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(C) Let the carbon atom be at the origin $(0, 0)$. The position of the carbon atom is $x_1 = 0$ with mass $m_1 = 12 \, a.m.u.$The position of the oxyg

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At a distance of $0.63 \, \mathring A$ from the carbon atom

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(C) Let the carbon atom be at the origin $(0, 0)$. The position of the carbon atom is $x_1 = 0$ with mass $m_1 = 12 \, a.m.u.$The position of the oxygen atom is $x_2 = 1.1 \, \mathring A$ with mass $m_2 = 16 \, a.m.u.$The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$X_{cm} = \frac{12 	imes 0 + 16 	imes 1.1}{12 + 16}$$X_{cm} = \frac{17.6}{28} \, \mathring A$$X_{cm} \approx 0.6285 \, \mathring A \approx 0.63 \, \mathring A$Thus,the center of mass is at a distance of $0.63 \, \mathring A$ from the carbon atom.

The distance between the carbon and oxygen atoms in a carbon monoxide molecule is $1.1 \, \mathring A$. Given that the mass of the carbon atom is $12 \, a.m.u.$ and the mass of the oxygen atom is $16 \, a.m.u.$,find the position of the center of mass of the carbon monoxide molecule.

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