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

Question

46271 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-07-16T05:21:03

#"Li"_2"CO"_3#

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#