in connexion with Chemical Formula. 373 



of perissad atoms in a compound, the number of nodes (n) ex- 

 ceeds the number of loops (x) by 



P 



or 



a-" 1 ' 



p 



x = n~2 +1. 



But since triad and tetrad atoms count as ordinary nodes, while 

 pentads and hexads count as double nodes (§ 5), 



n= (III) + (IV) + 2(V) + 2(VI) + . . . , 



where (III) denotes symbolically the number of triad atoms in 

 the compound, (IV) the number of tetrads, and so on ; hence 



*=1-| + (III) + (IV)+2(VH..., 



or 



or 



*=J + 1-(I) + (IV) + (V)+2(VI) + 



2ff=2-(I)+(in) + 2(IV) + ...=2 + 2(r.&-2). 



Each of these expressions has its convenience ; the last tells us 

 that if there are r atoms of atomicity k, s atoms of atomicity I, 

 &c. (that is, in the compound K r L s . . .), the number of loops is 



r(*-2) »(/-2) ,, 



— g— +— j— +... + 1. 



(9) This general theorem shall now be applied to a special 

 case. Suppose L is a monad like hydrogen, 1= 1, and the num- 

 ber of loops in K n H m is — ^ — — tt + 1. Calling this num- 



ber x, it is evident that m = n(Jc — 2) + 2 — 2a?, or that there are x 

 loops in the compound K n H M(t _ 2 )+ 2 -2a;« If now K be a tetrad 

 (such as carbon), we shall have x loops in C n H 2w+2 _ 2x , which is 

 Clifford's rule (§ 1). It also follows that the only hydrocarbon 

 with no loops is C n H 2n+2 ; and more generally, that if an A-dro- 

 c-bon with w-atoms of atomicity c has no loops, it must contain 



n ( c 2) +2 



— ~ — -r — atoms of atomicity h. This shows that in binary 



linkages with no loops one of the elements must be a monad, 

 while the other may have any atomicity whatever. 



(10) A simple way of proving the general formula (§ 8) for 

 the number of loops in a compound K,.L 5 . . . consists in taking a 

 ring of any number of dyads, turning them into &-ads by adding 



