108 
IV. 0*1805 grm. (ethylic ale.) gave 0*4960 grm, carbonic 
acid and 0*1330 grm. water. 
These numbers lead to the following composition : 
I. 
ii. 
III. 
IV. 
c... 
... 74*50 .. 
.... 75-10 ... 
.... 74*97 .., 
.... 74-95 
H... 
... 8*90 .. 
.... 8*85 .., 
.... 8*95 ... 
.... 8*26 
O... 
... 16*60 
.... 16*05 
.... 16*08 ... 
... 16*79 
100*00 
100*00 
100*00 
100*00 
The formula C 12 H 8 0 2 with which they correspond requires 
in 100 parts 
C 75*00 
H 8*40 
0 16*60 
100*00 
The formation of the resins from the alcohols might per- 
haps be explained by assuming that aldehyde is formed by 
the oxidation of alcohol, and then by simple loss of water the 
aldehyde would be converted into resin, since 
3(C 4 H 4 0 2 ) - 4HO = C 12 H 8 0 2 . 
But the formic acid in this case would not stand in any 
relation to the resin, and even the formation of the resin from 
methylic alcohol could not easily be explained. 
If we adopt the formula C 24 H 18 0 4 , which requires 74*28 
per cent carbon and 9*10 per cent hydrogen, we must assume 
that carbonic and formic acid are taken up by the alcohol, 
though it is very doubtful whether these acids could separate 
from a strong base like soda in order to form a neutral resin. 
The following equations will show what m&y be imagined to 
take place in this case : 
5(C 4 He 0 2 ) + C 2 0 4 + C 2 H 2 0 4 = C 24 H 18 0 4 + 14 HO. 
10(C 3 H 4 0 2 ) + C 2 0 4 + C 2 H 2 0 4 = C 21 H 18 0 4 + 24 HO. 
