Question

Prin arderea in oxigen a 178 g substanta organica,A, se obtin 126 g H2O si 156,8 g L(c.n) gaze . Dupa condensarea apei si trecerea prin solutie de NaOH raman 22,4 L N2. Se cere :
a) determina formula moleculara a substantei organice stiind ca are in molecula de doua ori mai multi atomi de oxigen decat de azot
b)daca A este un aminoacid ,scrie o formula de structura posibila a acestei substante organice si denumeste-o.va rog

Answer 1

Notam formula: CxHyNzO2z
VCO2=Vgaze-VN2=156,8-22,4=134,4L
2*M............................2x*22,4....18y.......22,4z
2CxHyNzO2z + O2 -> 2xCO2 + yH2O + zN2
178g...........................134,4L....126g.....22,4L
44,8x*126=134,4*18y => y=2,3333x
134,4*22,4z=22,4*44,8x => z=0,33333x
f.m: CxH2,33333xN0,3333xO0,66666x => (CH2,33N0,33O0,66)x
presupunem ca x=3 => C3H7NO2
b)
CH3-CH(NH2)-COOH
acidul-2-aminopropanoic numit uzual alfa alanina