An Unknown compound contains only, C, H, and O. Combustion of 6.40g of this compound produced 15.6g CO2 and 6.40g H2O. what is the empirical formula of the unknown compound?

Answer 1

The first step is to convert everything into moles:

6.40 g H2O = 0.355 moles H2O

15.6 g CO2 = 0.355 moles CO2

Then, add up how many moles of each element we have (we'll deal with oxygen later, since combustion is a reaction with oxygen gas we need to backtrack it):

H: 0.355*2 = 0.711 moles

C: 0.355 moles

Then, we find out the mass of each of these elements in our original compound:

0.711 moles H * 1 g/mol = 0.711 g H

0.355 moles C * 12 g/mol = 4.26 g C

Find oxygen from the original mass of the compound:

6.40 - 0.711 - 4.26 = 1.429 g O

Convert g to mol for oxygen:

1.429 g O = 0.089 mol O

Divide the mol amounts by the least common denominator, in this case O with 0.089 mol. Round to the nearest whole number, or multiply by 2 if you get numbers that end in .5

C=0.355/0.089=4

H=.711/0.089=8

O=0.089/0.089=1

So, our formula is:

C4H8O