Chemistry, asked by Prajapatimeet, 10 months ago

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

Answered by jefferson7
2

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

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