A compound is known to contain only carbon, hydrogen, and oxygen. If the complete combustion of a 0.150-
G sample of this compound produces 0.225 g of CO2 and 0.0614 g of H2O, what is the empirical formula of
This compound?A compound is known to contain only carbon, hydrogen, and oxygen. If the complete combustion of a 0.150-
G sample of this compound produces 0.225 g of CO2 and 0.0614 g of H2O, what is the empirical formula of
This compound?

I got the moles of CO2 and H2O ,
I did not know how to continue..
It was the only question of the 150 questions I did not solve right ! Haha

I got it...
!

0.225 g CO2 x (12.01/44.01) = 0.0614 g C
0.0614 g H2O x (2.02/18.02) = 0.00688g H

0.150 -(0.0614 + 0.00688)g = 0.0817g O

Now, divide each weight by the atomic mass of the respective element.This will give a set of numbers proportional to the empirical formula.

0.0614/12.01 = 0.00511
0.00688/1.01 = 0.00681
0.0817/16.00 = 0.00511

Now, we need to get in whole numbers we will divide all of them by the smallest value (it is a good starting point)

0.00511/0.00511 = 1.00 C
0.00681/0.00511 = 1.33 H
0.00511/0.00511 = 1.00 O

Now we will multiply all of them by 3 to get all as whole numbers:

So the empirical formula is C3H4O3

;)

Good,work!