Question

Prin arderea a 7,2 g hidrocarbura se obtin 22 g CO2.Determinati f moleculara,stiind ca densitatea vaporilor sai in raport cu aerul e 2,49.

Answer 1

MCxHy=2,49*28,9=72g/mol

n=m/M=7,2/72=0,1 moli CxHy

44g CO2.................12g C
22g CO2.................x=6g C

mH2O=7,2-6=1,2g H2

n=m/M=6/12=0,5 moli C

n=m/M=1,2/2=0,6 moli H2

0,1 moli CxHy............0,5 moli C..........0,6 moli H2
1 mol CxHy................x=5 atomi C.....y=12 atomi H

F.M => C5H12 - n-pentan