70 THE ATOMIC WEIGHTS. 



to calculate the atomic weight of nitrogen. Let us first collect and num- 

 ber these ratios : 



(i.) Ag : AgNO 3 : : ioo : 157-479, .o3 



(2.) AgNO 3 : AgCl : : ioo : 84-3743, - OO2 5 



(3.) AgNO 3 : KC1 : : ioo : 43- 8 7i5> .0004 



(4.) AgNO 3 : NH 4 C1 : : ioo : 31.488, .0006 



(5.) Ag : NH 4 C1 : : ioo : 49.5983, dr .00031 



(6.) Ag : NH 4 Br : : ioo : 90.8299, .0008 



(7.) KC1 : KNO 3 : : ioo : 135.636, .0007 



(8.) KC1O 3 : KN0 3 : : ioo : 82.500, .0012 



(9.) NaCl : NaNO 3 : : ioo : 145 418, .001 1 

 (10.) NaClO 3 : NaNO 3 : : ioo : 79.8823, .0029 

 (n.) NH 3 : HC1 : : i.oo : 2.1394, d= .000053 



From these ratios we are now able to deduce the molecular weight of 

 ammonium chloride, ammonium bromide, and three nitrates. For these 

 calculations we must use the already ascertained atomic weights of oxy- 

 gen, silver, chlorine, bromine, sodium and potassium, and the molecular 

 weights of sodium chloride, potassium chloride, and silver chloride. The 

 following are the antecedent values to be employed : 



Ag = 107.108, d= .0031 

 K = 38.817, =b .0051 

 Na 22.881, .0046 

 Cl = 35.179, =b .0048 

 Br = 79.344, .0062 

 O 3 = 47.637, -0009 

 AgCl 142.287, .0037 

 KC1 = 74.025, .0019 

 NaCl = 58.060, .0017 



Now, from ratio number five we get the molecular weight of NH 4 C1 = 

 53.124, .0016, and N = 13.945, .0051. 



From ratio number six, NH 4 Br = 97.286, .0029, and N = 13.942, 

 .0077. 



From ratio number eleven, NH 3 = 16.911, .0048, and N = 13.911, 

 .0048. 



From ratio number four, which involves an expression of the type 

 A : B : : C + x : D + x, an independent value is deducible, N = 13.935, 

 .0073. 



For the molecular weight of silver nitrate there are three values, 

 namely : 



From (i) AgNO 3 == 168.673, .0049 



From (2) " = 168.634, .0066 



From (3) " = 168.731, .0046 



General mean A gNO 3 = 168.690, .0030 



Hence N= 13.945, .0044. 



