The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of "C" to "H" to "O".
Your compound contains 54.53 % "C", 9.15 % "H", and 36.32 % "O".
Assume that you have 100 g of sample.
Then it contains 54.53 g of "C", 9.15 g of "H", and 36.32 g of "O".
"Moles of C" = 54.53 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "4.540 mol B"
"Moles of H" = 9.15 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "9.077 mol H"
"Moles of O" = 36.32 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "2.270 mol O"
From this point on, I like to summarize the calculations in a table.
bb("Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×3"color(white)(mm)"Integers")
color(white)(m)"C" color(white)(XXXmm)54.53 color(white)(Xm)4.540
color(white)(mmll)2.000color(white)(mml)6.000color(white)(mmml)6
color(white)(m)"H" color(white)(XXXXmll)9.15 color(white)(mm)9.077 color(white)(Xmll)4.304 color(white)(mll)12.91color(white)(mmml)13
color(white)(m)"O"color(white)(XXXmm)36.32 color(white)(Xm)2.270color(white)(mmll)1color(white)(mmmml)3color(white)(mmmmml)3
The empirical formula is "C"_6"H"_13"O"_3.
