A49.3 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.)

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