How to find the empirical formula of a compound containing C, H, O?

Problem: A 5.000 g sample of an organic compound containing C, H, and O atoms was combusted and produced 14.26 g CO2 and 3.335 g H2O. Its molar mass is 108.1 g/mol.



```
 Solution:
      Nc = 14.26 g/44.01 g/mol 
         = 0.3240 mol C
      m  = Nc (12.01) 
         = 3.8912 g
      Nh = 3.335 g/18.02 g/mol
         = 0.1851 mol H
      m  = Nh (1.008)
         = 0.1866 g
      Mo = 5.000 g - 3.891 g - 0.1866 g 
         = 0.9224 g
      No = 0.9224 g/16.00 g/mol
         =0.05765 mol O

  C = 0.3240 mol C/0.05765    
    = 5.6 (5)
    = 28 
  H = 0.1851 mol H/0.05765
    = 3.2 (5)
    = 16
  O = 0.05765 mol O/0.5765
    = 1 (5)
    = 5

```

Comment: In finding the empirical formula of the compound, I keep getting C28H16O5 which has a molar mass of approx. 400+ that is not equal to the given molar mass above. I followed the process of finding the empirical formula, but I'm getting the same result. I don't know what else to do. Oh, and I also considered to get the mass percent composition first, but also the same result. I'm guessing that the formula could be either C7H8O (with 108.1 g/mol), I forgot the other one. What should I do?




Your sample contains $0.324$ mol $\ce{C}$ (weighing $3.888$ g) and $0.371$ mole $\ce{H}$. Total mass : $4.259$ g. Out of $5.000$ g, the sample contains still $5.000 - 4.259 = 0.741$ g $\ce{O}$, or $0.741/ 16 = 0.0463$ mole $\ce{O}$. Compared to the $\ce{O}$ atoms, there is nearly $7$ $\ce{C}$ atoms and $8$ $\ce{H}$ atoms. This is $\ce{C7H8O}$.


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