Question

A 48.6 g ball of copper has a net charge of 2.4 µc. what fraction of the copper's electrons have 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 = (46.6 g) / (63.5 g per mol of atoms) = 0.73385 mol
# of atoms = (0.73385 mol) (6.02 × 10^23 atoms per mol) = 4.4178 × 10^23 atoms

There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
normal # electrons = (4.4178 × 10^23 atoms) (29 electrons per atom) = 1.2811 × 10^25 electrons

Currently, the charge in the ball is 2.5 µC, which means -2.5 µC worth of electrons have been removed.
# removed electrons = (-2.5 µC) / (1.602 × 10^-13 µC per electron) = 1.5604 × 10^13 electrons removed

# removed electrons / normal # electrons =
(1.5604 × 10^13 electrons removed) / (1.2811 × 10^25 electrons) = 1.22 × 10^-12

That's 1 / 1.22 × 10^12