We can calculate the masses of #"C"# and #"H"# from the masses of their oxides (#"CO"_2# and #"H"_2"O"#).

#"Mass of C" = 0.2829 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.077 20 g C"#

#"Mass of H" = 0.1159 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.012 97 g H"#

#"Mass of O" = "Mass of menthol - mass of C - mass of O" = "0.1005 g - 0.077 20 g - 0.012 97 g" = "0.01 033 g"#

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)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(mml) "Ratio" color(white)(m)"Integers"#

#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)"0.077 20" color(white)(mll)"0.006 428"
color(white)(Xlll)9.956color(white)(Xm)10#

#color(white)(ll)"H" color(white)(XXXXl)"0.012 97" color(white)(mll)"0.012 87" color(white)(mlll)19.93 color(white)(XXll)20#

#color(white)(ll)"O" color(white)(mmmml)"0.010 33"color(white)(mll)"0.000 6456"color(white)(ml)1color(white)(mmmmll)1#

The empirical formula is #"C"_10"H"_20"O"#.