Question

O proba de masa 15,5 g ce contine 0.25 moli de substanta x formeaza la ardere 11,2 L (c.n) de CO2 si 13.5 g de H2O. Substanta X contine 51,612% S. Care este formula moleculara a subst X?

Answer 1

15,5·51,612/100 = 8 g S
M = 15,5 ·4 = 62 g/mol
mC + mH + 8·4 = 62      mC + mH = 30 g (intr-un mol de X)
nCO2 = 11,2/22,4 = 0,5 moli (in 0,25 moli X) ⇒  nC = 2 atomi (in 1 mol X)
n H2O = 13,5/18 = 0,75 moli ( in 0,25 moli X) ⇒ nH = 2·0,75·4 = 6 atomi
X = C2H6S