236 
SELECTED ARTICLES. 
of the sulphate of baryta (2,) take 58 times the weight of the 
mixed alkaline sulphates (1) and divide the difference by 29. 
2d. For the weight of the potassa. From 87 times the 
weight of the mixed sulphates (1) take 54 times the weight of 
the sulphate of baryta (2,) and divide the difference by 29. 
In the example referred to by Dr. Thomson, m = 29 and 
n = 43.5, whence, 
44x43.5—58x29 1914 
z= — =— — 58 = 8;andw=S7— 81 = 6grs. 
The following arithmetical process may be substituted for 
the algebraic one. 
Having found the weight of the mixed alkaline sulphates 
(1) and the sulphuric acid which they contain (2) the weight 
of the mixed alkalies is known. 
The oxygen which these mixed alkalies contain is known, 
being equal to that of the baryta in the precipitate of sulphate 
of baryta (2,) or -/-gths of the weight of that sulphate. 
Potassa contains J ( T 2 ¥ ths,) and soda ith (^Ihs,) of its 
weight of oxygen. The oxygen of the mixed alkalies, found 
as just stated above, is equivalent to T 2 ^ ths of the weight of the 
potassa, and -j^ths of the weight of the soda; that is to y^ths, 
or -J-th of the weight of the mixed alkalies, together with 
T ^th of the weight of the soda. Therefore taking from the 
weight of the oxygen in the mixed alkalies, -J- of the weight of 
the alkalies themselves, the remainder will be -fa of the weight 
of the soda. 
From which is deduced a very simple rule. 
Find, from the steps of the analysis, the united weights 
of the alkalies, and the weight of the oxygen which they 
contain. Take \th of the former weight from the latter, 
and multiply the difference thus found, by twelve. The 
result will be the weight of the soda. Subtract this from 
the weight of the mixed alkalies, the remainder will be the 
potassa. 
To apply this to the example in which the weight of the 
mixed sulphates is 29, and of the sulphate of baryta, equivalent 
to their acid, is 43.5 grains. 
