3S8 Mr. Children's Summary View of [MaV, 



/ Let us next take an analysis of copper pyrites, and suppose 

 that it has given us 



Copper 34 



Iron 30 



Sulphur 36 



100 

 The atom of copper by the table is 791*39 ; that of iron 

 678-43. Therefore y^^ = 429 atoms of copper ; ^^^ = 442 



36 



atoms of iron, and = 1789 atoms of sulphur. Now these 



numbers are nearly as 1, 1 and 4, and consequently the sulphur 

 must be equally divided between the two metals, so as to fonh 

 bisulphurets, each containing 1 atom of metal, and 2 atoms of 

 sulphur. If we calculate the composition of the pyrites accord- 

 ing to these numbers, we shall have 



Bisulphuret of copper , 52*48 



Bisulphuret of iron 47*52 



100*00 

 Or if we take the elements separately, 



Copper 34*79 



Iron 29*82 



Sulphur 35*39 



100*00 



which agrees very nearly with the experimental residts, and 

 confirms their accuracy. 



Let us now take the analysis of a quaternary compound, a 

 variety of emerald, which gave 



Atoms. 



Sihca 68*64 or oxygen 34*52 = 8 



Alumina 17*96 8*38 = 2 



Glucina 13*40 4*17 = 1 



100*00 



By the tables, we find the respective quantities of oxygen in 

 the three elements of the mineral as stated above. Now we 

 may consider this compound (says M. Beudant) in two ways, 

 either as consisting of one base (glucina) united to a double acid 

 (silica and alumina), or as a double salt formed of the silicate of 

 alumina and silicate of glucina ; both views lead to the same 

 conclusion. In the first case the mineral is supposed to consist 

 of 2 atoms of acid (composed of 4 atoms of silica and 1 atom of 

 alumina) combined with 1 atom of glucina. In the second 



