Question

A 11.75-gram sample of an organic compound containing C, H, and O is analyzed by combustion analysis, and 20.95 grams of CO₂ and 3.578 grams of H₂O are produced. In a separate experiment, the molecular weight is found to be 74.08 amu. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O.

Empirical formula =
Molecular formula =

Answer 1

The empirical formula of the organic compound is CH₂O, and the molecular formula is C₂H₄O₂.

Step-by-step explanation:

To determine the empirical formula of the organic compound, we need to calculate the molar amounts of carbon and hydrogen in the sample. From the given masses of carbon dioxide and water produced, we can calculate that the molar amount of carbon is 0.5365 mol and the molar amount of hydrogen is 3.538 mol. To find the empirical formula, we divide these molar amounts by their smallest common factor, which is 0.5365.

Thus, empirical formula of the mentioned compound is CH₂O.

To find the molecular formula, we need to compare the molecular weight of the compound to the empirical formula weight. The molecular weight of the compound is 74.08 amu, and the empirical formula weight is 30.03 amu.

Dividing the molecular weight by the empirical formula weight gives us a factor of 2.47. Multiplying the empirical formula by this factor, we get the molecular formula: C₂H₄O₂.