An **empirical formula** represents the lowest whole number ratio of elements in the compound. In order to determine the empirical formula, the moles of each element must be determined, and then the moles are divided by the least number of moles. If the results are all whole numbers, those are the lowest whole number ratios for the formula. If the results are not whole numbers, then they must be multiplied by a number that will make all of them whole.

Since the percentages add up to 100%, we can assume a 100 g sample and the percentages will become the masses in grams.

#color(blue)("Given Masses"# (from the question)

#"C":##"75.7 g"#

#"H":##"8.80 g"#

#"O":##"15.5 g"#

#color(blue)("Molar Masses"# (from the periodic table)

#"C":##"12.011 g/mol"#

#"H":##"1.008 g/mol"#

#"O":##"15.999 g/mol"#

#color(blue)("Determine Moles of Each Element"#

Multiply the given mass of each element by the inverse of its molar mass.

#75.7color(red)cancel(color(black)("g C"))xx(1"mol C")/(12.011color(red)cancel(color(black)("g C")))="6.30 mol C"#

#8.8color(red)cancel(color(black)("g H"))xx(1"mol H")/(1.008color(red)cancel(color(black)("g H")))="8.73 mol H"#

#15.5color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="0.969 mol O"#

#color(blue)("Determine the Empirical Formula"#

Divide the moles of each element by the least number of moles.

#"C":##(6.30color(red)cancel(color(black)("mol")))/(0.969color(red)cancel(color(black)("mol")))=6.50#

#"H":##(8.73color(red)cancel(color(black)("mol")))/(0.969color(red)cancel(color(black)("mol")))=9.01#

#"O":##(0.969color(red)cancel(color(black)("mol")))/(0.969color(red)cancel(color(black)("mol")))=1.00#

Since #6.5# is not a whole number, divide all of the results by #2# in order to get all whole numbers.

#"C":##6.50xx2=13#

#"H":##9.01xx2=18.02~~18#

#"O":##1.00xx2=2#

Now all of the ratios are whole numbers, which makes the empirical formula:

#"C"_13"H"_18"O"_2"#

This is the formula for ibuprofen: both empirical and molecular.

https://www.ncbi.nlm.nih.gov/pccompound?term=C13H18O2