A compound is known to contain only C, H, and O what is the empirical formula of this compound?

Question

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?

Answer 1

CH7O

Answer 2

0.225 g CO2 x 12 g C/44 g CO2=0.0614 g C

0.0614 g H2O x 2 g H/18 g H2O=0.00682 g H

0.0614 g+0.00682 g=0.0682 g

0.150 g-0.0682 g=0.08178 g O

0.0614 g C x 1 mol C/12 g C=0.00512 mol C

0.00682 g H x 1 mol H/1 g H=0.00682 mol H

0.08178 g O x 1 mol O/16 g O=0.00511 mol O

C=0.00512/0.00511=1

H=0.00682/0.00511=1.33

O=0.00511/0.00511=1

3(CH1.33O)=C3H4O3

Empirical formula is C3H4O3

Answer 3

First, you need to find the g of C and H that were in the sample then you can get the O by difference, then you can determine % by moles to get empirical formula.

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