Lindgren and Hillebrand — Minerals from Arizona. 449 



Mn0 2 ._ 56-13* 



MnO 6-56 



PbO •_ 26-48 



ZnO 010 



CuO 0-05 



Mo0 3 ..__._ 0-34 



Al o 3 0-63f 



Fe 2 3 t I'Ol 



H 2 l-03§ 



Insol. and silica 7*22 



CaO, MgO, Alk., and loss 0*45 



100-00 

 The material available did not admit of determining quanti- 

 tatively the vanadium, which maybe present in rather more 

 than a mere trace, but neither it nor the phosphorus can influence 

 materially the ratios given below. The vanadium would be 

 effective in two ways : (1) by requiring a base for its neutrali- 

 zation, if existing as an acid constituent, and (2) by liberating 

 chlorine when acted on by hydrochloric acid, and thus affecting 

 the values found for peroxide oxygen. If the iron exists in 

 the ferrous state, it too would affect the values found for the 

 peroxide oxygen and consequently for both the oxides of man- 

 ganese. Assuming it to so exist and applying the proper cor- 

 rections, also deducting from the lead oxide an equivalent for 

 the molybdenum, assuming its existence as molybdate of lead, 

 the following are the results : 



Mn0 2 56.68 -f- 87 = '6515 = 3'00 



MnO 6-11 -v- 71 = -0861 



PbO 2696-T-222-9 = '1165 



FeO... , 0-91 -T- 72- = -0126 }■ 0-217=1-00 



ZnO 0-10-7- 81 = -0012 | 



CuO 0-05-7- 79 = -0006 J 



H 2 _.. 1-03-7- 18 = -0572 =0-264 



If the mineral is to be regarded as anhydrous, the compara- 

 tively simple formula R" (Mn,0 7 )* satisfies the above ratio, and 

 it may be written structurally : 



O = Mn"" - O 



i 

 O 



O = Mn"" 



O 



O = Mn"" - 



*Mean of 56-10 and 5613. Total Mn as MnO from MnS0 4 , 5238 per cent. 

 Peroxide oxygen 10 "31 per cent. 



f With a little Ti0 2 , P 2 5 and V 2 5 . % State of oxidation not known. 

 § Nothing at 100°, only 0-14 per cent below 200°. 



