
Find the distance of centre of mass of ${{\text{H}}_{\text{2}}}{\text{O}}$ molecule.
Answer
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Hint: In order to solve this question, it is necessary to know the structure ${{\text{H}}_{\text{2}}}{\text{O}}$ molecule. The mass of one hydrogen atom is 1 unit and the mass of an oxygen atom is 16 units. Using these data, the centre of mass can be calculated.
Formula Used: The formulae used in the solution are given here.
${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$, ${Y_{CM}} = \dfrac{{{m_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Complete Step by Step Solution: The ${{\text{H}}_{\text{2}}}{\text{O}}$ molecule is formed by the two hydrogen atoms and an oxygen atom. The mass of one hydrogen atom is 1 unit and the mass of an oxygen atom is 16 units.
The structure is in the form of a triangle, with the oxygen atom forming the apex of the triangle and the angle between the two $O - H$ bonds, the two arms of the triangle is ${104^ \circ }$.
The distance between the hydrogen atom and oxygen atom is ${\text{0}}{{.96 \times 1}}{{\text{0}}^{{\text{ - 10}}}}{\text{m}}$.
By the formula of Centre of Mass, ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Using the values we have in the above formula, we have, ${X_{CM}} = \dfrac{{ - d\sin {{52}^ \circ } + d\sin {{52}^ \circ } + 16 \times 0}}{{1 + 1 + 16}} = 0$
${Y_{CM}} = \dfrac{{{m_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Using the values we have in the above formula, we have,${Y_{CM}} = \dfrac{{0 + 0 + 16 \times 0.96 \times {{10}^{ - 10}}\cos {{52}^ \circ }}}{{1 + 1 + 16}} = \dfrac{8}{9} \times 0.96 \times {10^{ - 10}}\cos {52^ \circ } = 6.4 \times {10^{ - 12}}m$.
Centre of mass of a water molecule is $0,6.4 \times {10^{ - 12}}$.
Note: In this question we have the answer in form of coordinates, it means the centre of mass is at a distance of $6.4 \times {10^{ - 12}}m$ away from the centre of the line joining the two hydrogen atom.One interesting fact about the water molecule is that is forms hydrogen bond, which is a special type of bond only formed when a hydrogen atom is covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.
Formula Used: The formulae used in the solution are given here.
${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$, ${Y_{CM}} = \dfrac{{{m_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Complete Step by Step Solution: The ${{\text{H}}_{\text{2}}}{\text{O}}$ molecule is formed by the two hydrogen atoms and an oxygen atom. The mass of one hydrogen atom is 1 unit and the mass of an oxygen atom is 16 units.
The structure is in the form of a triangle, with the oxygen atom forming the apex of the triangle and the angle between the two $O - H$ bonds, the two arms of the triangle is ${104^ \circ }$.
The distance between the hydrogen atom and oxygen atom is ${\text{0}}{{.96 \times 1}}{{\text{0}}^{{\text{ - 10}}}}{\text{m}}$.
By the formula of Centre of Mass, ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Using the values we have in the above formula, we have, ${X_{CM}} = \dfrac{{ - d\sin {{52}^ \circ } + d\sin {{52}^ \circ } + 16 \times 0}}{{1 + 1 + 16}} = 0$
${Y_{CM}} = \dfrac{{{m_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$.
Using the values we have in the above formula, we have,${Y_{CM}} = \dfrac{{0 + 0 + 16 \times 0.96 \times {{10}^{ - 10}}\cos {{52}^ \circ }}}{{1 + 1 + 16}} = \dfrac{8}{9} \times 0.96 \times {10^{ - 10}}\cos {52^ \circ } = 6.4 \times {10^{ - 12}}m$.
Centre of mass of a water molecule is $0,6.4 \times {10^{ - 12}}$.
Note: In this question we have the answer in form of coordinates, it means the centre of mass is at a distance of $6.4 \times {10^{ - 12}}m$ away from the centre of the line joining the two hydrogen atom.One interesting fact about the water molecule is that is forms hydrogen bond, which is a special type of bond only formed when a hydrogen atom is covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.
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