78 GEORGE I. FIN LAY 



Let:x;=the number of albite molecules 

 Let ;y=the number of nephelite molecules 

 then 5(;+;y=Na20 = AL03 

 and 6:v+23'=Si02 

 ^+^'=125 

 6:x;+2>'=53i 

 2x-\-2y = 2^o 

 subtracting 4X=28i 



.T= 70 molecules of albite 

 ;)' = 55 molecules of nephelite. 



It is clear that with nephehte in the norm we shall not have quartz. 



Analysis M proceeds on the same lines as L except that the minor 

 inflexible molecule zircon (Z), is first introduced, taking ZrOj^ 

 Si02 in accordance with its formula ZrO^.SiOa. 



In N we have a case with SiOa still lower than in M. If we 

 attempt to calculate it in the same way as L and M we find that we 

 have run over on SiOa by 21 units, holding out Na20.Al203, 45, 

 for albite and nephelite, and making the allotments for apatite, 

 orthoclase, anorthite, magnetite, diopside, and olivine. We have 

 therefore no Si02 with which to make even nephelite with the 45 

 Na20.Al203 held out in the beginning. 



There is not enough SiOa therefore to begin the calculation by 

 making orthoclase. This case is analogous to the situation in L 

 and M where Na2 is distributed between albite and nephelite. We 

 proceed by holding out all the K2O and equal AI2O3 for a certain 

 amount of orthoclase, and a certain amount of leucite (K2O.AI2O3. 

 48102), which calls for less Si02 than orthoclase does. These 

 minerals will each use up K2O and AI2O3 in the ratio 1:1. The 

 NajO is allotted with AI2O3 and Si02 to nephelite. This is much 

 lower in Si02 than albite is. Anorthite, magnetite, diopside, and 

 olivine are then made. The SiOg remaining over is 369. This is 

 given to the K2O.AI2O3, previously set aside, for orthoclase and 

 leucite, by means of the equations where 



x—\ht number of molecules of orthoclase 

 and y=i\\& number of molecules of leucite 

 x-\-y=lL,0 

 and 6x+43'=Si02 

 Here x+y=2)6 

 6x-\-4y = 2)()g 



X=24 



and y=s^- 



