A compound contains 4.07% hydrogen,24.27% car | vedclass

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(A) Step $1.$ Conversion of mass percent to grams.Assuming $100\ g$ of the compound,we have $4.07\ g$ hydrogen,$24.27\ g$ carbon,and $71.65\ g$ chlori

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$CH_2Cl$ and $C_2H_4Cl_2$

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(A) Step $1.$ Conversion of mass percent to grams.Assuming $100\ g$ of the compound,we have $4.07\ g$ hydrogen,$24.27\ g$ carbon,and $71.65\ g$ chlorine.Step $2.$ Convert into number of moles of each element.Moles of $H = 4.07 / 1.008 = 4.04$Moles of $C = 24.27 / 12.01 = 2.021$Moles of $Cl = 71.65 / 35.453 = 2.021$Step $3.$ Divide by the smallest mole value $(2.021)$.Ratio $H:C:Cl = 4.04/2.021 : 2.021/2.021 : 2.021/2.021 = 2:1:1$.Step $4.$ Empirical formula is $CH_2Cl$.Step $5.$ Molecular formula.Empirical formula mass $= 12.01 + (2 	imes 1.008) + 35.453 = 49.48\ g/mol$.$n = 	ext{Molar mass} / 	ext{Empirical formula mass} = 98.96 / 49.48 = 2$.Molecular formula $= n 	imes (CH_2Cl) = C_2H_4Cl_2$.

$A$ compound contains $4.07\%$ hydrogen,$24.27\%$ carbon,and $71.65\%$ chlorine. Its molar mass is $98.96\ g/mol$. What are its empirical and molecular formulas?

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