Question

Calculate the position of the center of mass of (a) \(^{1} \mathrm{H}^{19} \mathrm{F},\) which has a bond length of \(91.68 \mathrm{pm},\) and (b) \(\mathrm{HD}\) which has a bond length of \(74.15 \mathrm{pm}\)

Short answer

The position of the center of mass for (a) \(^1\text{H}^{19}\text{F}\) is \(x_{CM} = \frac{(1 \cdot 0) + (19 \cdot 91.68)}{1 + 19}\) and for (b) \(\text{HD}\) is \(x_{CM} = \frac{(1 \cdot 0) + (2 \cdot 74.15)}{1 + 2}\).

Step-by-step solution

Determine the masses of the atoms

We use the atomic mass units (amu) for hydrogen (H) and fluorine (F): H = 1 amu F = 19 amu

Set the initial positions of the atoms

Since we have a linear molecule, let's set the position of hydrogen (H) as the origin (0): \(x_1 = 0\) Now, the position of fluorine (F) is equal to the given bond length: \(x_2 = 91.68 pm\)

Apply the formula for the center of mass

Using the formula for the center of mass and the given masses and positions, we find the center of mass for ^1H^19F: \(x_{CM} = \frac{(1 \cdot 0) + (19 \cdot 91.68)}{1 + 19} \) (a) Calculate the center of mass for HD:

Determine the masses of the atoms

In this case, our molecule consists of hydrogen (H) and deuterium (D). The atomic masses are: H = 1 amu D = 2 amu

Set the initial positions of the atoms

As we did before, let's set the position of hydrogen (H) as the origin (0): \(x_1 = 0\) The position of deuterium (D) is equal to the given bond length: \(x_2 = 74.15 pm\)

Apply the formula for the center of mass

Using the formula for the center of mass and the given masses and positions, we find the center of mass for HD: \(x_{CM} = \frac{(1 \cdot 0) + (2 \cdot 74.15)}{1 + 2}\)

Key concepts

Atomic Mass Units

Atomic mass units (amu) are pivotal in understanding the intricacies of molecular structures and are essential for calculating the center of mass of molecules. This standardized unit is equivalent to one twelfth of the mass of a carbon-12 atom which is approximately equal to 1.66 x 10^-24 grams. In the context of the exercise, hydrogen (H) has an atomic mass of 1 amu, and fluorine (F) has a mass of 19 amu, indicating that fluorine is 19 times heavier than hydrogen on an atomic scale.

Knowing the atomic masses of individual atoms allows chemists and physicists to determine how mass is distributed within a molecule and to calculate properties such as the molecule's center of mass. This is crucial when examining the behavior of the molecule in different conditions such as reactions, phase changes, or even simply its orientation in space.

Molecular Bond Length

Molecular bond length is a measure of the distance between the nuclei of two bonded atoms in a molecule. It's a crucial factor in calculating the center of mass, particularly in linear molecules where all the atoms lie along a straight line. For the calculations in the given exercise, bond lengths are provided in picometers (pm), a sub-unit ideal for expressing atomic-scale lengths (1 pm = 1 x 10^-12 meters).

In the case of ¹H¹⁹F, the bond length is 91.68 pm, whereas for HD (a molecule consisting of hydrogen and deuterium), it's 74.15 pm. These bond lengths indicate the distances over which the molecule's mass is distributed, which in turn affects the position of the center of mass. It's important to have accurate bond length measurements since they impact the molecular geometry and physical properties of the molecule.

Linear Molecule

A linear molecule is a chemical species where all the constituent atoms are aligned in a straight line. This arrangement has a profound impact on the calculation of the center of mass because it simplifies the consideration of atomic positions. When atoms in a molecule form a straight line, their positions can be described using a one-dimensional coordinate system.

In our exercise, both ¹H¹⁹F and HD are linear molecules, which allow us to use a single axis to represent the positions of atoms for center of mass calculations. The simplicity of this molecular geometry makes it easier to conceptualize and calculate properties like moment of inertia and polarization, and it's especially convenient for clarifying the center of mass concept for students.

Center of Mass Formula

The center of mass formula is a mathematical expression used to locate the average position of mass in a system. For a system of particles, it's represented by: \[ x_{CM} = \frac{\sum{m_ix_i}}{\sum{m_i}} \] where \(m_i\) is the mass of the i-th particle, and \(x_i\) is the position of the i-th particle. This formula accounts for the differing masses and distances of each particle from a reference point, providing a weighted mean of their positions.

In the context of our exercise involving ¹H¹⁹F and HD molecules, students need to strategically choose a reference point (like setting the position of hydrogen as 0) and apply the masses and positions into the formula to calculate the center of mass. Correct application of this formula allows for accurate descriptions of the molecule's behavior under forces and contributes to deeper insights into molecular dynamics.