(a) How many electrons would have to be removed from a penny to leave it with a charge of 1.0×10^-7 C1.0×10−7C? (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
Matt
Jul 11, 2016

6.25times10^116.25×1011 electrons and 7.31times10^-137.31×10−13

Explanation:

The charge on an electron is 1.6times 10^-19C1.6×10−19C. So for part (a) we can determine that the number of electrons would be:
(1.0times10^-7)/(1.6times10^-19)=6.25times10^111.0×10−71.6×10−19=6.25×1011 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^-16.02×1023mol−1.

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

3.11/63.546 times (6.02times10^23)=2.95times10^223.1163.546×(6.02×1023)=2.95×1022 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^-136.25×101129×(2.95×1022)=7.31×10−13

Related questions
Impact of this question
38281 views around the world
You can reuse this answer
Creative Commons License