Question

stabiliti formula bruta si formula moleculară a substanței ce contine 60,87%C, 4,35%H, 34,78%O masa molara a substantei M=138

Answer 1

100%-60,87%C - 4,35%H - 34,78%O = 0

C = 60,87/12 = 5,0725

H = 4,35/1 = 4,35

O = 34,78/16 = 2,174     : 2,174

Fb = C2,33H2O

Fm = (C2.33H2O)n

n = miu.Fm/miu.Fb, miu.Fb= 46

= 138/46 = 3

=> Fm = C7H6O3