avem a moli C6H6 si b moli C7H8
M.C6H6 = 78 g/mol
M.C7H8 = 92 g/mol
1 mol C6H6 ........................... 6g H
a moli ................................... 6a g H
1 mol C7H8 ......................... 8g H
b moli .................................. 8b g H
100% ................................ (78a+92b) g amestec
8,31% H ................................. (6a+8b) g H
78a+92b = 12,03(6a+8b)
=> 5,8a = 4,26b => 1,36a = b
=> raport C6H6 : C7H8 = a : b = 1,36 : 1 /x3 = 4 : 3
