# What fraction of its electrons have been removed?

A 2.20mm -diameter copper ball is charged to + 90.0nC. What fraction of its electrons have been removed?

I have tried answering this question four times and I can’t get the answer right. I tried to follow the instructions given by these two links:

Answers I have gotten are 1.310^-9, 4.110^-8, and 4.1*10^-9.

I really thought the last answer was correct so I tried changing my significant figures to 4.09*10^-9, but that was also wrong. Please help.

I also tried following this link:

The problem is on page 13.

• e = charge on electron = 1.602*10^-19 C

9010^-9 C represents the charge of 9010^-9/1.602*10^-19 removed electrons.

= 56.18010^10 electrons = 5.618010^11 electrons.

Mass of copper ball = volumedensity = (4/3)pir^3density

r = 1.1 mm = 1.1*10^-3 m

r^3 = 1.331*10^-9 m^3

density = 8,940 kg/m^3

mass of ball = (4/3)pi1.33110^-98,940 = 4984310^-9 kg = 4.984*10^-5 kg

molar mass of copper = 63.546 g/mol = 63.546*10^-3 kg/mol.

Number of mols of copper in 4.98410^-5 kg = 4.98410^-5 kg/(63.54610^-3) = 7.84310^-4.

1 mol of a substance contains Avogadros number of particles = 6.02214*10^23

7.84310^-4 mols contains 6.0221410^237.84310^-4 = 47.231*10^19 atoms.

A neutral copper atom has 29 electrons.

Number of electrons in ball = 2947.23110^19 = 1368.7110^19 = 1.368710^22 electrons.

Fraction of electrons removed = 5.618010^11/1.368710^22 = 4.1046*10^-11

Answer is 4.10*10^-11 to 3 sig figs