Calculate the total number of electrons present in one mole of methane

Question.

(i)Calculate the total number of electrons present in one mole of methane

(ii) Find (a) the total number and (b) the total mass of neutrons in $7 \mathrm{mg}$ of ${ }^{14} \mathrm{C}$.

(Assume that mass of a neutron $=1.675 \times 10^{-27} \mathrm{~kg}$ ).

(iii) Find (a) the total number and (b) the total mass of protons in $34 \mathrm{mg}$ of $\mathrm{NH}_{3}$ at STP.

Will the answer change if the temperature and pressure are changed?

Solution:

(i) Number of electrons present in 1 molecule of methane $\left(\mathrm{CH}_{4}\right)$

$\{1(6)+4(1)\}=10$

Number of electrons present in 1 mole i.e., $6.023 \times 10^{23}$ molecules of methane

$=6.022 \times 10^{23} \times 10=6.022 \times 10^{24}$

(ii)(a) Number of atoms of ${ }^{14} \mathrm{C}$ in 1 mole $=6.023 \times 10^{23}$

Since 1 atom of ${ }^{14} \mathrm{C}$ contains $(14-6)$ i.e., 8 neutrons, the number of neutrons in $14 \mathrm{~g}$ of

${ }^{14} \mathrm{C}$ is $\left(6.023 \times 10^{23}\right) \times 8 .$ Or, $14 \mathrm{~g}$ of ${ }^{14} \mathrm{C}$ contains $\left(6.022 \times 10^{23} \times 8\right)$ neutrons.

Number of neutrons in $7 \mathrm{mg}$

$=\frac{6.022 \times 10^{23} \times 8 \times 7 \mathrm{mg}}{1400 \mathrm{mg}}$

$=2.4092 \times 10^{21}$

(b) Mass of one neutron $=1.67493 \times 10^{-27} \mathrm{~kg}$

Mass of total neutrons in $7 \mathrm{~g}$ of ${ }^{14} \mathrm{C}$

$=\left(2.4092 \times 10^{21}\right)\left(1.67493 \times 10^{-27} \mathrm{~kg}\right)$

$=4.0352 \times 10^{-6} \mathrm{~kg}$

(iii) (a) 1 mole of $\mathrm{NH}_{3}=\{1(14)+3(1)\} \mathrm{g}$ of $\mathrm{NH}_{3}$

$=17 \mathrm{~g}$ of $\mathrm{NH}_{3}$

$=6.022 \times 10^{23}$ molecules of $\mathrm{NH}_{3}$

Total number of protons present in 1 molecule of $\mathrm{NH}_{3}$

$=\{1(7)+3(1)\}$

$=10$

Number of protons in $6.023 \times 10^{23}$ molecules of $\mathrm{NH}_{3}$

$=\left(6.023 \times 10^{23}\right)(10)$

$=6.023 \times 10^{24}$

$\Rightarrow 17 \mathrm{~g}$ of $\mathrm{NH}_{3}$ contains $\left(6.023 \times 10^{24}\right)$ protons. Number

of protons in $34 \mathrm{mg}$ of $\mathrm{NH}^{3}$

$=\frac{6.022 \times 10^{24} \times 34 \mathrm{mg}}{17000 \mathrm{mg}}$

$=1.2046 \times 10^{22}$

(b) Mass of one proton $=1.67493 \times 10^{-27} \mathrm{~kg}$

Total mass of protons in $34 \mathrm{mg}$ of $\mathrm{NH}_{3}$

$=\left(1.67493 \times 10^{-27} \mathrm{~kg}\right)\left(1.2046 \times 10^{22}\right)$

$=2.0176 \times 10^{-5} \mathrm{~kg}$

The numberof protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.