Empirical & Molecular Formula?

Mass fraction of C in CO₂ = 12.01/(12.01 + 16.00×2) = 12.01/44.01
Mass fraction of H in H₂O = 1.008×2.(1.008×2 + 16) = 2.016/18.018

In 2.607 g of CxHy:
Mass of C = Mass of C in CO₂ produced = (7.637 g) × (12.01/44.01) = 2.084 g
Mass of H = Mass of H in H₂O produced = (4.687 g) × (2.016/18.016) = 0.524 g

Mole ratio C : H
= (2.084/12.01) : (0.524/1.008)
= 0.174 : 0.520
= (0.174/0.174) : (0.520/0,174)
≈ 1 : 3

Empirical formula = CH₃
Let molecular formula = (CH₃)ₙ

Molar mass of (CH₃)ₙ in g/mol:
(12.01 + 1.008×3)n = 30.07
15.034n = 30.07
n = 2

Molecular formula = C₂H₆

moles of C in the compound = moles of C from the CO2 
= 7.631 grams / 44 g/ mol = 0.17343 moles of C in the sample of compound
moles of H2O = 4.687 grams / 18 g/ mol = 0.260 molesmoles of H in the sample of compound = 0.520molar ratio C:H = 0.17343: .520 dividing by the smallest number we have 1:3empirical formula is the molar ratio of the elements we have CH3empirical formula mass = 15 g/ unitwhich is half the molecular mass so the molecular formula =  2 * empirical formula or C2H6 the empirical formula of the hydrocarbon = CH3molecular formula of the hydrocarbon = C2H6in this case we don't need to calculate the masses of the C and H .. the ratios are enough