What is empirical formula for Chloroform (89.1% Chlorine, 0.84% Hydrogen, 10.06% Carbon)?

Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.

To make the calculations easier, you can pick a #"100.0-g"# sample of chloroform and use its percent concentration by mass to find that it contains

• #"89.1 g " -># chlorine
• #"0.84 g " -># hydrogen
• #"10.06 g " -># carbon

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

#"For Cl: " 89.1 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453 color(red)(cancel(color(black)("g")))) = "2.513 moles C"#

#"For H: " 0.84 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.8334 moles H"#

#"For C: " 10.06 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.8376 moles C"#

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

#"For Cl: " (2.513 color(red)(cancel(color(black)("moles"))))/(0.8334 color(red)(cancel(color(black)("moles")))) = 3.0154 ~~ 3#

#"For H: " (0.8334 color(red)(cancel(color(black)("moles"))))/(0.8334color(red)(cancel(color(black)("moles")))) = 1#

#"For C: " (0.8376 color(red)(cancel(color(black)("moles"))))/(0.8334color(red)(cancel(color(black)("moles")))) = 1.005 ~~ 1#

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

#"C"_1"H"_1"Cl"_3 implies color(green)("CHCl"_3)#