Step 1. Calculate the empirical formula
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"C to "H"H to "Cl"Cl.
Your compound contains 24.27 % "C"C, and 4.07 % "H"H.
Assume that you have 100 g of sample.
Then it contains 24.27 g of "C"C and 4.07 g of "H"H.
"Mass of O = (100 - 24.27 - 4.07) g = 71.66 g"Mass of O = (100 - 24.27 - 4.07) g = 71.66 g
"Moles of C" = 24.27 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "2.021 mol C"
"Moles of H" = 4.07 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "4.038 mol H"
"Moles of Cl" = 71.66 color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)( "g Cl")))) = "2.021mol Cl"
From this point on, I like to summarize the calculations in a table.
ulbb("Element"color(white)(m) "Mass/g"color(white)(Xl) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers")
color(white)(m)"C" color(white)(XXXmml)24.27 color(white)(Xml)2.021color(white)(mml)1color(white)(mmmmmll)1
color(white)(m)"H" color(white)(XXXXmll)4.07 color(white)(mml)4.038 color(white)(Xmll)1.998 color(white)(mmmll)2
color(white)(m)"Cl"color(white)(XXXmm)71.66 color(white)(Xml)2.021color(white)(mml)1.000color(white)(mmmll)1
The empirical formula is "CH"_2"Cl".
Step 2. Calculate the molecular formula of the compound
The empirical formula mass of "CH"_2"Cl" is 49.48 u.
The molecular mass is 98.96 u.
The molecular mass must be an integral multiple of the empirical formula mass.
"MM"/"EFM" = (98.96color(red)(cancel(color(black)("u"))))/(49.48 color(red)(cancel(color(black)("u")))) = 2.000 ≈ 2
The molecular formula must be twice the empirical formula.
"MF" = ("EF")_2 = ("CH"_2"Cl")_2 = "C"_2"H"_4"Cl"_2
