Problem 107 - Chemistry A Molecular Approach

Question

Carbon-12 contains six protons and six neutrons. The radius of the nucleus is approximately \(2.7 \mathrm{fm}\) (femtometers), and the radius of the atom is approximately \(70 \mathrm{pm}\) (picometers). Calculate the volume of the nucleus and the volume of the atom. What percentage of the carbon atom's volume is occupied by the nucleus? (Assume two significant figures.)

Step-by-Step Solution

Answer

The volume of the nucleus is approximately \(1.23 \times 10^{-44} \mathrm{m^3}\), the volume of the atom is approximately \(1.44 \times 10^{-30} \mathrm{m^3}\), and the nucleus occupies about \(8.5 \times 10^{-15}\)% of the carbon atom's volume.

Step 1: Calculate the Volume of the Nucleus

Use the formula for the volume of a sphere, which is \( V = \frac{4}{3}\pi r^3 \) where \(r\) is the radius. The radius of the nucleus is \(2.7 \mathrm{fm} = 2.7 \times 10^{-15} \mathrm{m}\). Plug this value into the volume formula to get the volume of the nucleus.

Step 2: Calculate the Volume of the Atom

Similarly, calculate the volume of the atom using the same volume formula. The radius of the atom is \(70 \mathrm{pm} = 70 \times 10^{-12} \mathrm{m} \). Convert picometers to meters and then compute the volume of the atom.

Step 3: Calculate the Percentage Occupied by the Nucleus

The percentage occupied by the nucleus is given by the formula \(\text{Percentage} = \frac{\text{Volume of Nucleus}}{\text{Volume of Atom}} \times 100 \)%. Use the volumes calculated in steps 1 and 2 to find this percentage. Since only two significant figures are required, round off the percentage to two significant figures.

Key Concepts

Atomic Volume Calculation

Understanding the size of an atom relative to its nucleus is quite fascinating when we think about the mostly empty space that makes up matter. To calculate the atomic volume, we must view the atom as a spherical space where the electrons orbit around the nucleus. Visualize it like a mini solar system, where the nucleus is the sun and the electrons are planets in wide orbits. The volume of the atom is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3}\pi r^3 \]
where \( r \) is the radius of the atom. For a carbon atom with a radius of about \(70 \mathrm{pm}\), we first convert picometers to meters (since the standard unit for volume will be cubic meters). One picometer is \(10^{-12}\) meters. Now, the radius would be \(70 \times 10^{-12}\) meters. Substituting this into the volume formula yields the atomic volume. This calculation gives us an appreciation for how tiy atoms actually are and the huge difference between atomic and human scales.

Nuclear Volume Calculation

Contrary to the vastness of the atom, the nucleus is incredibly dense and occupies a much smaller volume. To calculate the nuclear volume, we still use the spherical volume formula:
\[ V = \frac{4}{3}\pi r^3 \].
However, the radius we use here is at the femtometer scale, where one femtometer \( \mathrm{fm} \) is \( 10^{-15} \) meters. The radius of a carbon-12 nucleus is approximately \( 2.7 \mathrm{fm} \). Again, we convert femtometers to meters and then apply the volume formula. It’s through this nuclear volume calculation that we can really see how minuscule the nucleus is compared to the cloud of electrons that make up the bulk of the atom's perceived size.

Volume Ratio in Atoms

When we talk about the 'volume ratio in atoms', we're comparing the amount of space the nucleus takes up within the overall volume of the atom. It gives us a sense of the 'emptiness' of an atom. After calculating the volumes of both the nucleus and the atom, we can determine this ratio or percentage. Using the formula:
\[ \text{Percentage} = \frac{\text{Volume of Nucleus}}{\text{Volume of Atom}} \times 100\% \].
This calculation will invariably result in a very small number, underscoring the fact that atoms are mostly empty space, with the dense nucleus making up just a tiny fraction of their volume. This concept is fundamental in understanding the structure of matter and has profound implications in fields ranging from chemistry to solid-state physics.