Answer to Question #194817 in Molecular Physics | Thermodynamics for Kevin Albert

Question

A 5.0 ng particle of pure carbon has a net charge of +12 µC.

What fraction of electrons on this particle of carbon have been removed? ( give your answer as a decimal number to 2 s.f)

[note: M_C = 12 g mol⁻¹ and one neutral atom of carbon has 6 protons and 6 electrons.] 

Answer 1

Let us determine the number of atoms in the particle

$N = \dfrac{m}{\mu}\cdot N_A = \dfrac{5.0\cdot10^{-9}}{12}\cdot 6\cdot10^{23} = 2.5\cdot10^{14}$ atoms.

The total number of electrons is $N_{\text{tot}}=6N = 1.5\cdot10^{15}.$

The amount of removed electrons is

$N_e = \dfrac{12\cdot10^{-6}\,\mathrm{C}}{1.6\cdot10^{-19}\,\mathrm{C}} = 7.5\cdot10^{13}$ .

The fraction of removed electrons is $\dfrac{N_e}{N_{\text{tot}}} = \dfrac{7.5\cdot10^{13}}{1.5\cdot10^{15}} = 0.05$