1822.] Method of analyzing the Ores of Nickel. ' 441 



a. 15*1 parts of the pulverized ore (deprived of its carbonate 

 of lime) were dissolved in nitromuriatic acid, which left 

 0*56 residuum, of which 0*4 were sulphur, and 0*16 silica. 



h. The solution, precipitated by muriate of barytes, gave 

 18*18 parts of sulphate of barytes, equivalent to 2*5 parts 

 of sulphur : the whole quantity of sulphur was, therefore, 

 2*9 that is to say, 19*29 hundredths of the weight of the ore. 



c. The liquid having been filtered, was saturated with am- 

 monia until a precipitate began to appear ; acetate of lead was 

 then poured in as long as a precipitate formed. This precipi- 

 tate was then washed in boiling water, and as after an edul- 

 coration continued for a long time, the washings continued to 

 re-act with nitrate of silver. I dried it, without attempting to 

 deprive it entirely of its muriate of lead. Heated in the fire, 

 it became yellow, and weighed 45*5 parts. 



I dissolved 42*53 of this in diluted nitric acid, which left 

 undissolved 0*58 part of red oxide of iron, which gave out 

 the odour of arsenic in a heated tube. The solution, mixed 

 with sulphate of soda, gave 40*73 parts of sulphate of lead, 

 and afterwards, with nitrate of silver, 3*77 parts of chloride of 

 silver. The liquid was deprived of the superfluous nitrate of 

 silver, by the addition of muriatic acid ; and afterwards filtered 

 and evaporated to dryness. The dried mass dissolved ia 

 water left a white powder, which weighed 1*25 : this was 

 arseniate of-lead. The liquid neutralized as nearly as possible 

 by means of caustic potash, deposited 0*1 part of a light 

 and whitish substance ; this was the neutral arseniate of iron. 

 These quantities were obtained from the 42*53 parts dissolved 

 by the nitric acid ; calculating for the 45*5 parts, the entire 

 weight of the precipitate obtained by means of the acetate of 

 lead, we have the following quantities : 



Sulphate of lead . . 43*58 = 32*06 parts of oxide of lead. 

 Chloride of silver . . 4*03 = 0*77 of muriatic acid. 

 Arseniate of lead . , 1*34 = 0*89 of oxide of lead. 

 Arseniate of iron . . 0*11 = 0*03 of oxide of iron. 

 Oxide of iron = 0*62 



34*37 



This 3437 must be taken from the 45*5 parts of the above 

 mentioned precipitate, to obtain the weight of the arsenic acid, 

 which is 11*13, and which corresponds to 7*268 parts of ar- 

 senic. 



We must not expect that a result obtained by so intricate 

 an experiment can be exact ; because, if the small errors 

 of each of the determinations, for example, small losses, are 

 added together, the sum will occasion a great error in the ge- 

 neral result. And this really happens. 



d. The liquid remaining after the separation of the ^sejii^c 



