Results of Crystal Analysis. 73 



the three substances, there can be no doubt that also for 



kassiterite the ratio -^ is equal to c'la' . 

 dioo 1 ' 



The elementary cell has a volume 



"100 "001 — ^100 ~> 



a 



and the number of molecules n associated with this volume 

 is 





(5) 



,p is the density, N the number of molecules in a gram- 

 molecule, M is the molecular weight; and in order to 

 preserve the analogy between the three substances, we write 

 the molecular formula for rutile Ti 2 4 and for kassiterite Sn 2 4 . 

 The number n is given in the last column of Table III. ; 

 and we see that within the limit of experimental error 

 the number n is equal to 1/8 for all three substances, or 

 in a rectangular prism with sides 2d 1Q0 , 2d 10 o« and 2d oi there 

 should be just one molecule. 



Table III. 



Substance. 



^lOO - ^001' 



1 



^001. 



^100 



c 



a' 



n. 



ZrSi0 4 



cm. | cm. 

 2-30 xl0~ 8 T47 XlO -8 



2-26 „ 1-46 „ 



2-335 „ 



0-640 

 0646 



0-639 

 0-644 



0-124 

 0-123 

 0-123 



TLO, 



Sxi.,0 





Before proceeding further in our attempt to arrange the 

 atoms, we shall make a few remarks regarding our 

 elementary lattice, and the formation of compound lattices 

 and some of their properties. 



Our elementary lattice will be a prism with one atom in each 

 •corner. Let the side of its square base be a, and height c ; 

 then the spacing of a simple elementary lattice would be 



^100 = #, 



'001 





'101 



din = 



\A+(°y' \A + cy 



(6a) 



