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{ \underline{ \underline { \sf \pink{Solution :- }}}}

The smallest value we got is 2.01

The empirical formula BECOMES CH₂Cl

Molecular WEIGHT of Empirical formula = 12 + 2(1) + 35.5 = 49.5

We are ALREADY given that molecular weight as 99.

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Relation between Empirical formula and Molecular formula is given by ,

\star\large{\boxed{ \sf{ \purple{Molecular\:Formula=Empirical\:formula\times\:n}}}}⋆

\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}

We know the values for calculating the value of n . So , by substituting we GET ;

\begin{gathered}\begin{gathered}\\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}\end{gathered}\end{gathered}

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Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;

\begin{gathered}\begin{gathered}\\ \implies \sf Molecular\:Formula=Empirical\:formula \times \: n \\ \\ \\ \implies \sf Molecular\:Formula=CH_2Cl \times 2 \\ \\ \\ \implies \sf \: Molecular\:Formula=C_2H_4Cl_2\end{gathered} \end{gathered}

HENCE , The molecular formula of the given compound is C₂H₄Cl₂.


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