Question

Calculate the total number of electrons present in $1.4{\text{ g}}$ of dinitrogen gas.

Answer 1

To solve this we must first calculate the number of moles of dinitrogen gas in $1.4{\text{ g}}$ of dinitrogen gas. Then calculate the number of molecules of dinitrogen gas in the calculated number of moles. Then calculate the total number of electrons. Remember one molecule of dinitrogen gas contains 14 electrons.

Complete step by step solution:
We are given $1.4{\text{ g}}$ of dinitrogen gas. Dinitrogen gas has a chemical formula ${{\text{N}}_{\text{2}}}$.
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $1.4{\text{ g}}$ for the mass of nitrogen gas, $28{\text{ g/mol}}$ for the molar mass of dinitrogen gas. Thus,
${\text{Number of moles of }}{{\text{N}}_2} = \dfrac{{1.4{\text{ g}}}}{{28{\text{ g/mol}}}}$
${\text{Number of moles of }}{{\text{N}}_2} = 0.05{\text{ mol}}$
Thus, the number of moles of dinitrogen gas in $1.4{\text{ g}}$ of dinitrogen gas are $0.05{\text{ mol}}$.
We know that one mole of any substance contains $6.022 \times {10^{23}}$ molecules of the substance. $6.022 \times {10^{23}}$ is known as Avogadro’s number.
Thus, $1{\text{ mol}}$ of dinitrogen gas contains $6.022 \times {10^{23}}$ molecules of dinitrogen gas. Thus, the molecules of dinitrogen gas in $0.05{\text{ mol}}$ are,
${\text{Number of molecules of }}{{\text{N}}_2} = 0.05{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}}}{{{\text{1 mol}}}}$
${\text{Number of molecules of }}{{\text{N}}_2} = 3.01 \times {10^{23}}{\text{ molecules}}$
Thus, the molecules of dinitrogen gas in $0.05{\text{ mol}}$ are $3.01 \times {10^{23}}{\text{ molecules}}$.
We know that one molecule of dinitrogen gas contains 14 electrons. Thus, the number of electrons in $3.01 \times {10^{23}}{\text{ molecules}}$ of dinitrogen gas are,
${\text{Number of electron}} = 3.01 \times {10^{23}}{\text{ molecules}} \times \dfrac{{14{\text{ electrons}}}}{{1{\text{ molecule}}}}$
${\text{Number of electron}} = 4.214 \times {10^{23}}{\text{ electrons}}$

Thus, the total number of electrons present in $1.4{\text{ g}}$ of dinitrogen gas are $4.214 \times {10^{23}}{\text{ electrons}}$.

Note: The number $6.022 \times {10^{23}}$ is known as Avogadro’s number. The number of molecules of a compound is Avogadro’s number $\left( {6.022 \times {{10}^{23}}{\text{ molecules}}} \right)$ for 1 mole of compound. We know that the atomic number of nitrogen is 7. Thus, nitrogen has 7 electrons. Thus, dinitrogen has 14 electrons.