214 SMITH'S INTERMEDIATE CHEMISTRY 



then all the information we require for making the needed formulae. 

 Suppose we know that the atomic weight of aluminium is trivalent 

 Al m (see next section). Making the total valences of each half 

 of the compound alike, we get the formulae: 



When we know the valences of the elements and radicals, we can 

 make the formula of any required compound. 



The reader must therefore make a special effort always to learn 

 the valences of each element and radical, and always to use them 

 in making formula. 



The reader must also always check every formula he writes from 

 memory, to make sure that it is correct. Thus, if he thinks the 

 formula of zinc nitrate is ZnN0 3 , he must count the valences, 

 Evidently, the correct formula is Zn(N0 3 ) 2 . 



How to Learn the Valence of an Element. To find out 

 the valence of an element, we must obtain the formula of one sim- 

 ple compound of the element, containing another element of known 

 valence. Thus, what is the valence of carbon? Its oxide is CO 2 . 

 The total valence of oxygen here is 2 X 2 = 4. Carbon C IV is 

 therefore quadrivalent. Hence its chloride must be CVCU 1 (car- 

 bon tetrachloride), and its compound with hydrogen C IV H4 r 

 (methane, composing a large part of natural gas). When car- 

 bon combines with a trivalent element, equi-valent amounts of 

 each element must be used, as in Al 4 in C 3 IV (aluminium carbide), 

 where Al 4 m and C 3 IV contain 3 X 4, or 12 units of 'valence each. 



Again, when we know the formula of sodium iodide to be Na 1 !, 

 or that of hydrogen iodide to be H 1 !, we infer that iodine is univa- 

 lent. The formula of silica (sand) SiCV 1 shows silicon to be 

 quadrivalent, and indicates that the chloride must be SiCl 4 . Simi- 

 larly the formula of calcium carbonate Ca n C0 3 shows that the 

 radical CO 3 , which is common to all carbonates, must be bivalent. 



The chemist does not memorize the valences themselves; he 



