(39) What is empirical formula for a compound which contains 0.0130 mol C, 0.390 mol H ;0.065 mol O CHO CH5O2 CH30O5 CH39O6 C2H60O10?

Question

Update:

[1]       CHO    [2]       CH5  O2  [3]       CH30  O5  [4]       CH39  O6  [5]        C2H60  O10 

Answer 1

Empirical formula ....

Bo gets CH3O5 .... which is correct for the numbers given, but I don't think that is really what the question writer had in mind.  Instead, I think there is a typo, and the moles of oxygen should be 0.0065 mol.

0.0130 mol C / 0.0065 = 2 mol C

0.0390 mol H / 0.0065 = 6 mol H

0.0065 mol O / 0.0065 = 1 mol O

Empirical formula = C2H6O ............ CH3CH2OH ... ethanol

I think you are going to be hard-pressed to find a molecule with so many oxygen atoms when using CH3O5.

And the choices given .... They're nuts.

Answer 2

If you divide by the smallest value in order to get a whole number, we obtain

0.0130 mole C/0.0130 = 1 mole C

0.0390 mole H/0.0130 = 3 moles H (assuming you meant 0.0390 moles and not 0.390 moles as written)

0.065 mol O/0.0130 = 5 moles O

Empirical formula = CH3O5