When 6.270 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 19.67 grams of CO2 and 8.055 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Question

Mary Anthony

Chemistry

12 October, 18:23

When 6.270 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 19.67 grams of CO2 and 8.055 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

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Destinee Estrada · Answer 1

6.270/56.11 = 0.1117447870 / 0.1117 ... ≈ 1

19.67/44.01 = 0.446943876 / 0.1117 ... ≈ 4

8.055/18 = 0.4475 / 0.1117 ... ≈ 41 mole of compound forms 4 moles of both CO2 and H2O, so the empirical formula will be C4H8

As this gives molar mass of 56.11, this is also the molecular formula

This could be cyclobutane, or butene