A combustion 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

Question

Lilly Villarreal

Chemistry

26 April, 22:46

A combustion 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

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Answers (1)

Know the Answer?

Kenyon Calhoun · Answer 1

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.