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(a) How many electrons would have to be removed from a penny to leave it with a charge of #1.0×10^-7 C#? (b) To what fraction of the electrons in the penny does this correspond? [A penny has a mass of 3.11g; assume it is made entirely of copper.]

1 Answer
Jul 11, 2016

Answer:

#6.25times10^11# electrons and #7.31times10^-13#

Explanation:

The charge on an electron is #1.6times 10^-19C#. So for part (a) we can determine that the number of electrons would be:
#(1.0times10^-7)/(1.6times10^-19)=6.25times10^11# electrons

For part b, we need to know how may electrons are in the penny. The relative atomic mass of copper is 63.546. By definition, 63.546g of copper would contain a number of atoms equal to Avogadro's constant which is #6.02 times 10^23 mol^-1#.

So in 3.11g there would be the following number of atoms:

#3.11/63.546 times (6.02times10^23)=2.95times10^22# atoms

The atomic number of copper is 29, so a neutral copper atom has 29 electrons.

Hence the fraction of electrons that would need to be removed would be:

#(6.25times10^11)/(29times(2.95times10^22))=7.31times10^-13#