# What is the percentage of lithium in lithium carbonate (Li_2CO_3)?

Apr 20, 2016

Acquiring the percentage requires the usage of "unified" units, meaning units that are equivalent across the board, because a percentage is really a unitless fraction (all the involved units must cancel out).

So, we can use $\text{g/mol}$, given that all atoms have an atomic mass, to compute the following:

\mathbf(%"Li" = "quantity Li"/("quantity Li"_2"CO"_3)xx100%)

= \mathbf("quantity Li"/("quantity Li" + "quantity C" + "quantity O")xx100%)

where the "quantity" can be any chosen unified unit, as long as we use the same units for both lithium and lithium carbonate.

The atomic masses needed are

• $\text{Li}$: we only know a range (6.938~6.997) with decent certainty, so let's choose "6.94 g/mol";
• $\text{C}$: $\text{12.011 g/mol} ,$
• $\text{O}$: $\text{15.999 g/mol} .$

Therefore, the total mass of lithium carbonate (${\text{Li"_2"CO}}_{3}$) is

$= 2 \cdot 6.94 + 12.011 + 3 \cdot 15.999$

$=$ $\textcolor{g r e e n}{\text{73.888 g/mol}}$.

Now, we use the mass of lithium in the compound to determine the percentage of it in the compound.

Given that a subscript $\setminus m a t h b f \left(2\right)$ indicates that there are two $\text{Li}$ atoms in ${\text{Li"_2"CO}}_{3}$, we simply have

color(blue)(%"Li") = "quantity Li"/("quantity Li"_2"CO"_3)xx100%

= (2*6.94 cancel"g/mol" "Li")/(73.888 cancel"g/mol" "Li"_2"CO"_3)xx100%

= color(blue)(18.8%),

rounded to three sig figs.