Question

Un compus organic A contine 54.54 c ,9.09H , restul O in procente masice si are masa moleculara 44g/mol.Determinati formula moleculara

Answer 1

Po (procentul de oxigen)=100-54,54-9,09=36,37
(1)C 54,54/12=4,54 |:2,27=>C=2
(2)H 9,09/1=9,09 |:2,27=>H=4
(3)O 36,37/16=2,27 |:2,27=>O=1
(1), (2), (3)=>(C2H4O)n formula brută
M(C2H4O)=44 => 44n=44 => n=1 => C2H4O formula moleculara
/=supra
:=împărțire