478  The  Examination  of  Chaulmoogra  Oil.  |Aq( 
Jour.  Pharm. 
ctober,  1915. 
free  fatty  acids  (see  infra).  This  test  can  also  be  used  to  dis- 
tinguish the  two  oils. 
(D)  Free  Fatty  Acids:  Acid  Value. — The  reaction  of  both 
samples  of  oil,  as  previously  stated,  was  acid  to  litmus;  the  acid 
value  was  determined  according  to  Lewkowitsch  method. 
(1)  Smith  oil. — 5.4204  grammes  of  oil  taken.  Required  for 
neutralization  21.5  Cc.  ^  NaOH  (corrected).  Therefore  milli- 
grammes of  KOH  required  for  neutralizing  1  gramme  of  oil  or  its 
acid  value  =  215  X  5-61   __  221    Acid  number  or  number  of  cubic 
5-4204 
centimetres  of  decinormal  alkali  required  to  neutralize  free  acid  in 
1  gramme  of  oil  —     215     ==  3.94.   Free  fatty  acids  expressed  as 
oleic  acid  per  cent.  =  "-5 X -0282  x  100  =  ll  lS 
5-4204 
(2)  B.  K.  Paul  oil. — 3.1762  grammes  of  oil  required  for  neutral- 
ization 50  Cc.  ~  NaOH  (corrected). 
Therefore  acid  value  (mgrms.  KOH)  ==  50X5  61  =  88.3.  Acid 
3-1762 
number  =      50      =  1*5.7.    Free  fatty  acids  as  oleic  acid  per  cent. 
3-1762         D/  J 
=  50  X  -0282  X  100  = 
3-1762         m  ^ 
(3)  Chaulmoogra  fat. — 2.8886  grammes  of  acid  required  for 
neutralization  60.6  Cc.  —  NaOH.   Therefore  acid  value  = 
^fmr  =  "7-69-    "cid  number  =  Jgfg  =  2o.9S.    Free  fatty 
acid  as  oleic  acid  per  cent.  =  6o-6X'°l*X 100  =  59.16. 
2-8886 
(E)  Saponification  Value. — The  saponification  value  was 
determined  according  to  the  method  described  in  Lewkowitsch. 
(1)  Smith  oil. — 1.2396  grammes  of  oil  were  taken  after  saponifi- 
cation with  25  Cc.  Kottstorfer  solution  required  for  neutralization 
of  excess  of  alkali  23.5  Cc.  half-normal  HC1  (cor.).  25  Cc. 
Kottstorfer  solution  =  33.5  Cc.  ^*  HC1.  Therefore,  saponifica- 
tion value  -  (33-5-  23-5)X28-o6  =    280-6  =  22(S  6 
1-2396  1-2396 
Another  sample  gave  224.55. 
(2)  B.  K.  Paul  Oil. — 1. 25 10  grammes  of  oil  taken.  After  sapon- 
ification 23.8  Cc.  semi-normal  acid  was  required  to  neutralize. 
Therefore,  saponification  value  =  (33'5  ~  23'8)  X  28,06  =  = 
r  1-2510  1-2510 
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