Dieldrin, an insecticide, contains ...
Question
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 Dieldr
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Dieldrin, an insecticide, contains ( mathrm{C}, mathrm{H}, ) Cl and O. Combustion of 29.72 mg of Dieldrin gave ( 41.21 mathrm{mg} mathrm{CO}_{2} ) and ( 5.63 mathrm{mg} ) of ( mathrm{H}_{2} mathrm{O} . ) In a separate analysis ( 25.31 mathrm{mg} ) of Dieldrin was converted into ( 57.13 mathrm{mg} ) AgCl. What is the empirical formula of Dieldrin?
(a) ( mathrm{C}_{6} mathrm{H}_{4} mathrm{Cl}_{3} mathrm{O} )
(b) ( mathrm{C}_{8} mathrm{H}_{8} mathrm{ClO} )
(c) ( mathrm{C}_{12} mathrm{H}_{8} mathrm{Cl}_{6} mathrm{O} )
(d) ( mathrm{C}_{6} mathrm{H}_{4} mathrm{Cl}_{3} mathrm{O}_{2} )

JEE/Engineering Exams
Chemistry
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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?
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On combustion,
( 29.72 mathrm{mg} ) of dieldrin gives
( 41.21 mathrm{mgCO}_{2} )
( 5.63 mathrm{mg} mathrm{H}_{2} mathrm{O} )
Separately, ( 25.31 mathrm{mg} ) dieldrin was converted to ( 57.13 mathrm{mg} mathrm{AgCl} ) No. of moles ( left(C O_{2}right)=frac{41.21 g}{44.01 g / m o l}=0.9363 mathrm{mol} )
No. of moles ( (C)=0.9363 mathrm{mol} ) No.of moles( left(H_{2} Oright)=frac{5.63 g}{18.016 g / m o l}=0.3125 mathrm{mol} )
Moles of ( H=0.3125 times 2=0.625 mathrm{mol} )
Moles of ( C l=0.9363 / 2=0.46815 mathrm{mol} )
( operatorname{Mass}(C)=12.01 mathrm{g} / mathrm{mol} times 0.9363 mathrm{mol}=11.2356 mathrm{g} )
Mass of ( (H)=1.008 g / operatorname{mol} times 0.625 mathrm{mol}=0.63 mathrm{g} )
Mass of ( (C l)=35.45 g / operatorname{mol} times 0.46815 ) mol( =16.59 g )
Mass of ( (O)=29.72 g-(11.2356 g+0.63 g+16.59 g)=1.27 g )
No. of moles of ( (C)=11.2356 mathrm{g} / 12.01 mathrm{g} / mathrm{mol}=0.93 mathrm{mol} / 0.079=11.77=12 )
No. of moles of ( (H)=0.63 g / 1.008 g / m o l=0.625 ) mol ( / 0.079=7.9=8 ) No. of moles of ( (C l)=16.59 / 35.45 g / m o l=0.467 mathrm{mol} / 0.079=5.9=6 )
No. of moles ( (O)=1.27 g / 16.00 g / m o l=0.079 m o l / 0.079=1 )
The ratio of ( C: H: O: C l ) atoms ( =12: 8: 1: 6 )
the empirical formula is thus ( C_{12} mathrm{H}_{8} mathrm{OCl}_{6} )

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