Combustion Analysis

Question

When 6.508 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 20.42 grams of CO₂ and 8.361 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 28.05 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Emperical Formula:

Molecular Formula:

Answer 1

Since the hydrocarbon only contains carbon (C) and hydrogen (H), we need only figure out moles of each.

20.42 g CO2 x 1 mol CO2/44 g = 0.464 moles CO2. Since each mole CO2 contains ONE mole C...

moles C = 0.464

8.361 g H2O x 1 mole H2O/18 g = 0.465 moles H2O. Since each mole of H2O contains TWO moles H...

moles H = 0.929

We want whole numbers so divide both by 0.464 (ratio is 1:2) and get CH2 = empirical formula

Molar mass of the empirical formula = 12 + 2 = 14 g/mole

Molar mass of the molecular formula = 28.05 which is twice that of the empirical formula meaning it takes 2 empirical formulae to make 1 molecular formula.

Molecular formula = C2H4

Answer 2

Hello, Nicole,

We don't have enough information to start with a balanced equation. So I'll work backwards from what we do know. Calculate the moles of CO₂ and H₂O produced by dividing their masses by their molar masses. A mole is just a counting unit equal to Avogadro's Number. We have 0.464 moles of CO₂, but we can also conclude we have 0.464 moles of C and 0.928 moles of O (2 O's per CO₂). The 0.464 moles of H₂O account for 0.929 moles of H (2 H's per H₂O). Since the carbon and hydrogen only come from the mystery compound, I can relate the two quantities to get the ratio of H's to C's. I find a ratio that is exactly 2.00, so the C_xH_y empirical formula is C_xH_2y. Possible actual formulas are CH₂, C₂H₄, C₃H₆, . . . .and so on. These are all formulas consistent with a double-bonded carbon somewhere, and a name that ends in -ene.

I have a table for the calculations, but it won't fit here. Hopefully, you'll see what the approach is and duplicate the results.

I hope this helps,

Bob