Question

The distance between the oxygen molecule and each of the hydrogen atoms in a water
(H2
O) molecule is 0.96 Å and the angle between the twooxygen-hydrogen bonds is
105°. Treating the atoms as particles, find the centre of mass of the system.

Answer 1

Answer on Question #41550 - Chemistry - Inorganic Chemistry

Question:

The distance between the oxygen molecule and each of the hydrogen atoms in a water $(\mathsf{H}_2\mathsf{O})$ molecule is 0.96 Å and the angle between the two oxygen-hydrogen bonds is $105^{\circ}$ . Treating the atoms as particles, find the centre of mass of the system.

Solution:

A water molecule may be represented as follows

Thus, we have an isosceles triangle $\Delta$ HOH, where

HO = 0.96 Å

angle $\mathsf{L}\mathsf{HOH} = 105^{\circ}$

Let us designate the centre of mass of the system in figure above as $C$ .

Since water molecule is symmetric the center of mass lies on the axis of symmetry, i.e. it is equidistant from the hydrogen atoms (H). Let us designate the midpoint between H atoms as $D$ (HD = HH/2).

Thus, in given case to find the centre of mass of the system means to find values of HD, DC and CO.

In rectangular triangle $\Delta$ DOH angle $\mathsf{L}$ DOH = $\mathsf{L}$ HOH/2 = 105°/2 = 52.5°

OD = HO·cos(L·DOH) = 0.96·cos(52.5°) = 0.96 · 0.609 = 0.58 Å

HD = HO·sin(L·DOH) = 0.96·sin(52.5°) = 0.96 · 0.793 = 0.76 Å

DC + CO = OD = 0.58 Å

Molar mass of O atom $M_{\mathrm{O}} = 16.00 \, \mathrm{g/mol}$ , molar mass of H atom $M_{\mathrm{H}} = 1.01 \, \mathrm{g/mol}$ .

In consideration of law of the lever:

CO·Mo = DC·2MH

Assigning DC = x, CO = y and substituting the known values we get the system of two equations:

$\begin{array}{l}\left\{ \begin{array}{l}x + y = 0.58\\ 16.00\cdot y = 2\cdot 1.01\cdot x \end{array} \right.\\ y = 2.02\cdot x / 16.00 = 0.126\cdot x\\ x + 0.126\cdot x = 0.58\\ 1.126\cdot x = 0.58\\ x = 0.58 / 1.126 = 0.51\\ y = 0.58 - x = 0.58 - 0.51 = 0.07 \end{array}$

So, DC = 0.07 Å and CO = 0.51 Å.

Answer

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