Chemistry Homework

Question

A 6.082 gram sample of an organic compound containing, and is analyzed by combustion analysis and 10.84 grams of and 4.438 grams of are produced.

In a separate experiment, the molar mass is found to be 74.08 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

The question is incomplete. Please be more careful when entering your question. If we assume that you meant that the compound contains carbon (C) and hydrogen (H), and that upon combustion 10.84 grams of CO₂ and 4.438 grams of H₂O are produced, then we can proceed as follows...

Calculate moles of carbon: 10.84 g CO₂ x 1 mol CO₂ / 44 g x 1 mol C / mol CO₂ = 0.246 mols C

Calculate moles of hydrogen: 4.438 g H₂O x 1 mol H2O/18g x 2 mol H/mol H₂O = 0.493 mols H

Divide both by 0.246 to get the ratio of C to H: 0.246 mol C/0.246 = 1; 0.493 H /0.246 = 2 mols H

Empirical formula = CH₂

To find the molecular formula, we need to determine the molar mass of the empirical formula.

Molar mass CH₂ = 12.01 + 2(1.00) = 14.01

For the molecular formula, we have 74.08 / 14 = 5

Molecular formula = C₅H₁₀