What is the empirical formula of the unknown compound?

Question

A 0.464g sample of a compound containing the elements carbon, hydrogen, oxygen and sulphur was burned completely, yielding 0.969g CO2 and 0.198g H₂O. In a separate experiment, all the sulphur in another 0.464g sample was converted to 0.858g BaSO₄. What is the empirical formula of the unknown compound?

 

So what I have done so far is calculate the % of C, H and S in the sample. The remainder being the % of O.

%C in CO₂ = 12.01/44.01 x 100% = 27.3%

%H in H₂O = 1.008/18.02 x 100% = 5.59%

%S in BaSO₄ = 32.07/233.37 x 100% = 13.7%

I added the three percentages to get 46.6%, then subtracted that from 100% to get the percentage of O = 53.4%.

 

Then using the percentages I found the mol of each:

#moles C = 27.3% x 0.969g CO₂ = 0.265g C x 1 mol/12.01g C = 0.0221 mol C

followed those same steps for H and S to get :

H= 0.0109 mol

S= 0.00368 mol

 

Now for O, since it is present in the CO₂, H₂O and the BaSO₄, do I multiply the 53.4% O by each of the masses given and then add together to find the total mol of O?

It's the mol of O I'm stuck on...

Answer 1

It's a bit unusual that they didn't give you the sulfur products of combustion, such as SO₂ and SO₃ but be that as it may,

you simply take the % oxygen (53.4%) and multiply it by the original mass.  That is 0.0534 x 0.464 g = 0.248 g O and converting to moles you have 0.248 g x 1 mole/15.999 g = 0.00155 moles O.  Continuing with finding the empirical formula by dividing all moles by 0.00368 you would obtain

moles C = 6

moles H = 3

moles S = 1 

moles O = 4

giving an empirical formula of C₆H₃SO₄