We must calculate the masses of #"C, H, Cl"# and #"O"# from the masses given.
#"Mass of C" = 58.67 color(red)(cancel(color(black)("g CO"_2))) Γ "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "16.01 g C"#
#"Mass of H" = 15 color(red)(cancel(color(black)("g H"_2"O"))) Γ "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "1.68 g H"#
#"Mass of Cl" = 40.17 color(red)(cancel(color(black)("g compd" )))Γ "22.06 g Cl"/(75.00 color(red)(cancel(color(black)("g compd")))) = "11.82 g Cl"#
#"Mass of O" = "Mass of compound -Mass of C - Mass of H - Mass of Cl" = "(40.17 - 16.01 - 1.68 - 11.82) g" = "10.67 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.
#ulbb("Element"color(white)(X) "Mass/g"color(white)(X) "Amt/mol"color(white)(m) "Ratio"color(white)(m)"Integers")#
#color(white)(mm)"C"color(white)(mmmll)16.01color(white)(mmml)"1.333"color(white)(Xml)4.000color(white)(Xmmll)4#
#color(white)(mm)"H" color(white)(XXXml)1.68 color(white)(mmml)1.66 color(white)(mmll)4.995color(white)(Xmmm)5#
#color(white)(mm)"Cl" color(white)(XXXll)11.82 color(white)(mmm)0.3334 color(white)(mm)1color(white)(Xmmmmll)1#
#color(white)(mm)"O" color(white)(XXXll)10.67 color(white)(mmm)0.6669 color(white)(mm)2.001color(white)(Xmmll)2#
The empirical formula is #"C"_4"H"_5"ClO"_2#.