Chapter 20, Problem 20.8QP

Interpretation Introduction

Interpretation:

The total mass of nitrogen, oxygen and carbon dioxide in atmosphere has to be calculated.

Concept Introduction:

• Molar mass is the ratio of mass of given substance to that number of moles of compound.
• $\text{Molar}\text{mass}\text{=}\frac{\text{Amount}\text{of}\text{given}\text{substance}}{\text{Number}\text{of}\text{moles}}$

Answer to Problem 20.8QP

The mass of $78.03\%$ nitrogen present in atmosphere is $\text{3}{\text{.96×10}}^{\text{18}}\text{kg}$

The mass of $\text{20}\text{.99\%}$ oxygen present in atmosphere is $\text{1}{\text{.22×10}}^{\text{18}}\text{kg}$

The mass of $\text{0}\text{.033\%}$ carbon dioxide present in atmosphere is $\text{2}{\text{.63×10}}^{\text{15}}\text{kg}$

Explanation of Solution

Given,

Molar mass of air is $\text{29}\text{.0g/mol}$

Total mass of air is $\text{5}\text{.25}\text{×}{\text{10}}^{\text{18}}\text{kg}$

The total moles of gases is calculated using molar mass of air which is given as,

$\begin{array}{l}\text{Moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{air}}{\text{Molar}\text{mass}\text{of}\text{air}}\\ \\ \text{=}\text{5}\text{.25}\text{×}{\text{10}}^{\text{18}}\text{kg}\text{×}\frac{\text{1}\text{mol}}{\text{29}\text{.0g}}\\ \\ \text{=}\text{1}\text{.81}\text{×}{\text{10}}^{\text{20}}\text{mol}\end{array}$  

The mass of $78.03\%$ nitrogen present in atmosphere is calculated as

 $\begin{array}{l}\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{of}{\text{N}}_{\text{2}}\text{×}\text{Molar}\text{mass}{\text{ofN}}_{\text{2}}\\ \\ \text{Mass}\text{=}\left(\text{0}\text{.7803}\right)\left(\text{1}{\text{.81×10}}^{\text{20}}\text{mol}\right)\text{×}\frac{\text{28}\text{.02g}}{\text{1mol}}\\ \\ \text{=}\text{3}{\text{.96×10}}^{\text{21}}\text{g}\\ \\ \text{=3}{\text{.96×10}}^{\text{18}}\text{kg}(\text{1g}{\text{= 10}}^{\text{-3}}\text{kg)}\end{array}$

The mass of $\text{20}\text{.99\%}$ oxygen present in atmosphere is calculated as,

$\begin{array}{l}\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{of}{\text{O}}_{\text{2}}\text{×}\text{Molar}\text{mass}\text{of}{\text{O}}_{\text{2}}\\ \\ \text{Mass}\text{=}\left(\text{0}\text{.2099}\right)\left(\text{1}{\text{.81×10}}^{\text{20}}\text{mol}\right)\text{×}\frac{\text{32g}}{\text{1mol}}\\ \\ \text{=}\text{1}{\text{.22×10}}^{\text{21}}\text{g}\\ \\ \text{=1}{\text{.22×10}}^{\text{18}}\text{kg}(\text{1g}{\text{= 10}}^{\text{-3}}\text{kg )}\end{array}$

The mass of $\text{0}\text{.033\%}$ carbon dioxide present in atmosphere is calculated as,

$\begin{array}{l}\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{of}{\text{CO}}_{\text{2}}\text{×}\text{Molar}\text{mass}\text{of}{\text{CO}}_{\text{2}}\\ \\ \text{Mass}\text{=}\left(\text{3}{\text{.3×10}}^{\text{-4}}\right)\left(\text{1}{\text{.81×10}}^{\text{20}}\text{mol}\right)\frac{\text{44}\text{.01g}}{\text{1mol}}\\ \\ \text{=}\text{2}{\text{.63×10}}^{\text{18}}\text{g}\\ \\ \text{=2}{\text{.63×10}}^{\text{15}}\text{kg}(\text{1g}{\text{= 10}}^{\text{-3}}\text{kg)}\end{array}$