Dieldrin, an insecticide, contains C, H, Cl and O. Combustion of 29.72 mg of Dieldrin gave 41.21 mg CO2 and 5.63 mg of H2O. In a separate analysis 25.31 mg of Dieldrin was converted into 57.13 mg AgCl. What is the empirical formula of Dieldrin ? I have got the solution but please explain how you are finding moles of chlorine?
Answers
Answer:
On combustion,
29.72mg of dieldrin gives
41.21mgCO2
5.63mgH2O
Separately, 25.31mg dieldrin was converted to 57.13mgAgCl No. of moles (CO2)=41.21g44.01g/mol=0.9363mol
No. of moles (C)=0.9363mol No.of moles(H2O)=5.63g18.016g/mol=0.3125mol
Moles of H=0.3125×2=0.625mol
Moles of Cl=0.9363/2=0.46815mol
Mass(C)=12.01g/mol×0.9363mol=11.2356g
Mass of (H)=1.008g/mol×0.625mol=0.63g
Mass of (Cl)=35.45g/mol×0.46815 mol=16.59g
Mass of (O)=29.72g−(11.2356g+0.63g+16.59g)=1.27g
No. 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 (Cl)=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:O:Cl atoms =12:8:1:6
the empirical formula is thus C12H8OCl6
Explanation:
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Answer:
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