Question

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

Answer 1

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 `"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

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

Therefore, the total mass of lithium carbonate (`"Li"_2"CO"_3`) is

`= 2*6.94 + 12.011 + 3*15.999`

`=` `color(green)("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 `\mathbf(2)` indicates that there are two `"Li"` atoms in `"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.