66 PROCEEDINGS OF THE AMERICAN ACADEMY 



0.9945 gr. lost by ignition witli W0 4 Na 2 0.23G2 gr. = 23.75% water. 

 1.4588 gr. gave 0.0713 gr. P 2 7 Mg 2 = 3J2 % P 2°5- 



The analysis leads to the formula 



24 MoO s . P 2 6 . 3 H 2 -f- 59 aq, 



which requires : — 



The phosphoric oxide was determined by double precipitation and 

 treatment with ammonic sulphide. The molybdic oxide was estimated 

 by difference. The crystallized acid effloresces so readily that the pre- 

 cise determination of the water is difficult. In a portion of the crystals 

 which had effloresced in a very marked degree, — 



0.9873 gr. lost on ignition with W0 4 Na 2 0.1760 gr. = 17.82% water 

 2.2472 gr. gave 0.1163 gr. P 2 7 M g 2 = 3.31 % P 2 5 



The ratio of the molybdic to the phosphoric oxide is in this analysis 

 also 24 : 1 ; and, if we compute the results of both analyses for an an- 

 hydrous compound of the two oxides, we find : — 



Calc'd. 



24Mo0 3 3456 96.06 95.91 95.97 



P 2 5 142 3.94 4.09 4.03 



3598 100.00 100.00 100.00 



The analyses leave, I think, no reasonable doubt as to the ratio of 

 the two oxides. Phospho-molybdic acid therefore corresponds in com- 

 position with phospho-tungstic acid, the ratio of the two oxides being 

 24:1, as given by Finkener,* and not 20:1, as stated by Debray. 

 With respect, however, to the number of atoms of water in the crys- 

 tallized octahedral hydrate, 1 may remark that, while the analysis 

 agrees best with the formula given, 



24 MoO a . P 2 5 . 3 H 2 + 59 aq, 



* Loc. cit. 



