Bayley — On Complex Oxides of Cobalt and Nicliel. 105 



The question, which is the true formula of the oxide I obtained, is 

 determined by the amount of iodine liberated by the oxide on treat- 

 ment with potassic iodide and hydrochloric acid. According to the 

 formula Coo O3 3 H2O, there should be liberated -402 gram by '1865 

 gram of cobalt; according to the formula C03O5, 4 HoO, -5343 grams 

 should be liberated. I found in three experiments "SSSS, -5380, and 

 •5328. 



When the oxide C03O5, 4 HoO, obtained as described above, is 

 boiled for an hour or two in the solution in which it is precipitated, 

 and the amount of iodine liberated then estimated, the result points to 

 the formation of the oxide C010O19 intermediate between C03O5 and 

 C00O3. 



Co. taken. 



Theory of I. 

 C03O5 



for 



Theory of I. 

 C02O3 



for 





Found. 



gram. 



gram. 





gram. 







gram. 



■1865 



•5343 





•4007 





(1) 



•453 







Mean. 







(2) 



•463 







•4676 







(3) 



•462 





Theory for C012O19 

 lod. 







Found, 

 gram. 



•0620 





•1566 



gram. 







•1556 



In the last experiment a fresh solution of cobalt, and a fresh solu- 

 tion of potassium bichromate (to standardise the thiosulphate), were 

 used. 



A quantity of the oxide of cobalt prepared by precipitating with 

 potash and sodic hypochlorite, and hoiling for some hours, then washing 

 and drying over sulphuric acid in vacuo, was submitted to a current of 

 air at a low red heat, and the water collected and weighed in a calcium 

 chloride tube. The oxide was afterwards ignited to bright redness in 



Oxide dried over H2SO4, -7455 



Oxide after ignition in tube (C012O19), -6255 



Oxide after ignition in air (CosOi), ^5975 



Calcium chloride tube + OH2, 65^5900 



Calcium chloride tube, 65-4645 



•1255 = OH, 



i2 



