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

Jul 16, 2014

${\text{Li"_2"CO}}_{3}$

Explanation:

Solution:

1) Assume $\text{100 g}$ of the compound is present. $\text{100 g}$ of this compound on decomposition gives me

• "Li" => 18.7 % or $\text{18.7 g}$ of $\text{Li}$
• "C" => 16.3 % or $\text{16.3 g}$ of $\text{C}$
• "O" => 65.0% or $\text{65 g}$ of $\text{O}$

2) Calculate the number of moles of each element.

$\text{no. of moles" = "mass of the element"/"molar mass of the element}$

So

$\text{moles of Li" = "18.7 g" /"7 g mol"^(-1) = "2.6 moles}$

$\text{moles of C" = "16.3 g"/"12 g mol"^(-1) = "1.3 moles}$

$\text{moles of O" = "65 g"/"15.9 g mol"^(-1) = "4.0 moles}$

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

$\text{Li: " "2.6 moles" /"1.3 moles} = 2$

$\text{C: " "1.3 moles"/"1.3 moles} = 1$

$\text{O: " "4.0 moles"/"1.3 moles} = 3$

4 ) The empirical formula is ${\text{Li"_2 "C"_1"O}}_{3}$, or ${\text{Li"_2"CO}}_{3}$