The answer is something like this:
mass of C in the compound = (81.3g CO2)x (12.0 g C/44.0 g CO2) = 22.17 g C
mass of H in compound = (16.6 g H2O) x (2.0 g H /18.0 g H2O) = 1.84 g H
mass of O in compound = 29.9 g total - 22.17 g C - 1.84 g H = 5.89 g O
now convert mass of each element to moles
22.17 g C x (1 mole C/12.0 g C) = 1.85 moles C
1,84 g H x (1 mole H/1.0 g H) = 1.84 moles H
5.89 g O x (1 mole O/a6.0 g O) = 0.368 moles O
divide each by 0.389 to get C5H5O, which has a mass of 81 g
double that to get C10H10O2
This is a "classic" example of a combustion analysis problem. Hope that helped.

J.R. S.
09/02/19