Question

Calculate the moment(s) of inertia of methyl fluoride, CH₃F. The C-H bond length is 108.2 pm. The C-F bond length is 136.5 pm. The HCH bond angle is 110.2°. mH = 1.008mu, mC = 12.01mu, mF = 19.00mu

Answer 1

We have mH = 1.008 mu, mC= 12.01 mu and mF= 19.00 mu

So the reduced mass of methyl fluoride (CH3F) $\mu$ = m1 x m2 x m3/ m1+m2+m3

i.e. mC x 3 mH x mF/ mC + 3 mH + mF = 12.01 x (3 x 1.008) x 19.00/ 12.01 + (3 x 1.008) + 19.00 = 20.27 mu

So now to calculate moment of inertia (I) the following equation is used = I = $\mu$ x r2

Now we calculate r = 108.2 + 3 x (136.5) = 517.7 pm

Now to simplify the value we can convert mu into kilogram so we have 20.27 = 3.3659 x 10-26 kg

r = 517.7 pm = 5.177 x 10-10 m

now putting these values into equation we have 3.3659 x 10-26 x (5.177 x 10-10)2 = 9.021 x 10-47 kgm2