A 50.0 g ball of copper has a net charge of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathit{C}$. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5 .)

• Mass of ball m= 50.0 g .
• Charge on ball $Q=2.00\mu C$
• One copper atom has 29 protons

• Atomic mass of copper is 63.5

One mole is defined as the amount of substance that contains $6.022\times {10}^{23}$elementary particles. The number of moles of any substance can be calculated by taking the ratio of given mass of the substance to its molar mass.

$N=\frac{m}{M}$

Here, N is the number of moles, m is the mass, and M is the molar mass of the substance.

The number of moles of copper is,

$N=\frac{m}{M}$

Here, m is the mass of the copper ball (m = 50.0 g) , and M is the molar mass of the copper (M = 63.5 g) .

Substituting all known values,

$$\begin{gathered} N=\frac{50.0g}{63.5g} \\ \approx 0.787 \end{gathered}$$

Since, one mole of substance contains $6.022\times {10}^{23}$elementary entities of the given substance. Therefore, total number of atoms of copper in 50.0 g of copper ball is,

role="math" localid="1653644359589" $$\begin{gathered} N_a=\left(0.787\right)\times \left(6.022\times {10}^{23}\right) \\ =4.739\times {10}^{23} \end{gathered}$$

We know that one atom of copper contains 29 electrons. Therefore, total number of electrons in 50.0 g of copper ball is,

$$\begin{gathered} N_e=\left(29\right)\times \left(4.739\times {10}^{23}\right) \\ =1.374\times {10}^{25} \end{gathered}$$

According to quantization of charge,

$Q=n\left|q_e\right|$

Here, Q is the net charge created on the object due to removal of electron $\left(Q=2.00\mu C\right)$, is the number of electrons removed, and $\left|q_e\right|$is the fundamental unit of charge $\left(\left|q_e\right|=1.6\times {10}^{-19}C\right)$.

The number of electrons removed is,

$$\begin{gathered} n=\frac{\left(2.00\mu C\right)}{\left(1.6\times {10}^{-19}C\right)} \\ =\frac{\left(2.00\mu C\right)\times \left(\frac{{10}^{-6}C}{1\mu C}\right)}{\left(1.6\times {10}^{-19}C\right)} \\ =1.25\times {10}^{13} \end{gathered}$$

The fraction of copper’s electrons removed is,

$f=\frac{n}{N_e}$

Here, n is the number of electrons removed $\left(n=1.25\times {10}^{13}\right)$ , and N_e is the total number of electrons in 50.0 g of copper ball $\left(N_e=1.374\times {10}^{25}\right)$.

Substituting all known values,

$$\begin{gathered} f=\frac{1.25\times {10}^{13}}{1.374\times {10}^{25}} \\ =9.09\times {10}^{-13} \end{gathered}$$

Hence, the fraction of electrons has been removed is $9.09\times {10}^{-13}$.