# What is the percentage composition of PbCl_2?

Dec 14, 2015

$\text{74.504% Pb}$
$\text{25.496% Cl}$

#### Explanation:

For a given compound, percent composition can be found by using the number of moles of each element that makes up said compound, the molar masses of these elements, and the molar mass of the compound itself.

As you know, a compound's molar mass tells you what the mass of one mole of that substance is. The molar mass of lead(II) chloride is equal to $\text{278.106 g/mol}$, which means that one mole of lead(II) chloride will have a mass of $\text{278.106 g}$.

Now, one mole of lead(II) chloride will contain

• one mole of lead(II) cations, ${\text{Pb}}^{2 +}$
• two moles of chloride anions, ${\text{Cl}}^{-}$

The molar masses of these two ions are equal to $\text{207.2 g/mol}$ for ${\text{Pb}}^{2 +}$ and $\text{35.453 g/mol}$ for ${\text{Cl}}^{-}$.

This means that one mole of lead(II) chloride will contain

1 color(red)(cancel(color(black)("mole Pb"^(2+)))) * "207.2 g"/(1color(red)(cancel(color(black)("mole Pb"^(2+))))) = "207.2 g Pb"^(2+)

and

2 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = "70.906 g Cl"^(-)

This means that the percent composition of lead(II) chloride will be

( 70.906 color(red)(cancel(color(black)("g"))))/(278.106color(red)(cancel(color(black)("g")))) xx 100 = color(green)("25.496% Cl")

( 207.2 color(red)(cancel(color(black)("g"))))/(278.106color(red)(cancel(color(black)("g")))) xx 100 = color(green)("74.504% Pb")

So, every mole of lead(II) chloride will contain $\text{70.906 g}$ of chlorine and $\text{207.2 g}$ of lead.

This is equivalent to saying that every $\text{100 g}$ of lead(II) chloride will contain $\text{25.496 g}$ of chlorine in the form of chloride anions and $\text{74.504 g}$ of lead in the form of lead(II) cations.