Question

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Partile de masa intr-un acid carboxilic sunt C-62% , H-10,4% , O-27,6%.Densitatea relativa a vaporilor acidului fata de hidrogen este 58.Determina formula moleculara a acidului carboxilic.

Answer 1

drelativa=Msubst/MH2=> Msubst=d*MH2=58*2=116g/mol
116g..........100%
xgC..............62%
ygH...........10,4%
zgO...........27,6%
x=116*62/100=72gC
y=116*10,4/100=12gC
z=116*27,6/100=32gO
C:H:O=72:12:32
impartim fiecare raport masic la masa atomica, obtinand raportul atomic
C:72/12=6
H:12/1=12
O:32/16=2
C:H:O=6:12:2 => C6H12O2