Compound Formulas

Question

A 4.561-gram sample of an organic compound containing C, H  and O is analyzed by combustion analysis and 10.37 grams of CO2 and 4.246 grams of H2O are produced.

In a separate experiment, the molar mass is found to be 116.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Unlike your previous question about empirical and molecular formulae, this one has oxygen (O) in the compound, so the approach is just a tad different.

mols C = 10.37 g CO2 x 1 mol CO2 / 44 g x 1 mol C / mol CO2 = 0.236 mol C

mols H = 4.246 g H2O x 1 mol H2O / 18 g x 2 mol H / mol H2O = 0.472 mol H

Now, to find the mols O, we need to first find the mass of O. To do this, we find the mass of C and H and subtract that from the 4.561 g of original sample:

mass C = 0.236 mol x 12 g / mol = 2.832 g

mass H = 0.472 mol x 1 g / mol = 0.472 g

mass O = 4.561 g - 2.832 g - 0.472 g = 1.257 g O

mols O = 1.257 g O x 1 mol / 16 g = 0.0786 mol O

SUMMARY

0.236 mol C

0.472 mol H

0.0786 mol O

Divide all by 0.0786 in an attempt to get whole numbers

mol C = 0.236 / 0.0786 = 3 mol C

mol H = 0.472 / 0.0786 = 6 mol H

mol O = 0.0786 / 0.0786 = 1 mol O

EMPIRICAL FORMUAL = C₃H₆O (molar mass of C3H6O = 58 g / mol)

Molar mass of unknown = 116.2 g / mol

116 g/mol / 58 g/ mol = 2 so the molecular formula is 2 x the empirical formula

MOLECULAR FORMULA = C₆H₁₂O₂