Question

La analiza a 1,18 g de substanta organica s-au obtinut 1,344 dm³ CO₂, 2,016 L H₂O si 223 cm³ N₂ (toate volumele au fost masurate in conditii normale ) Determina formula moleculara a substantei organice, stiind ca densitatea vaporilor sai fata de azot este de 2,1.

Answer 1

m/M=V/Vm=>m=M*V/Vm=44*1,344/22,4=2,64g CO2

m/M=V/Vm=>m=18*2,016/22,4=1,62g H2O

n=V/Vm=0,224/22,4=0,01 moli N2

mN2=28*0,01=0,28g

44g CO2...............12g C
2,64g CO2...........x=0,72g C

18g H2O...............2g H2
1,62g H2O.............x=0,18g H2

mO=1,18-0,72-0,18-0,28=0 => nu contine O2

n=m/M=0,72/12=0,06 moli C

n=m/M=0,18/2=0,09 moli H2

dgaz=M/Mgaz=>M=dgaz*Mgaz=28*2,1=59g/mol

n=m/M=1,18/59=0,02 moli s.o

0,02 moli s.o.............0,06 moli C.....0,09 moli H2......0,01 moli N2
1 mol........................x=3 atomi C.....y=9 atomi H,,,,,,z=1 atom N

F.M => C3H9N