168 Dr. Thomson on the [March, 



The three atoms of sulphuric acid then must be combined with 

 three atoms of alumina. Consequently 6*74555875 must be 



equivalent to three atoms of alumina ; but = 2*2485, 



a number which would represent the weight of an atom of 

 alumina if my experiments had been perfectly accurate. But it 



is easy to show that my number is yr^th part too small, and that 



the true weight of an atom of alumina is 2*25. 



For this purpose let us take the constituents of 60*875 grains 

 of alum as determined by the preceding experiments. 



Sulphuric acid 20*000 or 4 atoms 



Water 28*125 25 atoms 



Potash 6*000 1 atom 



Alumina 6*745 3 atoms 



60*870 

 Loss 0*005 



Total 60*875 



There is obviously a loss amounting to 0*005 of a grain. If 

 Tve add this to the alumina, it will make the three atoms of it to 

 weigh 6*75 ; and consequently the weight of 1 atom will be 

 2*25. Now as the weight of an atom of sulphuric acid, potash, 

 juid water, is known with precision, it is obvious that the loss 

 can only fall upon the alumina. Hence there can be no doubt 

 that the true quantity of alumina contained in 60*875 grains of 

 alum is 6*75, and that an atom of alumina weighs exactly 2*25. 

 Alum then is composed of 



4 atoms sulphuric acid = 20*0 



3 atoms alumina 6*75 



1 atom potash 6*0 



25 atoms water , , . 28* 125 



So that the weight of an integral particle of alum is 60*875. 

 We may represent the composition of alum in a different way,. 



3 atoms sulphate of alumina 21*75 



1 atom sulphate of potash 11*0 



25 atoms water 28*125 



60*875 



These proportions are more convenient for calculation tha» 

 the usual mode of representing the constituents of 100 grains of 

 alum. However, for the sake of those who prefer that method, 

 I shall state the centesimal constituents of alum as follows : 



