444 Rev. W. Whewell on the Employment 



deviates so far from the other, as to suggest strong doubts of 

 the soundness of Berzelius's view. The latter analysis may be 

 putin this form : 11(38 + C + 2 q) + 5 S + C + K ; and the 

 former in this form : 8 (3 S + C + 2q) + 6 S + K : from which 

 it appears that by much the largest portion of the mineral may 

 be considered as 3 S + C + 2q, with an excess of S, and a 

 small quantity of K. If we consider C and K as isomorphous, 

 we may, by a rule which, will be given immediately, write the 

 formula thus, dividing by 9 in one case, and by 13 in the 

 other (see p. 446.) : 



neither of these differs much from 3 S + C, K, + 2q, |C in one 

 case, and T ^C in the other, being replaced by a corresponding 

 portion of K. Other analyses must be employed to determine 

 whether this is the true formula for apophyllite. 



It must, I think, be clear to the reader, that this reasoning 

 could not have been conducted with any tolerable facility or 

 clearness by employing the symbols of Berzelius. According to 

 these, the grouping first mentioned would be 8 C S 3 + K S 6 4. 

 IGq ; the next might be 12 C S 3 + K S 2 + 22q : the one in 

 which K is omitted would be C S 3 4- 2q ; and the one last 

 mentioned would be (C, K) S 8 + 2q. There is in these cases 

 no trace in the symbols themselves of that relation by which 

 their identity is to be tried, or of their dependence on two 

 resembling analyses. 



I will now exemplify the use of such formulae by writing in 

 this notation the results of the analyses of several minerals, 

 which approach to apophyllite in their constitution. These 

 minerals I will designate as follows : (1) Apophyllite, (2) 

 Heulandite, (3) Stilbite, (4) Harmotome, (5) Laumontite, (6) 

 Analcime, (7) Scolezite, (8) Mesotype. I will also, in addition 

 to the letters already introduced, use B for baryta (atom = 78). 



(1) = 308 + 8C + K + 16?=8(3S + C) + 6S + K)+16?. 



(2) = 15 S + 4 A+ C +6? = 4 (3 S + A) +3S-J- C) + 6?. 



(3) = 12 S + 3 A + C + 6 q = 3 (3 S + A) + (3S+C) + 6g. 



(4) = 12 S + 4A+ B +6 q = 4 (2S +A) + (4 S+ B) +69. 

 <5) =- 10 S + 4 A ;+ C + 6q = 4 (2S + A)-f (2S-J-C) + 6g. 



