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#.