fi04 



Mr. J. B. Jukes and the Rev. Samuel Haughton on the 



Solving these equations for I and m, I find, after some numerical reduc- 



tions, 



m=14'18, 

 I = 5-27. 



From (13), by the aid of (e) and (/), we find 



M= 00353, 

 L = 0-0047. 



(13) 



(14) 



From (14) combined with either (6) or (c), we obtain — 



F= 0-2025. (15) 



From equations (14) and (15), combined with (a), we find — 



Q= 0-6148; (16) 



with which combining {d), we find — 



(7 = 27-66 per cent. (17) 



From (13) and (17), with (h), we obtain — 



/= 52-94 per cent. (18) 



And, finally, from (18), (15), and {g), we find — 



(^ = 253. (19) 



Bringing together the values of the nine unknown quantities, and the atomic 

 weights known by observation, we have — 



9 = 27-66 Q= 0-6148 



/= 52-94 i^= 0-2095 



m = 14-18 Z = 0-0047 (20) 



Z= 5-27 ilf = 0-0353 



= atomic weight of felspar, 253. 



The unknown quantities just determined may be subjected to a severe test 

 by substitution in the original per-centage analysis of the granite, as follows, — 



Let/,;, /„,,/i,/„/„,/p, /.,/„, denote the unknown proportions of silica, alu- 

 mina, peroxide of iron, lime, magnesia, potash, soda, and water, belonging to 



