Question

A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

Answer 1

First, find how many copper atoms make up the ball: 
 moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.77638mol 
 # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23 atoms 

 There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
 normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) =  1.3554× 10^25 electrons 

 Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
 # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed
 
 # removed electrons / normal # electrons = 
(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 

 That's 1 / 9.21 × 10^13