20 MOLYBDENUM BLUE 



reduction and heating continued for four hours more. At the 

 end of that time the boat containing the residue was again 

 weighed. The boat was allowed to cool in a stream of hydrogen. 

 Weighing was done as rapidly as possible. It was found that 

 the Blue lost very little or no weight during the final four hours 

 heating. The gain in weight of the U tube served as a check. 



Three determinations were made. The results agreed very 

 well considering the possible experimental error, due to sub- 

 limination of Molybdenum, moisture absorbed by blue during 

 first weighing, and also variation in amount of water in hydrogel. 

 In discussing the various formulae that have been suggested 

 the average results obtained will be considered. 



Average time Average Average 



of reduction Wt. of hydrogel Wt. of residue Loss in Wt. 

 12.2 hrs. .2669 g .1785 g .0884 g 



The residue was blueish grey. It was taken to be Molyb- 

 denum with a trace of oxide. As no further loss in weight re- 

 sulted between weights at the end of eight and twelfth hours of 

 treatment the residue could not be M02O3M0O, M0O2, M0O3, 

 for these oxides are broken down below bright red heat. When 

 the heat was increased, a metallic mirror was deposited in the 

 tube. This would show that the residue was Molybdenum. 

 In the following calculations it will be treated as such: 



Let us consider the first formula suggested, M02O5. If 



Mo, Oj 192 



this were the formula residue would weigh ;7~X2669=— — : 



Mo, Oj 272 



X 2669 = 1884 g. 



This is^nore than S^% greater than actual weight of residue. 

 According to this formula the loss in weight would be: 



' X. 2669 = -^X. 2669 =.08045. 



Mo. O, 111 



This is 10% smaller than experimental result. The formula 

 MojOr must therefore be considered incorrect. 



