# Question dc235

Mar 19, 2016

"% Li" = 17.983%

#### Explanation:

A compound's percent composition essentially tells you how many grams of each constituent element you get per $\text{100 g}$ of compound.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% element" = "mass of said element"/"mass of the compound} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that in order to find the percent composition of lithium phosphate, ${\text{Li"_3"PO}}_{4}$, you need to know how many grams of each constituent element you get in a sample of the compound.

Now, notice that one mole of lithium phosphate contains

• three moles of lithium, $3 \times \text{Li}$
• one mole of phosphorus, $1 \times \text{P}$
• four moles of oxygen, $4 \times \text{O}$

The first thing to do here is use the molar mass of each element to find the molar mass of the compound. This will then allow you to calculate the percent composition of lithium by using the mass of lithium that corresponds to one mole of lithium phosphate.

So, the molar masses of lithium, phosphorus, and oxygen are

${\text{For Li: " M_M = "6.941 g mol}}^{- 1}$

${\text{For P: " M_M = "30.9738 g mol}}^{- 1}$

${\text{For O: " M_M = "15.9994 g mol}}^{- 1}$

The molar mass of lithium phosphate will thus be

overbrace(3 xx "6.941 g mol"^(-1))^(color(blue)("three atoms of Li")) + overbrace(1 xx "30.9738 g mol"^(-1))^(color(purple)("one atom of P")) + overbrace(4 xx "15.9994 g mol"^(-1))^(color(red)("four atoms of O"))

${M}_{M} = {\text{115.794 g mol}}^{- 1}$

This means that one mole of lithium phosphate will have a mass of $\text{115.794 g}$. But you also know that one mole of lithium has a mass of $\text{6.941 g}$, and that you get three moles of lithium per mole of lithium phosphate.

You can thus say that

"% Li" = (3 xx 6.941 color(red)(cancel(color(black)("g mol"^(-1)))))/(115.794color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"17.983%"color(white)(a/a)|)))#

Therefore, you will get $\text{17.983 g}$ of lithium for every $\text{100 g}$ of lithium phosphate.