Chapter 6, Problem 7P

To determine

The mass of oxygen gas in the atmosphere.

Answer to Problem 7P

The mass of oxygen gas in the atmosphere is $\underset{\_}{1.2\times {10}^{18}\text{kg}}$.

Explanation of Solution

Given that the total mass of Earth’s atmosphere is $5.1\times {10}^{18}\text{kg}$, the percentage abundance of ${\text{N}}_2$, ${\text{O}}_2$, and $\text{Ar}$ in the atmosphere is $78\%$, $21\%$, and $1\%$ respectively.

Let $X$ be the number of oxygen molecules in the atmosphere. The atomic mass of ${\text{N}}_2$ is $14\times 1.66\times {10}^{-27}\text{kg}$, the atomic mass of ${\text{O}}_2$ is $16\times 1.66\times {10}^{-27}\text{kg}$, and the atomic mass of $\text{Ar}$ is $18\times 1.66\times {10}^{-27}\text{kg}$.

Write the expression for finding the molecular mass of ${\text{N}}_2$, ${\text{O}}_2$, and $\text{Ar}$ can be computed as,

  $\begin{array}{l}{M}_{{\text{N}}_2}=2\times 14\times 1.66\times {10}^{-27}\text{kg}\\ M_{{\text{O}}_2}=2\times 16\times 1.66\times {10}^{-27}\text{kg}\\ M_{\text{Ar}}=1\times 18\times 1.66\times {10}^{-27}\text{kg}\end{array}\}$        (I)

Since the percentage abundance of ${\text{N}}_2$, ${\text{O}}_2$, and $\text{Ar}$ in the atmosphere is $78\%$, $21\%$, and $1\%$ respectively, for each molecule of ${\text{O}}_2$,there will be $78/21$ ${\text{N}}_2$ molecules, and $1/21$ $\text{Ar}$ atoms.

Thus, the total mass of the atmosphere can e expressed as,

  $\begin{array}{c}{M}_{\text{total}}=\left(\frac{78}{21}X\right)M_{{\text{N}}_2}+\left(\frac{1}{21}X\right)M_{\text{Ar}}+\left(X\right)M_{{\text{O}}_2}\\ =X\left[\frac{78}{21}{M}_{{\text{N}}_2}+\frac{1}{21}{M}_{\text{Ar}}+M_{{\text{O}}_2}\right]\end{array}$        (II)

Solve equation (II) for $X$.

  $X=\frac{M_{\text{total}}}{\left[\frac{78}{21}{M}_{{\text{N}}_2}+\frac{1}{21}{M}_{\text{Ar}}+M_{{\text{O}}_2}\right]}$        (III)

The total mass of ${\text{O}}_2$ in the atmosphere can be then computed as,

  $M_{{\text{total O}}_2}=X{M}_{{\text{O}}_2}$        (IV)

Use equation (III) in (IV).

  $M_{{\text{total O}}_2}=\frac{M_{\text{total}}{M}_{{\text{O}}_2}}{\left[\frac{78}{21}{M}_{{\text{N}}_2}+\frac{1}{21}{M}_{\text{Ar}}+M_{{\text{O}}_2}\right]}$        (V)

Conclusion:

Substitute $5.1\times {10}^{18}\text{kg}$ for $M_{\text{total}}$, $2\times 14\times 1.66\times {10}^{-27}\text{kg}$ for $M_{{\text{N}}_2}$, $2\times 16\times 1.66\times {10}^{-27}\text{kg}$ for $M_{{\text{O}}_2}$, and $1\times 18\times 1.66\times {10}^{-27}\text{kg}$ for $M_{\text{Ar}}$ in equation (V) to find $M_{{\text{total O}}_2}$.

  $\begin{array}{c}{M}_{{\text{total O}}_2}=\frac{\left(5.1\times {10}^{18}\text{kg}\right)\left(2\times 16\times 1.66\times {10}^{-27}\text{kg}\right)}{\left[\frac{78}{21}\left(2\times 14\times 1.66\times {10}^{-27}\text{kg}\right)+\frac{1}{21}\left(1\times 18\times 1.66\times {10}^{-27}\text{kg}\right)+\left(2\times 16\times 1.66\times {10}^{-27}\text{kg}\right)\right]}\\ =1.19\times {10}^{18}\text{kg}\\ \approx 1.2\times {10}^{18}\text{kg}\end{array}$

Therefore, the mass of oxygen gas in the atmosphere is $\underset{\_}{1.2\times {10}^{18}\text{kg}}$.