Acompound consists of 65.45% c, 5.492% h and 29.06%o. it has a molar mass of approximately 110 amu. determine the empirical and molecular formula of the compound

The  empirical  and  molecular  formula  of  the  compound is calculated  as  follows

the  empirical  formula  calculation   step

find the  mole of  each   element=  %  composition of an element/  molar  mass  of  an element
carbon(C)  = 65.45/12=5.454 moles
Hydrogen(H) = 5.492/1= 5.492 moles
oxygen(O)= 29.06/ 16= 1.816  moles

find  the  mole  ratio by  diving  each  mole  by the  smallest   number of  mole(1.816  moles) and round off to  the  nearest  integer

that  is  Carbon(C) = 5.454/1.816=3 
         Hydrogen(H)= 5.492/1.816= 3
            oxygen(O) =1.816/1.816=1

the  empirical  formula  is  therefore=  C3H3O

molecular  formula  calculation
110  amu  is the  total  molar  mass of C3H3O

therefore  (C3H3O)n =  110
find  for n

={(12  x3) + (1 x3)+ ( 16 x1)}n=  110

55n= 110
divide  both side by 55
n=2
the molecular  formula = (c3H3O)2 = C6H6O2

Mgo2 is the oxidation state of manganese oxide

And a) oxygen i believe