1856.] HAUGHTON — GRANITES OF IRELAND. 197 



a formula which is frequently found for the hornblende containing 

 alumina. 



The syenites of Carlingford are composed of variable proportions 

 of the anorthite and hornblende just described ; and a remarkable 

 instance of the formation of this syenite from granite occurs at 

 Grange Irish. The granite No. 2 there pierces the carboniferous 

 limestone in dykes, and in the dykes is converted into a coarse- 

 grained syenite, composed of anorthite and hornblende, in no way 

 distinguishable, physically or chemically, from the syenites of Car- 

 lingford Mountain. 



The metamorphic reaction of the granite and limestone upon each 

 other is strikingly shown at this point ; the limestone being con- 

 verted from an earthy bluish-brown stone into a crystalline marble 

 of a light bluish-white colour, and having garnets developed in 

 many places ; on the other hand, the granite is converted from a 

 ternary compound of quartz, orthoclase, and hornblende, into a 

 binary compound of anorthite and hornblende, of the following com- 

 position : — 



Granite converted into Syenite, from Grange Irish. 



Specific gravity =2*757. 



Per cent. Atoms. 



Silica 47*52 1-033 



Alumina 28*56 0-555 



Protoxide of iron. . . . 7*23"] 



Lime .^ 15'44 V. . . . 0'715 



Magnesia 1 •48J 



100-23 



Specific gravity of average specimen of medium-grained syenite 

 from Carlingford Mountain = 2*877. 



From the preceding analysis we can show that this rock is a 

 binary compound of anorthite and hornblende. From the formula 

 for anorthite already given, and assuming hornblende to be a simple 

 silicate of protoxides, as shown from the analysis of the hornblende 

 of this district, we have, denoting by A and H the number of atoms 

 of anorthite and hornblende in the rock, — 



1-033 = 3A -f a?H 

 0-555 = 2A+ (l-a?)H 

 0715 = 2A + H. 



In these equations, x denotes the proportion of silica, considered 

 as atomic quotient, which enters into the composition of each atom 

 of hornblende. 



Adding together the first two equations so as to eliminate x, and 

 solving from this sum and the third equation, we find 



1-588 = 5A -h H 

 0-715 = 2A -f H, 



