Dieldrin, an insecticide, contains C, H, Cl and O. Combustion of 29.72 mg of Dieldrin
41.21 mg CO, and 5.63 mg of H20. In a separate analysis 25.31 mg of Dieldrin was converted
into 57.13 mg AgCl. What is the empirical formula of Dieldrin ?
(a) C6H4Cl3O (b) C8H8ClO (C) C12H8Cl6O (d) C6H4Cl3O2
Answers
Dieldrin, an insecticide, contains C, H, Cl and O. Combustion of 29.72 mg of Dieldrin
41.21 mg CO, and 5.63 mg of H20. In a separate analysis 25.31 mg of Dieldrin was converted
into 57.13 mg AgCl. What is the empirical formula of Dieldrin ?
(a) C6H4Cl3O (b) C8H8ClO (C) C12H8Cl6O (d) C6H4Cl3O2
Explanation:
Upon combustion, 29.72mg of dieldrin gives 41.21mgCO2,
5.63mg H2O, In a separate analysis, 25.31mg dieldrin was converted to 57.
13mg AgCi No.of moles (CO2) = 24019140 = 0.9363 mol
No.of moles (C) = 0.9363 mol
No.of moles (H20) = 18.5163/mol = 0.3125 mol Moles of H=0.3125 x 2 =0.625mol
Moles of Cl = 0.9363/2 = 0.46815 mol 2.01g/mol x 0.9363 mol = 11.23569 Mass of (H) = 1.008g/mol x 0.625mol = 0.639
Mass of (CI) = 35.45g/mol x 0.46815 mol = 16.599 Mass of (0) = 29.729- ( 11.23569+0.639+16.599) = 1.279
Number of moles of (C)= 11.2356g/12.01g/mol = 0.93mol/0.079 = 11.77 = 12 No.of moles of (H) = 0.63g/1.008g/mol =0.625 mol/0.079 = 7.9= 8
No.of moles of (CI) = 16.59/35.45g/mol = 0.467mol/0.079 = 5.9=6 No.of moles (O)=1.27g/16.00g/mol = 0.079mol/0.079 = 1
The ratio of C:H:0:Cl atoms = 12:8:1:6 the empirical formula of Dieldrin will be C12H2OClo