What’s the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?

1 Answer

"Li"_2"CO"_3

Explanation:

Solution:

1) Assume "100 g" of the compound is present. "100 g" of this compound on decomposition gives me

  • "Li" => 18.7 % or "18.7 g" of "Li"
  • "C" => 16.3 % or "16.3 g" of "C"
  • "O" => 65.0% or "65 g" of "O"

2) Calculate the number of moles of each element.

"no. of moles" = "mass of the element"/"molar mass of the element"

So

"moles of Li" = "18.7 g" /"7 g mol"^(-1) = "2.6 moles"

"moles of C" = "16.3 g"/"12 g mol"^(-1) = "1.3 moles"

"moles of O" = "65 g"/"15.9 g mol"^(-1) = "4.0 moles"

3) Divide by the lowest, seeking the smallest whole-number ratio:

"Li: " "2.6 moles" /"1.3 moles" = 2

"C: " "1.3 moles"/"1.3 moles" =1

"O: " "4.0 moles"/"1.3 moles" = 3

4 ) The empirical formula is "Li"_2 "C"_1"O"_3, or "Li"_2"CO"_3