247 
In calculating the amount of alumina, which would be re- 
quired by soda and lime in the case of feldspar, we obtain ; 
NaOi = 16°8%, =, AlLO; > =) 21-41 
CiOve— 0180s. All Osi 0:38 
11-74 
which in connection with other facts, entitle us, I believe, in 
really considering these bases to have belonged to the albite 
of the rock. 
Al,O;s = 11:74 = SiO; = 30°79 
ISO) eS RY SOR 9.97 
CaO = O18 = SiO; = 0:29 
18°79 41:05 
18-79 
59°84 Albite. 
In subtracting the silica of the albite from the total silica 
found, less the part dissolved in hydrochloric acid, we should 
obtain the amount of quartz, a figure which will be found 
also in subtracting the albite from the part undissolved by 
the acid, making allowance for 0:29 per ct. Al,O,, which we 
counted dissolved ; thus we obtain 
95-79 Insoluble in Acid. 
59°84 Albite. 
35°95 77°28 Silica total. 
0°29 41:05 Silica of Albite. 
36°24 Quartz. 36°23 Quartz. 
The total composition may be then represented in the follow- 
ing way: 
59°84—A bite. 
36°23—Quartz. 
0:42—Silica soluble. 
0-82— Water. 
2°76—Bases dissolved (3°05 less 0-29.) 
100-07 
The other three are calculated in the same way. 
