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A hydrocarbon has an empirical formula of \[{\text{CH}}\] and a molar mass of \[{\text{26 g/mol}}\] . What is the molecular formula of this compound?
(A) \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\]
(B) \[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{3}}}\]
(C) \[{{\text{C}}_{\text{4}}}{{\text{H}}_{\text{4}}}\]
(D) \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\]

Answer
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Hint: From the empirical formula, write the molecular formula by giving suffix n. From the atomic weights of carbon and hydrogen and the molecular formula in terms of suffix n, calculate the molecular weight in terms of n and compare it to the given molecular weight.

Complete step by step answer:
From empirical formula, you can obtain the simplest whole number ratio of atoms of each element present in the compound. From the molecular formula, you can obtain the number of molecules of each element present in one molecule of the compound.
A hydrocarbon has an empirical formula of \[{\text{CH}}\]
Let \[{\left( {{\text{CH}}} \right)_n}\]Be its molecular formula.
The atomic masses of carbon and hydrogen are 12 and 1 respectively.
Calculate the molecular formula mass
\[{\left( {{\text{CH}}} \right)_n} = n\left( {12 + 1} \right) = 13n\]
But the molar mass of the hydrocarbon is \[{\text{26 g/mol}}\]
Molecular weight is the weight of one mole of the compound. One mole of a compound contains Avogadro’s number of molecules. Thus, the weight of the Avogadro’s number of molecules is equal to the molecular weight. Thus, \[{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\] molecules of the given hydrocarbon weighs \[26{\text{ g}}\] .
Hence, \[{\text{26 g/mol = 13}}n{\text{ g/mol}}\]
Or \[n = \dfrac{{26}}{{13}} = 2\]
Hence, the molecular formula of the hydrocarbon is \[{\left( {{\text{CH}}} \right)_n} = {\left( {{\text{CH}}} \right)_2} = {{\text{C}}_2}{{\text{H}}_2}\] .

Thus the correct option is (B).

Note: A molecular formula is obtained when you multiply empirical formula with a whole number. Thus, the molecular formula is the simple multiple of the empirical formula. Empirical formula gives us the simplest positive integer ratio of atoms in the compound.