ALUMINA. 293 



By comparing this analysis with the two follow- 

 ing, it will appear probable that the alkali, oxide 

 of iron, and carbonic acid, are merely accidental 

 ingredients ; and that silica, alumina, and lime, 

 are the only essential constituents. 



Let us consider meionite as a compound of 2 

 integrant particles of silicate of alumina and 1 

 integrant particle of silicate of lime ; and let us 

 calculate its constitution. 



(1.) 2-25 (atom of alumina) : 2 (atom of silica) : : 30-6 (alu- 

 mina in the mineral) : 27'2 = silica united to alu- 

 mina. 



(2.) 3-5 (atom of lime) : 2 : : 22-1 : (lime in the mineral) : 

 12-62 = silica united to the lime. 



But 27-2 + 12-62 = 39-82. Now, the silica 

 actually found in the meionite exceeds this by 

 0-98 grain. 



The silica united to the lime amounts to ~ 5 th 

 of the whole ; consequently, ^-jth of 0*98 or 

 0-311, belongs to the silica united to the lime. 

 There remains 0-669, which being added to the 

 27*2, makes 27-869 for the silica united to the 

 alumina in meionite by Gmelin's analysis. Now 



27-869 : 30-6 : : 2 : 2-196 = atomic weight of alumina. 



Professor Stromeyer analyzed some very pure 

 specimens of meionite from Monte Somma, and 

 found the constituents as follows:* 



* Untcrsuchungen, I. 580. 

 T 3 



