Question

3,7000g subastanta organica cu masa moleculara 74, formeaza prin ardere 6, 6000g dioxid de carbon si 2,7g apa. Care este formula moleculara a hidrocarburii? (Demonstrati)

Answer 1

s.o (s) = 3,7g 
M = 74g/mol

a = 6,6g CO2
b = 2,7g H2O

%C = 300a/11s = 300*6,6/11*3,7 = 48,64
%H = 100b/9s = 100*2,7/9*3,7 = 8,1
%O = 100-(%C+%H) = 43,251

nC = %C*M/AC*100 = 48,64*74/1200 = 3 atomi C
nH = %H*M/AH*100 = 8,1*74/100 = 6 atomi H
nO = %O*M/AO*100 = 43,251*74/1600 = 2 atomi O

F.M => C3H6O2

NE = (2*3+2-6)/2 = 1

74g..........................................3*44g ....3*18g
H3C-CH2-COOH + 9/2O2 => 3CO2 + 3H2O + Q
3,7g...........................................x.............y

x = 6,6g CO2
y = 2,7g H2O..

Answer 2

nCO2 = 6,6/44 = 0,15 moli ⇒ nC = 0,15 moli de atomi   mC = 1,8g
nH2O = 2,7/18 = 0,15 moli ⇒ nH = 0,3 moli de atomi      mH = 0,3g
                                                                                          -----------------
                                                                    mO = 3,7 - (1,8+0,3) = 1,6 g
n subst. = 3,7/74 = 0,05 moli
0,05 moli subst.............0,15 at.C.........0,3at.H............0,1atO
 1 mol.........................x = 3..................y = 6...............z = 2
C3H6O2