Acombustion analysis of a 0.503 g sample of an unknown hydrocarbon yields 1.53 g co 2 2 and 0.756 g h 2 2 o. what is the empirical formula of the sample? give your answer in the form c#h# where the number following the element’s symbol corresponds to the subscript in the formula. (don’t include a 1 subscript explicitly). for example, the formula ch 2 2 o would be entered as ch2o

C₅H₁₂.

Explanation:

The hydrocarbon is burned in excess of oxygen to give CO₂ and H₂O.The no. of moles of CO₂ produced = mass/molar mass = 1.53 g/44.0 g/mol = 0.03477 mol.

Which is corresponding to 0.03477 mol of C.

The no. of moles of H₂O produced = mass/molar mass = 0.756 g/18.0 g/mol = 0.042 mol.

which corresponds to (0.042 x 2) = 0.084 moles of hydrogen.

The ratio of the number of moles of hydrogen to carbon in the composition of the compound will be 0.084 / 0.03477 = 2.4 = 12/5.Therefore, the empirical formula of the compound under consideration is C₅H₁₂.

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