Finding the excess electrons per lead atom

[Physicslearner500039]

Homework Statement

Excess electrons are placed on a small lead sphere with mass 8.0g so that its net charge is -3.20*10^-9 C. a. Find the number of excess electrons. b. How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.

Relevant Equations

Total Charge q = no of electrons * charge of electron. q = ne

a. This is easy with $q = n*e$
$3.2 * 10^{-9} = n * 1.6*10^{-19}$
$n = 2*10^{10}$ electrons
b. Total Lead atoms are
$\frac {8 * 6.022*10^{23}} {207} = 2.3 *10^{22}$ I used the Avogadro number.
Total electrons = $2.3 * 10^{22} * 82 = 1.88 * 10^{20}$, here i multiplied with the atomic number of lead.
After that I don't understand how to calculate excess electrons per lead atom. Please advise.

 

[BvU]

You have $n$ and the total number of lead atoms ...

 

[Physicslearner500039]

Ok, Thank you, understood, the no of excess electrons per lead atom = $\frac {2*10^{10}} {2.3*10^{22}} = 8.69*10^{-13}$