Question

determinati formula moleculara a unei hidrocarburi in care elem componente sunt in raport de masa C:H=8:1 si are denditatea fata de azot 1,9285

Answer 1

MN2(gaz)=2AN=2*14=28g/mol

M-masa molara,g/mol
d-densitatea

dgaz=M/Mgaz => M=dgaz*Mgaz=1,9285*28 ≈ 54g/mol

C=8/12=0,666 moli C  ||:0,666 => 1 atom C

H=1/1=1 mol H ||:0,666  => 1,5 atomi H

F.B => (CH1,5)n

F.M => (CH1,5)n = 54 => 12n+1,5n=54 => 13,5=54 => n=4
 
⇒ C4H6 - butina

Answer 2

d = masa molara hidrocarbura/masa molara N2
masa molara hidrocarbura = 1.9285 * 28 = 54 g/mol
CxHy
12x/y = 8/1 => 8y = 12x => y = 12x/8y = 3x/2
FM: CxH3x/2
12x + 3x/2 = 54 => 13,5x = 54 => x = 4 => FM: C4H6