How to Find Empirical Formula

Question

A 0.7408 g sample of a unknown compound combusted and produced 1.3203 g CO₂ and 0.5405 g H₂O.

What is the empirical formula of this compound?

Answer 1

Answer 2

I offer an alternative method to that proposed by Erik D for solving this problem, not that there is anything wrong with the solution provided.

Assuming the unknown compound contains only C, H and O, we can proceed as follows:

moles of C: 1.3203 g CO₂ x 1 mole CO₂/44.00 g x 1 mole C/mole CO2 = 0.0300 moles C

mass of C: 0.0300 moles C x 12.01 g/mole = 0.3603 g C

moles H: 0.5405 g H₂O x 1 mole/18 g x 2 mole H/mole H₂O = 0.0600 moles H

mass H: 0.00600 moles H x 1.0079 g/mole = 0.06047 g H

To find mass of O and moles of O, we subtract the mass of C and H from the mass of the sample to get mass of O and then convert to moles of O.

0.7408 g - 0.3603 g - 0.06047 g = 0.3200 g O

moles O = 0.3200 g x 1 mole/16 g = 0.0200 moles O

Summarizing, we have...

0.0300 moles C

0.0600 moles H

0.0200 moles O

Dividing though by the smallest number (0.0200) in order to get whole numbers we obtain

1.5 moles C

3.0 moles H

1 mole O

and multiplying all by 2 to get whole numbers, we obtain

3 moles C

6 moles H

2 moles O

So empirical formula is C₃H₆O₂

Answer 3

Assuming there is only CO₂ and H₂O as products, the unknown compound must have reacted with the O₂ in the air. The equation must then be (with the given values written underneath):

ukn + O₂ --> CO₂ + H₂O

0.7408 g 1.3203 g 0.5405 g

You can then calculate by conservation of mass that O₂ must be 1.1200 g.

You can then calculate the moles given from their masses, 0.035 mol of O₂, 0.030 mol of CO₂, 0.030 mol of H₂O. Normalizing CO₂ and H₂O, gives 1 mol each and 1.16 mol of O₂. You need whole numbers of moles in an equation. Multiply each of these by 6, to get 7 mol O₂, 6 mol CO₂, and 6 mol H₂O.

Using the now balanced equation, you can determine the unknown compound to be C₆H₁₂O₄. By dividing the equation by 2, you get a final empirical formula of C₃H₆O₂