Combustion Analysis: Empirical and Molecular Formulas

Question

1) When 4.468 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 14.54 grams of CO₂ and 4.465 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

1. Enter the elements in the order presented in the question.

empirical formula =

molecular formula =

2) A 7.109 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 17.93 grams of CO₂ and 3.147 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 122.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

1. Enter the elements in the order C, H, O

empirical formula =

molecular formula =

3) A 2.538 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 5.070 grams of CO₂ and 2.076 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 88.11 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

1. Enter the elements in the order C, H, O

empirical formula =

molecular formula =

Answer 1

Find moles of carbon (C):

14.54 g CO₂ x 1 mole CO₂/44 g x 1 mole C/mole CO₂ = 0.330 moles C

Find moles of hydrogen (H):

4.465 g H2O x 1 mole H2O/18 g x 2 moles H/mole H2O = 0.496 moles H

To get whole numbers for moles, we can divide both by 0.330 to get 1.0 moles C:

moles C = 1.00

moles H = 0.496/0.330 = 1.50 and now we can multiply all by a factor of 2 to get whole numbers:

moles C = 1.00 x 2 = 2.00

moles H = 1.5 x 2 = 3.00

Empirical Formula = C₂H₃

To find molecular formula, we determine how many empirical formulae it takes to make 1 molecular formula:

molar mass C₂H₃ = 27 g/mole

54.09/27 = 2 empirical formula needed

Molecular formula = C₄H₆

Moles C: 17.93 g CO₂ x 1 mol CO₂/44 g x 1 mole C/mole CO₂ = 0.4075 moles C

Moles H: 3.147 moles H₂O x 1 mol H₂O/18 g x 2 mol H/mol H₂O = 0.3497 moles H

Moles O: mass of sample - mass of C - mass of H = mass of O. Then convert to moles of O...

0.4075 mol C x 12 g/mol = 4.89 g C

0.3497 mol H x 1 g/mol = 0.3497 g H

mass O = 7.109 g - 4.89 g - 0.3497 g = 1.869 g O

Moles O = 1.869 g O x 1 mol/16 g = 0.1168 moles O

To get whole numbers, we can divide all moles by 0.1168:

moles C = 0.4075/0.1168 = 3.5

moles H = 0.3497/0.1168 = 3

moles O = 0.1168/0.1168 = 1

Now, multiply all moles by 2 to get whole numbers...

moles C = 7

moles H = 6

moles O = 2

Empirical formula = C₇H₆O₂

molar mass empirical formula = 122 g/mole

Molecular formula = C₇H₆O₂

Do #3 in an identical fashion as #2 was done.