A= C6H5-CH3-xClx
100% ........................ 92+34,5x g A
28,06% Cl ................ 35,5x g Cl
=> x= 1 => A= C6H5-CH2-Cl, clorura de benzil
=> B= C6H5-CH2-OH, alcool benzilic
C= CH3-C6H5-y(NO3)y
avem 7x12 = 84 g C si 3x16y g O = 48y g O
C:O = 16,8:19,2
=> 84/48y = 16,8/19,2 => y = 2
=> C= CH3-C6H3(NO3)2 , o,p-dinitrotoluen
=> D= CH3-C6H3(NH2)2 , o,p-diaminotoluen
b)
36,8g m g
C6H5-CH3 --> ..... --> CH3-C6H3(NH2)2
92 122
=> m = 36,8x122/92 = 48,8 g D obtinut teoretic la 100%
100% ....................... 48,8g D
90% ............................ Cp = 43,92 g D