#color(blue)("Step 1: Assume 100 g sample to find mass of each element")#

#"C" = 34.6" g"#

#"H" = 3.85" g"#

#"O" = 61.55" g"#

#color(blue)("Step 2: Find number of moles from the mass of each element.")# *Use the periodic table to find molar masses of each element*

#"C" = (34.6 cancel"g")/1 *(1" mol")/(12 cancel"g") = 2.88" mol"#

#"H" = (3.85 cancel"g")/1 *(1" mol")/(1.00 cancel"g") = 3.85" mol"#

#"O" = (61.55 cancel"g")/1 *(1" mol")/(16 cancel"g") = 3.85" mol"#

#color(blue)("Step 3: Divide all the mole numbers by the smallest mole value of them all")#

#"C" = (2.88cancel"mol")/(2.88 cancel"mol") = 1#

#"H" = (3.85cancel"mol")/(2.88 cancel"mol") = 1.33#

#"O" = (3.85 cancel"mol")/(2.88 cancel"mol") = 1.33#

#color(blue)("Step 4: Multiply the number by a factor which will give out a whole number ratio.")#

#"C" = 1 * (3) = 3#

#"H" = 1.33 * (3) = 3.99~~4#

#"O" = 1.33 * (3) = 3.99~~4#

#color(blue)("Step 5: Combine elements and attach their corresponding mole ratios. Then, find the empirical formula mass")#

#C_(3)H_(4)O_(4)->(104" g")/"mol" = "empirical formula mass"#

#color(blue)("Step 6: Divide given molecular formula mass by the empirical formula mass to get a factor.")#

*This factor will tell you how many times to multiply the subscripts to get your mole ratios for the molecular formula*

#("Molecular formula mass")/("Empirical formula mass") ->[(208 cancel"g")/(cancel"mol")]/[(104 cancel"g")/(cancel"mol")]= 2#

#C_(3)H_(4)O_(4) -> color(orange)(C_(6)H_(8)O_(8) = "Molecular Formula")#

#"Answer": C_(6)H_(8)O_(8)#