216 Adams and Harrington — Alkali Hornblende and 



Here (SiO a )s = 3(56 + 18) = 222 



(R a + R0)2 = 3(56 + 18) = 222 



(R 2 + RQ ) _ 222X3 _ 



(R 2 + CaO)z =3 



= 95 



(MgO) A = 584-(MgO)s = 584 — 127 = 457 



(MgO)A 457 

 (CaO) A =- J ~- = — = I 52 



(CaO)2 = 222 — (CaO) A =222 — 152 = 70 

 (Na 2 0+K 2 + H 2 0)2=(R 2 + CaO)2-(CaO)s = 95 -70 = 25 

 But (Na 2 + K 2 0)2 =12 + 9 = 21 



.-.(H 2 0)2 = 4 

 Finally (Si0 2 ) A = (MgO + CaO) A = 457 + 152 = 609 



Having thus deduced the molecular ratios of the syntagma- 

 tite and actinolite, the numbers for each constituent are multi- 

 plied by the corresponding molecular weights, in order to 

 obtain the theoretical relative weights of the constituents of 

 the mixed hornblende. 



Syntagmatite. 



Actinolite. 



222X60 =13320 



609X60 =36540 



56X102*6= 5745 



. 



18X160 = 2880 







127 X 40 = 5080 



457X40 = 18280 



70X 56 = 3920 



152X56 = 8512 



12X 62 = 744 



_ . 



9X 94 = 846 



. ____ 



4X 18 = 72 







32607 



63332 



Then, 



(32607 + 63332): (13320 + 36540):: 100: x 



and x = 51*97 = p. c. of Si0 2 in the mixed hornblende. And 

 in like manner the percentages of the other constituents are 

 calculated. 



But 32607 : 63332 practically as 1:2, and therefore the 

 formula of the Edenville hornblende might be regarded as 



ir in 



R 3 R 2 Si 3 12 + 2(Mg 3 C a Si 4 O ia ) 



or as Scharizer gives it 

 11 in 

 10(U a U 2 Si a O 1 J + 20(Mg ! CaSi 4 O l2 ) 



