6.270/56.11 = 0.1117447870 /0.1117.. ≈ 1
19.67/44.01 = 0.446943876 /0.1117.. ≈ 4
8.055/18 = 0.4475 /0.1117.. ≈ 41 mole of compound forms 4 moles of both CO2 and H2O, so the empirical formula will be C4H8
As this gives molar mass of 56.11, this is also the molecular formula
This could be cyclobutane, or butene