# How do you convert 64.3 g PbBr_2 to moles?

Mar 18, 2017

$64.3$ $\text{g}$ of $P b B {r}_{\text{2}}$ is $0.175$ $\text{mols}$.

#### Explanation:

So the formula we are using is $n = \frac{m}{M}$.
=> Where $n$ is the amount in mols.
=> Where $m$ is the mass in grams.
=> Where $M$ is the molar mass in $\frac{g}{\text{mols}}$.

First off, we need to find the molar mass of $P b B {r}_{\text{2}}$.

Using a Periodic Table, lead has a molar mass of $207.20 \frac{g}{\text{mol}}$ and bromine has a molar mass of $79.90 \frac{g}{\text{mol}}$. Including the two atoms of bromine, when we add them all up, we get $367.00 \frac{g}{\text{mol}}$.

Now, we can plug them into the formula.

$n = \frac{m}{M}$

$n = \frac{64.3}{367}$

$n = 0.175204359$

$n = 0.175$

Therefore, $64.3$ $\text{g}$ of $P b B {r}_{\text{2}}$ is $0.175$ $\text{mols}$.

Hope this helps :)