What is the empirical formula for a compound that contains 18.8% Li, 16.3% C, and 64.9% O?

1 Answer
Jan 3, 2016



Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be #"Li"_2"CO"_3#. Two things to go by here

  • the position in the periodic table for the three elements, which gives you an idea about their molar masses
  • the fact that you're dealing with an Ionic compound tells you that its empirical formula will also be its chemical formula

So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be #"Li"_2"CO"_3#.

Let's do some calculations to see if the prediction turns out to be correct.

Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.

To make calculations easier, you can pic ka #"100-g"# sample. This will allow you to convert the percent composition directly to grams. Your sample will thus contain

  • #"18.8 g"# of lithium
  • #"16.3 g"# of carbon
  • #"64.9 g"# of oxygen

Next, use the molar mass of each element to determine how many moles of each you'd get in this sample

#"For Li: " 18.8 color(red)(cancel(color(black)("g"))) * "1 mole Li"/(6.941color(red)(cancel(color(black)("g")))) = "2.709 moles Li"#

#"For C: " 16.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "1.357 moles C"#

#"For O: " 64.9 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "4.056 moles O"#

Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.

To get the mole ratio that exists between these three elements, divide these three values by the smallest one.

#"For Li: " (2.709 color(red)(cancel(color(black)("moles"))))/(1.357color(red)(cancel(color(black)("moles")))) = 1.996 ~~ 2#

#"For C: " (1.357 color(red)(cancel(color(black)("moles"))))/(1.357color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (4.056 color(red)(cancel(color(black)("moles"))))/(1.357color(red)(cancel(color(black)("moles")))) = 2.989 ~~ 3#

Since this ratio, i.e. #2:1:3#, is also the smallest whole number ratio that can exist between these three numbers, the empirical formula of the compound will indeed be

#"Li"_2"CO"_3 -># lithium carbonate