We can calculate the masses of "C" and "H" from the masses of their oxides ("CO"_2 and "H"_2"O").
"Mass of C" = 3.03 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.8269 g C"
"Mass of H" = 1.55 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1734 g H"
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(m)×2color(white)(mm)"Integers"
stackrel(————————————————————)(color(white)(ml)"C" color(white)(XXXl)0.8269 color(white)(m)"0.068 85"
color(white)(Xl)1color(white)(Xmml)2color(white)(mmmml)2
color(white)(ml)"H" color(white)(XXXl)0.1734 color(white)(m)0.1720 color(white)(mlll)2.498 color(white)(Xl)4.996color(white)(mml)5
The empirical formula is "C"_2"H"_5.
The empirical formula mass is "(2×12.01 + 5×1.008) u = 29.06 u"
The molecular mass is "58 u".
"Molecular mass"/"Empirical formula mass" = (58 color(red)(cancel(color(black)("u"))))/(29.06 color(red)(cancel(color(black)("u")))) = 2.0 ≈ 2.
∴ The molecular formula is ("C"_2"H"_5)_2 = "C"_4"H"_10.
