502 
Estimation  of  Nicotine. 
Am.  Jour.  Pharm. 
Oct.,  1884. 
for  0-871  (the  entire  precipitate)  0-236  HgS  and  0*945  Agl  with 
0-510  I. 
0-236  HgS  correspond  to  0-461  Hgl2  with  0-258  I. 
0*461  Hgl2  deducted  from  the  amount  of  precipitate,  from  0*871, 
leave  0*410  for  acid  hydriodide  nicotine. 
0*258  I  deducted  from  the  entire  amount  of  I  found  in  precipitate, 
from  0*510  leave  0*252  I,  which  is  as  HI  in  combination  with 
nicotine. 
Calculation  requires  for  0*871  precipitate  .0*4534  Hgl2  and  0*4176 
and  hydriodide  of  nicotine  with  0*2537  I. 
From  these  investigations  it  appears  that  the  composition  of  the 
precipitate  is  HgI2,Ci0HuN2(HI)2. 
The  presence  of  nicotine  in  the  liquid,  filtered  off  the  precipitate 
after  M.  S.  has  ceased  to  give  a  precipitate  is  evidence  that  a  part  of 
the  combination  formed  is  kept  in  solution,  which  is  easily  ascribed  to 
the  excess  of  potassium  iodide  in  Mayer's  solution.  Dry  precipitate, 
which  is  very  little  souble  in  pure  or  acidulated  water,  dissolves  to  a 
greater  extent  when  iodide  potassium  is  added  to  the  water. 
But  as  with  a  certain  quantity  of  M.  S.  a  precipitate  ceases  to  be  pro- 
duced by  addition  of  more,  we  have  a  right  to  assume  that  the  nico- 
tine found  in  solution  is  in  the  same  combination  as  in  the  precipitate, 
and  from  the  quantity  of  mercury  in  solution  the  quantity  of  nicotine 
can  be  ascertained. 
By  calculating  the  combination  in  solution  from  the  amount  of  HgS 
found,  and  adding  the  amount  to  the  precipitate,  we  obtain  nearly 
double  the  amount  that  Dragendorff  claims  to  obtain. 
The  quantity  of  nicotine  indicated  by  1  cc.  of  Mayer's  solution  has 
to  be  taken  therefore  as  0*0081,  or  one-half  equivalent. 
From  among  the  many  perplexing  and  embarrassing  results  which 
were  obtained,  it  is  my  purpose  to  mention  only  a  few  to  prove  the 
assertion  that  when  a  resinous  precipitate  is  formed,  more  M.  S.  is 
required  to  finish  the  reaction. 
20  cc.  of  A.  required  22  cc.  of  M.  S.  when  the  precipitate  was  resin- 
ous at  first. 
20  cc.  of  B.  c.  required  25*7  cc.  M.  S.,  the  precipitate  having  formed 
resinous  except  towards  the  last. 
20  cc.  of  B.  c.  furnished  a  resinous  precipitate  until  23  cc.  of  M.  S. 
were  added ;  the  liquid  was  then  decanted  and  more  M.  S.  added  until 
it  ceased  to  give  precipitate ;  it  required  3  cc.  more. 
