# How to Find Empirical Formula

A 0.7408 g sample of a unknown compound combusted and produced 1.3203 g CO_{2} and 0.5405 g H_{2}O.

What is the empirical formula of this compound?

## 3 Answers By Expert Tutors

J.R. S. answered • 08/06/19

Ph.D. University Professor with 10+ years Tutoring Experience

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 CO_{2} x 1 mole CO_{2}/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 H_{2}O x 1 mole/18 g x 2 mole H/mole H_{2}O = **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 **C**_{3}**H**_{6}**O**_{2}

Erik D. answered • 08/06/19

Science and Math Tutor with Years of Experience

Assuming there is only CO_{2} and H_{2}O as products, the unknown compound must have reacted with the O_{2} in the air. The equation must then be (with the given values written underneath):

ukn + O_{2} --> CO_{2} + H_{2}O

0.7408 g 1.3203 g 0.5405 g

You can then calculate by conservation of mass that O_{2 }must be 1.1200 g.

You can then calculate the moles given from their masses, 0.035 mol of O_{2}, 0.030 mol of CO_{2}, 0.030 mol of H_{2}O. Normalizing CO_{2} and H_{2}O, gives 1 mol each and 1.16 mol of O_{2}. You need whole numbers of moles in an equation. Multiply each of these by 6, to get 7 mol O_{2}, 6 mol CO_{2}, and 6 mol H_{2}O.

Using the now balanced equation, you can determine the unknown compound to be C_{6}H_{12}O_{4}. By dividing the equation by 2, you get a final empirical formula of C_{3}H_{6}O_{2}

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