Given that the chlorine atom is 35.5 times as massive as a hydrogen atom, the mass of the chlorine atom is…

Question

Given that the chlorine atom is 35.5 times as massive as a hydrogen atom, the mass of the chlorine atom is mCl = 35.5 mH.

ScanSolve AI Tutor · Accepted Answer

Here's how to find the approximate location of the center of mass (CM) of the HCl molecule:

Define the masses and positions of the atoms.
Let $m_H$ be the mass of the hydrogen atom.
Given that the chlorine atom is 35.5 times as massive as a hydrogen atom, the mass of the chlorine atom is $m_{Cl} = 35.5 \, m_H$ .
Let's place the hydrogen atom at the origin of our coordinate system, so its position is $x_H = 0$ .
The separation between the nuclei is $r = 1.27 \, \text{Å}$ . Therefore, the position of the chlorine atom is $x_{Cl} = 1.27 \, \text{Å}$ .
Use the formula for the center of mass of a two-particle system.
The position of the center of mass ( $X_{CM}$ ) for two particles along a line is given by:
$X_{CM} = \frac{m_H x_H + m_{Cl} x_{Cl}}{m_H + m_{Cl}}$
Substitute the defined values into the formula.
$X_{CM} = \frac{m_H (0) + (35.5 \, m_H) (1.27 \, \text{Å})}{m_H + 35.5 \, m_H}$ $X_{CM} = \frac{0 + 35.5 \times 1.27 \, m_H \, \text{Å}}{36.5 \, m_H}$
The mass $m_H$ cancels out:
$X_{CM} = \frac{35.5 \times 1.27 \, \text{Å}}{36.5}$
Calculate the numerical value.
$X_{CM} = \frac{45.085}{36.5} \, \text{Å}$ $X_{CM} \approx 1.235 \, \text{Å}$
This value represents the distance of the center of mass from the hydrogen atom (our chosen origin).

The approximate location of the CM of the molecule is $\text{1.235 Å from the hydrogen atom}$ .

Given that the chlorine atom is 35.5 times as massive as a hydrogen atom, the mass of the chlorine atom is mCl = 35.5 mH.