﻿Measuring 
  large 
  Molecular 
  Masses. 
  505 
  

  

  It 
  is 
  to 
  be 
  noticed 
  that 
  four 
  out 
  of 
  the 
  five 
  values 
  of 
  v 
  are 
  

   nearly 
  integers, 
  while 
  that 
  for 
  the 
  first 
  specimen 
  of 
  pepsin- 
  

   fibrin 
  peptone 
  a 
  is 
  nearly 
  3'5. 
  Such 
  a 
  fractional 
  value 
  

   might 
  appear 
  if 
  a 
  peptone 
  was 
  split 
  up 
  by 
  NaOH 
  into 
  two 
  

   ions, 
  one 
  of 
  valency 
  3, 
  the 
  other 
  4. 
  That 
  there 
  is 
  a 
  splitting 
  

   of 
  the 
  peptone 
  molecule 
  we 
  shall 
  see 
  immediately. 
  But 
  for 
  

   the 
  four 
  cases 
  where 
  v 
  2 
  is 
  nearly 
  a 
  whole 
  number 
  we 
  shall 
  

   replace 
  its 
  value 
  in 
  the 
  table 
  by 
  the 
  nearest 
  whole 
  number, 
  

   with 
  which 
  we 
  shall 
  then 
  divide 
  the 
  tabulated 
  value 
  of 
  A 
  o2 
  

   to 
  get 
  the 
  following 
  amended 
  values 
  of 
  A 
  o2 
  jv 
  2 
  for 
  the 
  four 
  

   peptones 
  in 
  the 
  above 
  order: 
  — 
  

  

  vo. 
  2 
  2 
  3 
  3 
  

  

  Aq2/v 
  2 
  ... 
  27-5 
  36*5 
  273 
  27*2 
  

  

  These 
  valencies 
  are 
  the 
  same 
  as 
  Neumann 
  found 
  by 
  the 
  

   use 
  of 
  Ostwald's 
  empirical 
  valency 
  rule. 
  To 
  use 
  the 
  values 
  

   of 
  A 
  02 
  /v 
  2 
  in 
  (11) 
  for 
  finding 
  B 
  we 
  must 
  reduce 
  them 
  to 
  18° 
  

   by 
  dividing 
  by 
  1*09 
  and 
  then 
  we 
  get 
  

  

  Peptone. 
  

   Trypsinfibrin 
  or. 
  Trypsinfibrin 
  fi. 
  Pepsinfibrin 
  a. 
  Pepsinglutin. 
  

   B... 
  150 
  83 
  144 
  144 
  

  

  The 
  valencies 
  of 
  these 
  ions 
  must 
  represent 
  the 
  number 
  of 
  

   COO 
  groups 
  they 
  contain, 
  so 
  that 
  144/3 
  = 
  48 
  represents 
  the 
  

   volume 
  of 
  a 
  COO 
  group 
  in 
  a 
  trivalent 
  peptone 
  ion, 
  together 
  

   with 
  the 
  volume 
  of 
  the 
  atoms 
  associated 
  with 
  it. 
  But 
  B 
  for 
  

   COO 
  is 
  20, 
  so 
  that 
  the 
  volume 
  of 
  an 
  equivalent 
  of 
  a 
  trivalent 
  

   peptone 
  ion 
  is 
  about 
  2\ 
  times 
  the 
  volume 
  of 
  COO, 
  so 
  its 
  

   mass 
  will 
  be 
  roughly 
  2\ 
  times 
  that 
  of 
  COO, 
  which 
  is 
  44. 
  

   Thus 
  we 
  find 
  the 
  order 
  of 
  magnitude 
  of 
  an 
  equivalent 
  to 
  be 
  

   about 
  110 
  at 
  the 
  most, 
  and 
  this 
  is 
  much 
  smaller 
  than 
  248 
  

   and 
  320 
  found 
  by 
  Neumann 
  by 
  electric 
  titration 
  for 
  the 
  

   trivalent 
  peptones. 
  It 
  is 
  plain 
  then 
  that 
  we 
  have 
  to 
  do 
  with 
  

   a 
  splitting 
  up 
  of 
  the 
  peptone 
  molecule 
  by 
  NaOH. 
  Suppose 
  

   M 
  to 
  be 
  the 
  molecular 
  mass 
  of 
  peptone, 
  which 
  is 
  split 
  into 
  

   n 
  2 
  ions 
  of 
  valency 
  v 
  2 
  and 
  mass 
  p, 
  and 
  a 
  neutral 
  residue 
  of 
  

   mass 
  r, 
  then 
  M 
  = 
  ?i 
  2 
  ^>4- 
  ?% 
  and 
  the 
  equivalent 
  by 
  titration 
  is 
  

   ~M/n 
  2 
  v 
  2 
  =p/v 
  2 
  + 
  r/n 
  2 
  v 
  2 
  . 
  This 
  r/n 
  2 
  v 
  2 
  is 
  the 
  difference 
  between 
  

   our 
  rough 
  maximum 
  of 
  110 
  found 
  above 
  for 
  p/v 
  2 
  and 
  248 
  or 
  

   320 
  for 
  p\v 
  2 
  + 
  rjn 
  2 
  v 
  2 
  . 
  It 
  is 
  important, 
  then, 
  to 
  look 
  into 
  the 
  

   values 
  of 
  B 
  as 
  closely 
  as 
  we 
  can. 
  In 
  the 
  trivalent 
  ion 
  with 
  

   B 
  = 
  144 
  3000 
  would 
  contribute 
  60 
  to 
  B. 
  Knowing 
  the 
  

   ammo-acid 
  character 
  of 
  the 
  peptones 
  we 
  may 
  assume 
  that 
  

   possibly 
  an 
  NH 
  2 
  group 
  is 
  associated 
  with 
  one 
  COO 
  group 
  or 
  

   two, 
  if 
  two 
  they 
  contribute 
  32 
  to 
  B. 
  In 
  B 
  there 
  remains 
  

   only 
  52 
  to 
  account 
  for, 
  which 
  is 
  most 
  easily 
  done 
  by 
  assuming 
  

  

  