2 14 JVilliam E. Ford, 



cleavage rhombohedron of the mineral and that the carbon and 

 metallic atoms have the arrangement shown in Fig. i. They 

 have further been able to show the probable position of the 

 oxygen atoms, but for the sake of simplicity these are not shown 

 in the figure. 



Knowing the distance between the molecular layers and the 

 angles of the rhombohedral unit cell it is possible to calculate 

 its volume. A\'. L. Bragg has found this for calcite to be 31.25 X 

 10"-* cc. The specific gravity of pure calcite has been very 

 accurately determined as 2.715. Multiplying the volume of the 

 unit cell by the specific gravity will give the mass contained in 

 the cell. This equals 85.0 X lO"-* g. The mass of a molecule 

 of CaCOg is given by multiplying the molecular weight by the 

 weight of an atom of hydrogen as below, 



loo.i X 1-64 X io~-* = 164.0 X io~^* g. 

 From these two expressions it is evident that it takes two of the 

 unit cells to contain one molecule of CaCOs. Eight unit cells 

 will, therefore, contain four molecules of CaCO,. This is the 



case illustrated in Fig. i. It must be remembered in this con- 

 nection that the eight atoms of calcium placed at the corners of 

 the rhombohedron are each at the corners of eight similar figures 



