﻿12± 
  Clarke 
  and 
  Steiger 
  — 
  -Action 
  of 
  Ammonium 
  Chloride, 
  etc. 
  

  

  respond 
  to 
  differences 
  in 
  their 
  chemical 
  structure. 
  But 
  the 
  

   reaction 
  goes 
  further 
  than 
  this, 
  for 
  it 
  enables 
  us 
  to 
  substitute 
  

   in 
  many 
  minerals 
  a 
  volatile 
  base 
  for 
  fixed 
  bases, 
  yielding 
  sili- 
  

   cates 
  which 
  split 
  up 
  on 
  ignition 
  in 
  such 
  a 
  way 
  as 
  to 
  shed 
  light 
  

   upon 
  their 
  molecular 
  constitution. 
  

  

  For 
  example, 
  if 
  ammonium 
  leucite 
  is 
  a 
  true 
  metasilicate, 
  a 
  

   salt 
  of 
  the 
  acid 
  H 
  2 
  Si0 
  3 
  , 
  it 
  should 
  break 
  up, 
  when 
  ignited, 
  in 
  

   accordance 
  with 
  the 
  following 
  equation 
  : 
  

  

  2NH 
  4 
  Al(Si0 
  3 
  ) 
  2 
  = 
  Al 
  9 
  (SiO,), 
  + 
  2NH 
  3 
  + 
  H 
  2 
  + 
  Si0 
  2 
  ; 
  

  

  that 
  is, 
  one-fourth 
  of 
  the 
  silica 
  ought 
  to 
  be 
  set 
  free, 
  measurable 
  

   by 
  extraction 
  with 
  sodium 
  carbonate 
  solution. 
  No 
  such 
  split- 
  

   ting 
  off 
  occurs, 
  however 
  ; 
  an 
  ammonium 
  leucite 
  which 
  already 
  

   contained 
  1*97 
  per 
  cent 
  of 
  soluble 
  silica 
  gave 
  only 
  1*70 
  per 
  

   cent 
  after 
  ignition 
  ; 
  hence 
  no 
  additional 
  silica 
  had 
  been 
  liberated. 
  

   We 
  may 
  conclude, 
  therefore, 
  that 
  analcite 
  and 
  leucite 
  are 
  not 
  

   true 
  metasilicates 
  but 
  pseudo 
  compounds 
  ; 
  either 
  salts 
  of 
  a 
  

   polymeric 
  metasilicic 
  acid 
  or 
  mixtures 
  of 
  ortho-and 
  tri-silicates 
  

   as 
  suggested 
  in 
  our 
  former 
  paper. 
  The 
  ammonium 
  salt 
  corre- 
  

   sponding 
  to 
  such 
  a 
  mixture, 
  when 
  ignited, 
  might 
  be 
  expected 
  

   to 
  give 
  the 
  following 
  reaction 
  : 
  

  

  """-•2 
  v 
  ^ 
  lv 
  -'3 
  

  

  \Al-Si 
  3 
  8 
  = 
  Al 
  / 
  \Al-Si 
  8 
  8 
  =Al; 
  

  

  Si0 
  4 
  = 
  Am 
  2 
  SiCX 
  

  

  / 
  \ai_s; 
  r> 
  = 
  ai 
  / 
  

  

  Al— 
  SiO 
  = 
  Am 
  9 
  - 
  2 
  Am,0 
  = 
  Al 
  — 
  SiO. 
  

  

  Si 
  0=Al 
  Si.O=Al 
  

  

  a 
  reaction 
  which 
  is 
  in 
  harmony 
  with 
  our 
  experimental 
  results. 
  

   In 
  it 
  no 
  free 
  silica 
  appears 
  ; 
  and 
  many, 
  if 
  not 
  all, 
  conditions 
  of 
  

   the 
  problem 
  are 
  satisfied. 
  Other 
  formulae, 
  however, 
  are 
  possi- 
  

   ble 
  ; 
  so 
  that 
  these 
  offer 
  merely 
  one 
  explanation 
  of 
  the 
  phenom- 
  

   ena, 
  which 
  one 
  may 
  or 
  may 
  not 
  be 
  the 
  best. 
  

  

  