A gaseous hydrocarbon on analysis gave the following data : (i) It contains C = 82.7% and H = 17.3% (ii) The mass of 132 mL (measured at S.T.R) of it is 0.342 g.Find the molecular formula of the hydrocarbon.

A gaseous hydrocarbon on analysis gave the following data : (i) It co

1.	A gaseous hydrocarbon on analysis gave the following data : (i) It contains C = 82.7% and H = 17.3% (ii) The mass of 132 mL (measured at S.T.R) of it is 0.342 g.Find the molecular formula of the hydrocarbon.
Answer» SOLUTION :Calculation of empirical formula : `therefore` The empirical formula of the given hydrocarbon is `C_(2)H_(5)`. Calculation of molecular formula : Empirical formula mass = `(2 xx 12.01) + (5 xx 1.008) = 29.06` amu. As discussed earlier, the VOLUME occupied by one GRAM molecular mass of a substance at S.T.R is 22.4 L (22400 mL) `therefore` The gram molecular mass of the given hydrocarbon `=0.342/132 xx 22400 = 58.0` HENCE, its molecular mass = 58.0 amu `n = ("Molecular mass")/("Empirical formula mass") = 58.0/29.06 =2` `therefore` Molecular mass = 2 `xx` Empirical formula `=2 xx C_(2)H_(5) = C_(4)H_(10)` Hence, the molecular formula of the given hydrocarbon is `C_(4)H_(10)`.