Mixter — Formation of the Oxides of Molybdenum, etc. 491 



trioxide, heated in an electric furnace to expel moisture, 

 then the tube was closed with stoppers, allowed to cool 

 and weighed. The reduction was made with pure dry hydro- 

 gen at temperatures between 410° and 440° approximately. 

 It required 76 hours and in the last four hours the loss 

 was 20 milligrams. The original weight of the trioxide 

 was 49*795 grams and the total loss in weight 6*175 grams. 

 The composition calculated from these data is Mo0 2 99*11 per 

 cent and Mo0 3 0*89 per cent. It was not deemed best to try 

 to complete the reduction at a higher temperature than used 

 on account of danger of reducing some of the substance to the 

 metallic state. The following are the experimental data : 



6 7 8 



Molybdenum dioxide 10-123 gr. 12028 gr. 12*000 gr. 



Sodium peroxide 13* " 15* " 16* " 



Water equivalent of system 3,918* "3,989* " 4,037* " 



Temperature interval 1*802° 2*085° 2*08° 



Heat effect „. 7,060 c 8,317 c 8,397 c 



" of oxygen set free -fl,136 c +l,117 c +l,128 c 



" " oxidation of iron — 48 c — 48 c — 48 c 



8,148 c 9,386 c 9,477 c 



For 1 gram of MoO a 805 c 780 c 789° 



The fusions were good and dissolved in water with evolution 

 of oxygen and no black residue remained. The oxygen set free 

 in the bomb was collected in a flask over water and the volume 

 of it found by weighing the flask containing it, then filling 

 with water and weighing again. The number of cubic centi- 

 meters of oxygen at 0° and 760 mm multiplied by 1*73 gave the 

 number of calories lost by the change of sodium peroxide to 

 oxide. If the experimental data above are reduced, allowing 

 for the presence of 0*9 of 1 per cent of trioxide in the dioxide 

 used, the result is not essentially different from that given. 

 The mean for 1 gram of molybdenum dioxide is 791 c and for 

 128 grams it is 101,200 c . 



The following are the results of the experiments with 

 molybdenum : 



Na a O a + MoO„ = Na 2 Mo0 4 + 101,200 C 



Xa 2 + O = Xa 2 2 + 19,400 c 



Na a O + MoO„ + O = Na 2 Mo0 4 + 120,600 c 



Na a O -f Mo0 8 = Na 9 Mo0 4 + - 81,900 c 



MoO g 4- O = Mo0 3 + 38,700 c 



Mo + 30 = Mo0 3 + 181,500 c 



Mo + 20 = _ 142,800° 



