The microwave spectrum of 16O12O32S shows absorption lines separated by 12.16438

Question

The microwave spectrum of 16O12O32S shows absorption lines separated by 12.164385 GHz. That of 16O12C34S shows absorption lines separated by 11.866084 GHz. The determination of the bond distances involves solving of two coupled quadratic equations. Derive explicit expressions for the bond distances roc and res and calculate them. To obtain the correct results carry all figures throughout the calculations and round only at the end to get the final bond distances to 4 significant figures. As atomic masses use 12 amu for 12C, 15.9949 amu for 160, 31.9721 amu for 32S, and 33.9679 amu for 34S, and as conversion factor between kg and amu use 1.66083 Times 10-27 kg/amu. Calculate and use all masses to 4 significant figures, with the exception of the total mass of the second molecule, where you have to use 5 significant figures for its mass.. DO NOT USE SUCCESSIVE APPROXIMATIONS, but derive the solution in closed form without approximations in them. Solve the equations exactly.

Explanation / Answer

Let us consider the OCS molecule. ro, rc and rc represents the distances of atoms from the centre of gravity.

The moments can be represented as moro + mcrc = msrs....(1)

where mi is the mass of atom i. The moment of inertia is :

I = moro^2 + mcrc^2 + msrs^2 ......(2)

and we have further equations:

ro = rco + rc and rs= rcs - rc.....(3)

where rco and rcs are the bond lengths of the molecule. It is these we wish to determine. Substitute (3) in (1) we get.

(mc + mo + ms) rc = ms rcs - mo rco

or Mrc = msrcs - mo rco ....(4)

Where M is for the total mass of the molecule. Substitute (3) in (2) we get

I = mo (rco + rc) ^2 + mc rc^2 + ms (rcs - rc)^2

= M rc^2 + 2 rc (mo rco - ms rcs) + mo rco^2 + ms rcs^2

and finally substituting for rc from equation (4)

I = mo rco^2 + ms rcs^2 - ( mo rco - ms rcs)^2 / M ....(5) for 16O12C32S

Now if consider an isotopic molecule such as 16O12C34S, then we may write mo' for mo throughout the equation (5) and we get the following equation.

I' = mo rco^2 + ms' rcs^2 - ( mo rco - ms' rcs)^2 / M'

and we can now solve for rco and rcs

I is calculated using the formula , I = h/8?^2 B

substituting all values i.e h, ?, and B = 12.164385 GHz and 11.866084 GHz we can I and I' as follows

I = 6.905 x 10^-37

I' = 7.07 x 10^-37

I' - I = 0.165 = rcs^2 (ms' - ms) - ( mo rco - ms' rcs)^2 / M' + ( mo rco - ms rcs)^2 / M

   0.165 = rcs^2 (33.9679 - 31.9721) - (mo rco - ms' rcs)^2 / M' + ( mo rco - ms rcs)^2 / M

   M' and M can be calculated by adding the molecular weights of O , C and sulphure. And by solving the above equation we can find the values for rco and rcs.

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