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Deterninatu formula moleculara a unui compus organic care contine 61,01 C, 15,25% H si restul azot avand p=59 g/mol​

Răspuns :

Roberteladrianpb6zy7

miu.compus = 59 g/mol

100% - 61,01% C - 15,25% H = 23,74% N

notam compusul CaHbNc

100% ................ 61,01% C ............... 15,25% H ........... 23,74% N

59 g/mol .......... 12a g C .............. b g H ................. 14c g N

=> a = 3, b = 9, c = 1

=> Fm = C3H9N

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