AS for all these sorts of problems, we assume a 100⋅g mass of unknown compound, and we access the empirical formula......
Moles of C=54.5⋅g12.011⋅g⋅mol−1=4.54⋅mol.
Moles of H=9.3⋅g1.00794⋅g⋅mol−1=9.23⋅mol.
Moles of O=36.2⋅g15.999⋅g⋅mol−1=2.26⋅mol.
Note that normally %O would never be measured OR QUOTED. For a CnHmO compound, you would assume that the missing percentage i.e. 100%−%C−%H=%O.
And now we divide thru by the smallest molar quantity, that of oxygen, to get an empirical formula of......
C:4.54⋅mol2.26⋅mol=2.00
H:9.23⋅mol2.26⋅mol=4.08
O:2.26⋅mol2.26⋅mol=1.00
......to get an empirical formula, the simplest whole number ratio defining constituent atoms in a species, of......
C2H4O.
Now the molecular formula is a whole number mulitple of the empirical formula. And thus........
(empirical formula)×n=molecular formula....but we were quoted a molecular mass, and so.....
n×(2×12.011+4×1.00794+1×16.00)⋅g⋅mol−1=88⋅g⋅mol−1.
Clearly, n=2, and so our molecular formula is.....
2×{C2H4O}=C4H8O2.....
