Combustion Analysis!

Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of CO2 and 1.134 g of H2O. What is the molecular formula of the compound if the molecular mass was found to be about 240?
Empirical Formula

Please show work :) Thank you!

mass of C in the sample
= (2.086 g CO2) / (44.00964 g CO2/mol) x (1/1) x (12.01078 g C/mol) = 0.569295 g C

mass of H =
(1.134 g H2O) / (18.01532 g H2O/mol) x (2/1) x (1.007947 g H/mol) = 0.126893 g H

mass of O =
1.200 g - 0.569295 g C - 0.126893 g H = 0.503812 g O

moles of C =
(0.569295 g C) / (12.01078 g C/mol) = 0.0473987 mol C

moles of H =
(0.126893 g H) / (1.007947 g H/mol) = 0.125893 mol H

moles of O =
(0.503812 g O) / (15.99943 g O/mol) = 0.0314894 mol O

molar ratio of C :H:O = .0473987 : .125893: .0314894 or 1.50:3.99:1
or 3:8:2

empirical formula is C3H8O2 which has formula mass of 36+8+32 = 76 or 1/3 of 240

the molecular formula is 3 x the empirical formula or C9H24O6