We must first determine the empirical formula of the compound.
We must figure out the moles of #"C"#, #"H"#, and #"O"# and then calculate their ratio.
Step 1. Calculate the moles of each element.
#"Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C"#
#"Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C"#
#"Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O"#
Step 2. Calculate the empirical formula.
From this point on, I like to summarize the calculations in a table.
#"Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"#
#stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331
color(white)(Xm)1color(white)(mmmmml)1)#
#color(white)(m)"H" color(white)(XXXXm)0.667 color(white)(mml)0.6617 color(white)(Xm)1.987 color(white)(mmml)2#
#color(white)(m)"O" color(white)(XXXXm)5.33 color(white)(mmm)0.3331 color(white)(Xm)1.000 color(white)(mmml)1#
The empirical formula is #"CH"_2"O"#.
Step 3. Calculate the empirical formula mass
The empirical formula mass of #"CH"_2"O"# is 30.03 u.
Step 4. Calculate the molecular mass.
The molecular mass must be an integral multiple multiple of the empirical formula mass.
#"MM" = n × "EFM"#
#n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6#
Step 5. Calculate the Molecular Formula
The molecular formula is 6 times the empirical formula.
The molecular formula is #("CH"_2"O")_6 = "C"_6"H"_12"O"_6#.
