448 mL of a hydrocarbon, having C = 87.8%, H = 12.19%, weigh 1.64 g at S.T.P Determine the molecular formula of the compound.

448 mL of a hydrocarbon, having C = 87.8%, H = 12.19%, weigh 1.64 g at

1.	448 mL of a hydrocarbon, having C = 87.8%, H = 12.19%, weigh 1.64 g at S.T.P Determine the molecular formula of the compound.
Answer» SOLUTION :Calculation of EMPIRICAL formula: The empirical formula of hydrocarbon = `C_3H_5` Calculation of molecular formula : One gram mole of any gas occupies 22.4 L or 22400 mL at S.T.R `THEREFORE` 448 mL at S.T.R of hydrocarbon weigh = 1.64 g `therefore 22400` mL at S.T.R of hydrocarbon will weigh `=1.64/448 xx 22400 = 82.0 g` Hence, gram molecular mass of hydrocarbon = 82.0 g Empirical formula mass = `(3 xx 12.01) + (5 xx 1.008) = 41.07` `n=("gram molecular mass")/("Empirical formula mass") = 82/(41.07) = 2` `therefore` The molecular formula `=2 xx C_(3)H_(5)= C_(6)H_(10)`