(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?
 CHO  CH5 O2  CH30 O5  CH39 O6  C2H60 O10
- pisgahchemistLv 79 months ago
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.
- 9 months ago
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