Question

O proba cu masa de 15,5 grame ce contine 0,25 de substanta A formeaza la ardere 11,2L de co2 si 13,5 grame de h2o. Substanta A contine 51,612% S. Care este formula moleculara a acestei substante?

Answer 1

22,4 l CO2.......12 g C
11,2 l CO2...........x
x = 6 g C
18 g apa...........2 g H
13.5 g apa............y
y = 1.5 g H
n = m/masa molara => masa molara = 15.5/0,25 = 62 g/mol
%S = mS/m compus * 100 => mS = 51.612*62/100 = 32 g
62 g substanta.......32 g S
15.5 g substanta.........z
z = 8 g
8+6+1.5 = 15.5 => nu exista oxigen
6/12 = 0.5
1.5/1 = 1.5
8/32 = 0.25
Imparti la 0.25 =>(C2H6S)n
masa molara = 62 g/mol => 62n = 62 => n = 1 => FM: C2H6S