C\(_x\)H\(_y\)O + 5O\(_2\) →  4CO\(_2\)  +  4H\(_2\)O

On balancing the equation, we should have

X = 4 , y = 8  and O = 2 ⇒ C\(_4\)H\(_8\)O\(_2\)

Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.

We finally have C\(_2\)H\(_4\)O