278 DR. KANE ON THE CHEMICAL HISTORY OF ARCHIL AND LITMUS. 
Erythryline is insoluble in water, cold or hot, but by boiling with water it is gra- 
dually altered, and the liquor is then found to contain erythrine-bitter. It is easily 
soluble in alcohol and in ether. It is soluble in alkaline liquors, from which it is pre- 
cipitated on the addition of an acid. It combines with metallic oxides by way of 
double decomposition, forming salts or lakes, which are generally greenish coloured. 
It does not appear to possess acid properties, however ; its alkaline solutions cannot 
be obtained in any form neutral to test paper ; and I therefore think it better not to 
give to its name any form which acids are generally characterized by. 
When heated, erythryline fuses a little above 212°, and on a further application of 
heat it is decomposed without any trace of tendency to volatilize. 
Its analysis was effected as follows : 
A. 0*358 gramme of material gave 0*878 gramme of carbonic acid, and 0*262 of 
water. 
B. 0*407 gramme of material gave 0*987 gramme of carbonic acid, and 0*307 of 
water. 
These results give the composition in 100 parts. 
A. 
B. 
Carbon . . . 
67*83 
67*06'] 
Hydrogen . . 
8*13 
8*37 [>-100 
Oxygen . . . 
24*04 
24*57 J 
Leading to the formula C 22 H 16 O, 
5 , which gives 
22 Carbon . 
. . 134*2 
6771 
16 Hydrogen 
. . 16*0 
8-07 
6 Oxygen . 
. . 48*0 
24*22 
198*2 
100*00 
By mixing a solution of erythryline in ammonia, in which the alkali was as little as 
possible in excess, with a solution of nitrate of lead, and washing the precipitate until 
all impurities had been removed, there was obtained a pale greenish powder, which 
was dried at 212° Fahr. and then analysed. 
A. 0*739 gramme of material gave by Berzelius’ method 0*163 gramme of metallic 
lead, and 0*213 gramme of oxide of lead. 
B. 0*710 gramme of material gave 0*811 gramme of carbonic acid, and 0 251 
gramme of water. 
The formula which results is C 22 H 16 O s + 2 PbO. Thus 
Theory. 
Experiment. 
22 Carbon 
= 134*2 
31*85 
31*58 
16 Hydrogen 
= 16*0 
3*79 
3*92 
6 Oxygen 
= 48*0 
11*39 
11*90 
2 Oxide of lead 
= 223*2 
52*97 
52*60 
421*4 
100*00 
100*00 
