A 0.2795 g sample of an organic mixture containing only C6H6CI6(molar mass=290.83g/mol) and C14H9CI5(molar mass=354.49g/mol) was burned in ?
a steam of oxygen in a quartz tube. The products(CO2, H2O and HCI) were passed through a solution of NaHCO3. After acidification, the chloride in this solution yielded 0.7161 g AgCI(molar mass=143.22g/mol). Calculate the percentage of each halogen compound in the sample.
- Roger the MoleLv 72 months agoFavorite Answer
Let z be the mass (in grams) of C6H6Cl6 in the mixture.
Then the mass of C14H9Cl5 is (0.2795 - z) grams.
C6H6Cl6 + 6 O2 → 6 CO2 + 6 HCl
(z g) / (290.83 g C6H6Cl6/mol) x (6 mol Cl / 1 mol C6H6Cl6) =
(0.0206306 z) mol Cl from C6H6Cl6
C14H9Cl5 + 15 O2 → 14 CO2 + 2 H2O + 5 HCl
(0.2795 - z) g / (354.49 g C14H9Cl5/mol) x (5 mol C / 1 mol C14H9Cl5) =
(0.00394228 - 0.0141048 z) mol Cl from C14H9Cl5
(0.7161 g AgCI) / (143.22 g AgCl/mol) x (1 mol Cl / 1 mol AgCl) =
0.005000 mol Cl total
Set the sum of the two expressions in z for the moles of Cl equal to the total mass of Cl (all units in moles):
(0.0206306 z) + (0.00394228 - 0.0141048 z) = 0.005000
Solve for z algebraically:
z = 0.162083 g C6H6Cl6
(0.162083 g C6H6Cl6) / (0.2795 g total) = 0.5799 = 57.99% C6H6Cl6 by mass
100% - 57.99% = 42.01% C14H9Cl5