Question

Afla formula moleculara a substantei care:
a)contine doi atomi de azot in moleculara si alaturi de acesta 41,38% C,respectiv 10,34% H
b.are raportul de masa C:H:0=6:1:4 si are masa molara egala cu 88.Va rog sa-mi aratati si rezolvarea.Va multumesc!

Answer 1

Vezi poza
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Answer 2

a) C: 41,38% : 12 = 3,448     | : 3,448 = 1
    H : 10,34% : 1 = 10,34      | : 3,448 = 3
    N : 48,28% : 14 = 3,448    | : 3,448 = 1   formula bruta : CH3N
formula moleculara :(CH3N)n ⇔ CnH3nNn   n = 2 ⇒ C2H6N2
b) C:6 : 12 = 0,5     | :0,25 = 2
     H:1 :1 = 1           |:0,25 = 4
     O:4 :16 = 0,25    |:0,25 = 1    ( C2H4O)n
(2·12 + 4·1 + 1·16)n = 88     n = 2    C4H8O2