Dr. Schunck on Rubian and its Products of Decomposition. 867 



IV. 0-2480 grin, of the same preparation as the last, recry- 

 stallized from alcohol, gave 0-5290 carbonic acid and 0*1280 

 water. 



V. 0-3735 grm. of a third preparation gave 0-7925 carbonic 

 acid and 0-1785 water. 



VI. 0-3995 grm. of the same preparation gave 0-8450 car- 

 bonic acid and 0-1855 water. 



These numbers correspond in 100 parts to — 



I. II. III. IV. V. VI. 



Carbon . . 57'33 5726 57-2.9 58-17 57-86 5768 

 Hydrogen . 552 551 529 573 5'31 5-15 

 Oxygen . . 3745 3723 37'42 36-10 36-83 37-17 

 I have as yet been unsuccessful in my attempts to determine 

 the atomic weight of rubianine. The little affinity which it has 

 for bases is proved by the fact above mentioned, of its crystal- 

 lizing unchanged out of its alkaline solutions. The baryta com- 

 pound, which is obtained by adding chloride of barium to its 

 ammoniacal solution, is easily decomposed when it comes to be 

 washed with pure water, the baryta being dissolved by the water, 

 a yellow residue of rubianine being left at last. It is not ca- 

 pable of separating oxide of lead from acetic acid. In fact it 

 nearly approaches the character of a perfectly neutral body, a 

 circumstance which might be_a priori foreseen from its containing 

 more carbon and less oxygen than rubian itself, the properties 

 of which are not far removed from those of an indifferent sub- 

 stance. 



There are three formulae which all of them give for 100 parts, 

 numbers not widely differing from those found by experiment, 

 viz. C 28 H 17 O 13 , C 32 H 19 O 15 and C 44 H 24 O 20 . These formulae 

 require for 100 parts of substance the following amount of con- 

 stituents : — 



C 28 H'"0 13 . 



Carbon . . . 58-13 



Hydrogen . . 5-88 



Oxygen . . . 35-99 



It will be seen that the last formula is that with which the 



analyses agree best. 



If the first formula be the true one, then the formation of 

 this substance from rubian is easily explained. It would then 

 differ by 5 equivs. of water from 2 equivs. of rubiretine; and 

 1 equiv. of rubianine, 2 equivs. of verantine and 7 equivs. of 

 water added together would be equal to 1 equiv. of rubian, as 

 seen by tbe following equation : 



C 28 H 17 ,3 + 2(C 14 H 5 O 5 ) + 7110 = C 56 H 34 O 30 . 

 I shall presently show, however, that there is more probability 

 in favour of one of the two latter formulae. 



