# (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.]

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Matt Share
Jul 11, 2016

$6.25 \times {10}^{11}$ electrons and $7.31 \times {10}^{-} 13$

#### Explanation:

The charge on an electron is $1.6 \times {10}^{-} 19 C$. So for part (a) we can determine that the number of electrons would be:
$\frac{1.0 \times {10}^{-} 7}{1.6 \times {10}^{-} 19} = 6.25 \times {10}^{11}$ 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 $6.02 \times {10}^{23} m o {l}^{-} 1$.

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

$\frac{3.11}{63.546} \times \left(6.02 \times {10}^{23}\right) = 2.95 \times {10}^{22}$ 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:

$\frac{6.25 \times {10}^{11}}{29 \times \left(2.95 \times {10}^{22}\right)} = 7.31 \times {10}^{-} 13$

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