8 Bate MAN, Degrees of Freedom of a Molecule. 



Methylen Chloride 



Fig. I. « = 7 + 8 + 8-3-3-2-2-2-1-1= 9. 



Chloroform 



Fig. 2. « =7 + 4+12-3-6-1 = 1 3. 



Carbon tetrachloride 



Fig. 3. « = 7 + 16 - 8 = 15. 



CH^ group 



Fig. 4. « = 7 + 8- 3- 3-- 2 = 7. 



CH.^ group 



Fig. 5. « =7 + 12-9-2-1 = 7. 



Ethane 



CH., -CH„ « = 7 + 7-3=11. 



Propane 



CH.-CH.-CR, « = 7 + 7 + 7-6=i5. 



Butane 

 CH3-CH,-CH,-CH3 



« = 7 + 7 + 7 + 7-9=1 9. 



C,H„,+^ « = 7^-3(.y- i) = 4j + 3. 



Ammonia. // = 6+12-9-2-1= 6. 



In the COOH group, the four atoms can be supposed to 



set themselves in one plane. Taking then the structural 



formula to be 



0=C-0-H 



we get 



« = 7 + 5 + 5 + 4-5-3-3- i=9- 

 Acetic Acid. Assuming that the geometrical relations may be 

 roughly described by the formula 

 (CHJ-CCOOH) 

 we get 



« = 7 + 9-3=i3. 



Methyl Alcohol (CH., - OH). If we assume as an approximation 

 that the H in the OH group is equidistant from the 

 other three, we get 



« =7 + 6-3-2 = 8. 

 In the CHg group, when one H atom is replaced by an 

 atom or radicle having r degrees of freedom, the four 



