experiments and calculations with diverse measurement systems. It often involves factors across different metric units, like length (angstroms, nanometers, meters), volume, and others.
Successful conversion relies on understanding relationships between units and consistently applying these relationships, as done in earlier conversions. For instance:

• To convert length units, use established conversion factors like 1 Å = 0.1 nm.
• To handle volume, utilize formulas like \(V = \frac{4}{3} \pi r^3\) with appropriately converted metric units.

By mastering unit conversions, chemists ensure accuracy and standardization in various scientific computations, integral for research and applications.