Question

La analiza a 5.8g substanta organica sau obtinu 8.8 g CO2 , 3.6 g H20 si 1.121 g N2 gazos.Daca 0.5 kmoli cantaresc 58 kg care este formula moleculara a substantei

Answer 1

44g CO2-----------------12G C
8,8...................................Y
Y=2.4
18gH2O.......................2G H2
3,6..................................Z
Z=0.4 G H2
22,4 LN2.......................28g N2
1,121...............................T
T=1.401
SE ADUNA REZULTATELE OBTINUTE =4,2
Nu corespunde cu cantitatea data deci compusul contine O
5,8-4.2=1,6 g O2
0,5KMOLI........................................58KG
1MOL................................................M
M=58/0.5
M=116
se determina COMPOZITIA PROCENTUALA

5,8........2,4g C....0.4g H..........1,4g N2..............1.6g O2
100.........A..................B...............D.......................E
A=41,37% C
B=6,89% H2
D=24.13% N2
E=27,58% O2
SE DETERMINA ATOMII GRAM SI FORMULA BRUTA
C:41,37/12=3,44
H:=6.89/1=6,89
N=24,13/14=1,72
O=27,58/16=1,72
SE IMPART REZULTATELE OBTINUTE
LA NR CEL MAI MIC OBTINUT
C=2
H=4
N=1
O=1
FORMULA BRUTA (C1,2H4NO)n
116=(2x12+4+14+16)n
116=58n
n=2
deci formula este 
C4H8N2O2