Question

A 56.0 g ball of copper has a net charge of 2.10 μC. 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.)

Answer 1

The fraction of the cooper's electrons that is removed is
`8.5222* 10^(-11)`.

Step-by-step explanation:

An electron has a mass of
`9.1 * 10^(-31)\,kg` and a charge of
`-1.6 * 10^(-19)\,C`. Based on the Principle of Charge Conservation,
`-2.10* 10^(-6)\,C` in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

`n_(R) = (-2.10* 10^(-6)\,C)/(-1.6 * 10^(-19)\,C)`

`n_(R) = 1.3125 * 10^(13)\,electrons`

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

`n_A = (m_(ball))/(M_(Cu))\cdot N_(A)`

Where:

`m_(ball)` - Mass of the ball, measured in kilograms.

`M_(Cu)` - Atomic mass of cooper, measured in grams per mole.

`N_(A)` - Avogradro's Number, measured in atoms per mole.

If
`m_(ball) = 56\,g`,
`M_(Cu) = 63.5\,(g)/(mol)` and
`N_(A) = 6.022* 10^(23)\,(atoms)/(mol)`, the number of atoms is:

`n_(A) = \left((56\,g)/(63.5\,(g)/(mol) ) \right)\cdot \left(6.022* 10^(23)\,(atoms)/(mol) \right)`

`n_(A) = 5.3107* 10^(23)\,atoms`

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

`n_(E) = \left(29\,(electrons)/(atom) \right)\cdot (5.3107* 10^(23)\,atoms)`

`n_(E) = 1.5401* 10^(23)\,electrons`

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

`x = (n_(R))/(n_(E))`

`x = (1.3125* 10^(13)\,electrons)/(1.5401* 10^(23)\,electrons)`

`x = 8.5222 * 10^(-11)`

The fraction of the cooper's electrons that is removed is
`8.5222* 10^(-11)`.