Question

A 5.193 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.255 grams of CO₂ and 3.790grams of H₂O 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.

Enter the elements in the order C, H, O

empirical formula =

molecular formula =

Answer 1

First, We need to calculate the moles of C and of H in the CO2 and H2O formed, and then calculate the masses of those two elements:

9.2555 g CO2 / 44 g/mol = 0.210 mol CO2 X 1 mol C/ 1 mol CO2 = 0.210 mol C X 12 g/mol = 2.52 g C
3.790 g H2O / 18 g/mol = 0.210 mol H2O X 2 mol H/1 mol H2O = 0.420 mol H X 1 = 0.420 g H

Now, use the mass of the original sample and the masses of C and H in the sample to calculate the mass of O in the sample:
5.193 g -(2.52 g + 0.420 g) = 2.253 g O

Now, calculate the moles of O:
2.253 g / 16 g/mol = 0.141 mol O

So, your sample contained:
0.210 mol C
0.420 mol H
0.141 mol O

Divide all of these by the smallest to get the ratio of atoms:
C = 0.210 / 0.141 = 1.5 C
H = 0.420 / 0.141 = 3H
O = 0.141 / 0.141 = 1 O

Since C came out to be 1.5, multiply all of those final numbers by 2 to make them all whole numbers. That gives you:
C3H6O2 for the empirical formula of the compound.

This formula has a molar mass of: 3X12 + 6X1 + 2X16 = 74 g/mol. Since that is the same as the molar mass found experimentally, the molecular formula of the compound is C3H6O2.