251 
as chrysocolla, which the large amount of water seems to in- 
dicate, requires, 
CuO” = 6:51 
SiO; == 4-92 
HO = 295 (2Eq.) 
14°38 Chrysocolla, 
The total composition being perhaps represented thus: 
30°33 Albite. 
16°08 Clay. 
14°38 Chrysocolla. 
87.29 Quartz. 
0:48 Silica soluble. 
0:84 Water. 
0°58 Bases soluble. 
99°93 
5. Shale of a dark brownish red color, yrelding a brown red 
powder. 
The iron of this sample, which according to the views of 
Prof. Wurtz is in the form of limonite, would require exactly 
the amount of water found. 
Fe.0; = 16°46 (1 Eq.) 
Ho! A gs (3 Eq.) 
22:01 Limonite. 
In subtracting this and the soluble Bases from the total 
amounts found, we obtain 
SiOs = 44-81 = 60-51. cont, O11. — 82°27 
AlsOs = 2029 = 27-%2 a =" 12:02 
CaO 4478) ==) OOS dc aay alos 
MgO == 1 ee RI se OLS 
NaO = DEBT) ee BIS 6 =a OO 
73°72 100-00 
In considering this a true compound, we find the oxygen 
of the bases one half of that of silica, but whether we are en- 
titled to do so, as Prof. Wurtz thinks, he believing it to be 
disintegrated mica, muscovite, will not attempt to decide. 
