Chemical Composition of Amphibole. 



The attempt will now be made to show that by assuming 

 the existence of certain bivalent basic radicals of the type just 

 indicated, and regarding them as replacing the hydrogen atoms 

 of the amphibole acid, a similarity may be found between 

 these basic hornblendes and the members of the tremolite-acti- 

 nolite group. There will first be deducted from the ratios the 



— Al— (F,OH) 

 alumo, fluor-hvdroxyl radical ^>0 , the same as in 



_A1— (F,OH) 

 the tremolite group. Next there will be deducted a radical of 



the type 



— Al< 



-Na 

 >(Fe,Mg) 



and in most cases these two 



,0- 



— A1 <0— Na 



radicals may be so chosen as to satisfy or include all of the sesqui- 

 oxides, and leave the protoxide base in sufficient amount to 

 form RSi0 3 . In one analysis (X) a third radical seems neces- 



_A1— Ov 

 sary . ^>0 ^>R . To indicate the method of calculation ; 



_A1— CK 

 there will be deducted from the total ratio first the alumo, 

 fluor-hydroxyl radical, with its equivalent of silica, thus Si0 2 , 

 A1 2 3 and (F,OH) in the proportion 1:1:1, and next the 

 basic radical with A1 2 3 , (Fe,Mg)0 and JNTa 2 with its equiva- 

 lent of silica in the proportion 1:1:1:1. It is then possible 

 to calculate the proportion of the total hydrogen atoms 

 replaced by the several bivalent radicals and the different bases, 

 and attention may be called to the fact that calcium, helped 

 out at times by traces of sodium and potassium, satisfies 25 

 per cent, or one quarter of the total hydrogen atoms of the 

 amphibole acid, the same as in the tremolite group ; — that this 

 is a mere coincidence seems hardly possible, and it may be 

 taken, it is believed, in support of the theory advanced. The 

 results of calculation are as follows : — 



Analysis Vlf, Renfrew, Canada. 



Eatios 



[Si,Ti]0 2 

 [Al,Fe] 2 G 3 

 [Fe,Mn,Mg]0 

 CaO 



•738 

 •124 

 •468 

 •176 



[K,Na] 2 

 H„0 



•069 

 •OH 6 ) 



F 2 



•048 f 



Al— (F,OH)~ 



// 



>o 



Residue 



Al— (F,OH)_ 





-•065 

 -•065 



—•065 



A< 



A1<T 



O 



-Na" 

 >R 



•673 

 ■059 

 •468 



•176 

 •069 

 •019 



O— Na_ 



-•059 

 -•059 

 -•059 



-•059 



Residue 



•614 



•409 

 •176 

 •010 

 •019 



■186 



Total RO= -614 



