Chapter 21, Problem 21.8QP

Interpretation Introduction

Interpretation: The mass for the given individual gases present in given total mass of air should be determined.

Concept Introduction:

Partial pressure: The partial pressure for a gas is obtained by the multiplication of total pressure of the gas with the mole fraction of the gas. It is the pressure of individual gas present in the mixture of gases.

$\text{Partial}\text{Pressure}\text{for}\text{gas}\text{=}\text{Mole}\text{fraction}\text{for}\text{that}\text{gas}\text{×}\text{Total}\text{pressure}$

Mole Fraction: The mole fraction is given by dividing concentration of one component in the mixture by the total components present in the mixture.

Moles: It is the number of particles present in the given specific substance. The formula used to identify the moles present in given mass of a compound is as follows,

$\text{x}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{sample}}{\text{Molar}\text{mass}\text{of}\text{the}\text{sample}}$

Molar mass: It is the total mass of the substance including all the number and the mass of the elements present in substance.

To determine: The mass of gases present in the air with given total mass of air.

Answer to Problem 21.8QP

The mass of nitrogen present in total mass of air is $\text{3}{\text{.96×10}}^{\text{18}}\text{kg}$.

The mass of oxygen present in total mass of air is about $\text{1}{\text{.22×10}}^{\text{18}}\text{kg}$

The mass of carbon dioxide present in air is about $2.6{\text{×10}}^{\text{15}}\text{kg}$

Explanation of Solution

Determine the total moles of gases present in air.

$\begin{array}{l}\text{Total}\text{moles}\text{of}\text{gases}\text{present}\text{in}\text{air}\text{=}\left(\text{5}\text{.25}\times {\text{10}}^{\text{21}}\text{g}\right)\frac{\text{1}\text{mol}}{29\text{g}}\\ =1.81\times {\text{10}}^{\text{20}}\text{mol}\\ \end{array}$

The total moles obtained by dividing the total mass of air and molar mass of air.

It is given that the air is dry air hence the molar mass of dry air is about 29 g and dividing the given total mass value $\text{5}\text{.25}\times {\text{10}}^{\text{21}}\text{g}$ by molar mass of dry air results value equal to $1.81\times {\text{10}}^{\text{20}}\text{mol}$ serves as the total moles present in air.

Determine the mass of nitrogen gas present in air.

$\begin{array}{l}\text{x}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{sample}}{\text{Molar}\text{mass}\text{of}\text{the}\text{sample}}\\ \text{mass}\text{of}{\text{N}}_{\text{2}}={\text{moles of N}}_{\text{2}}\times \text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}\\ =\text{percentage}\text{of}{\text{ N}}_{\text{2}}\text{×}\text{Total}\text{moles}\text{of}\text{air}\text{×}\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}\end{array}$

$\begin{array}{l}\text{Mass}\text{=}\left(0.7803\right)\left(1.81\times {10}^{20}\text{mol}\right)\frac{28.02\text{g}}{1\text{mol}}\\ =3.96\times {\text{10}}^{\text{21}}\text{g}\\ \text{=3}\text{.96}\times {\text{10}}^{18}\text{kg}\\ \text{since, }\\ \text{1g}{\text{= 10}}^{\text{-3}}\text{kg}\\ \end{array}$

The mass of nitrogen present in the given mass of air is determined by using the data that is, of the total moles ( $1.81\times {10}^{20}\text{mol}$) nitrogen present about 78.03% composition in air which is divided by 100 then multiplied by total moles present in the air and the molar mass of nitrogen $\left({\text{N}}_{\text{2}}=28.02\text{g}\right)$

Determine the mass of oxygen gas present in air.

$\begin{array}{l}\text{x}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{sample}}{\text{Molar}\text{mass}\text{of}\text{the}\text{sample}}\\ \text{mass}\text{of}{\text{O}}_{\text{2}}={\text{moles of O}}_{\text{2}}\times \text{Molar}\text{mass}\text{of}{\text{O}}_{\text{2}}\\ =\text{percentage}\text{of}{\text{ O}}_{\text{2}}\text{×}\text{Total}\text{moles}\text{of}\text{air}\text{×}\text{Molar}\text{mass}\text{of}{\text{O}}_{\text{2}}\end{array}$

$\begin{array}{l}\text{Mass}\text{=}\left(0.2099\right)\left(1.81\times {10}^{20}\text{mol}\right)\frac{\text{32g}}{1\text{mol}}\\ =1.22\times {\text{10}}^{\text{21}}\text{g}\\ \text{=1}\text{.22}\times {\text{10}}^{18}\text{kg}\\ \text{since, }\\ \text{1g}{\text{= 10}}^{\text{-3}}\text{kg}\\ \\ \end{array}$

The mass of oxygen present in the given mass of air is determined by using the data that is, of the total moles ( $1.81\times {10}^{20}\text{mol}$)oxygen present about 20.99% composition in air which is divided by 100 then multiplied by total moles present in the air and the molar mass of oxygen $\left({\text{O}}_{\text{2}}=32\text{g}\right)$

Determine the mass of nitrogen gas present in air.

$\begin{array}{l}\text{x}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{sample}}{\text{Molar}\text{mass}\text{of}\text{the}\text{sample}}\\ \text{mass}\text{of}{\text{CO}}_{\text{2}}={\text{moles of CO}}_{\text{2}}\times \text{Molar}\text{mass}\text{of}{\text{CO}}_{\text{2}}\\ =\text{percentage}\text{of}{\text{ CO}}_{\text{2}}\text{×}\text{Total}\text{moles}\text{of}\text{air}\text{×}\text{Molar}\text{mass}{\text{of CO}}_{\text{2}}\end{array}$

$\begin{array}{l}\text{Mass}\text{=}\left(3.3\times {10}^{-4}\right)\left(1.81\times {10}^{20}\text{mol}\right)\frac{44.01\text{g}}{1\text{mol}}\\ =2.63\times {\text{10}}^{18}\text{g}\\ \text{=}2.63\times {\text{10}}^{15}\text{kg}\\ \text{since, }\\ \text{1g}{\text{= 10}}^{\text{-3}}\text{kg}\\ \end{array}$

The mass of ${\text{CO}}_{\text{2}}$ present in the given mass of air is determined by using the data that is, of the total moles ( $1.81\times {10}^{20}\text{mol}$) ${\text{CO}}_{\text{2}}$ present about 0.033% composition in air which is divided by 100 then multiplied by total moles present in the air and the molar mass of ${\text{CO}}_{\text{2}}$ is equal to 44.01g.

Conclusion

Conclusion

The mass of gases present in given mass of air is determined.