Question

were told oxygen is the most common element in the earths solid crust. but how much oxygen is there? lets compare to the amount of oxygen in the earths atmosphere, like this: the most common minerals in the earths crust are feldspars, and albite (naalsi3o8) is a typical feldspar. lets assume the entire crust is made of albite, with a density of 2.6 g/cm³. the radius of the earth is 6371 km and lets say the crust is the upper 50.0 km of it. lets model the earths atmosphere as a layer on top of the earth about 100. km thick with an average density of 0.99 g/m³, and which is 23.1% oxygen by mass. use this information to calculate the ratio of the mass of oxygen in the earths atmosphere to the mass of oxygen in the earths crust. round your answer to 2 significant digits. incorrect your answer is wrong. in addition to checking your math, check that you used the right data and did not round any intermediate calculations. 3.8×10⁻⁴

Explanation:

Step1: Calculate the volume of the Earth's crust

The Earth's crust is considered as a spherical - shell. The volume formula for a spherical - shell is $V=\frac{4}{3}\pi((R + h)^3 - R^3)$, where $R = 6371\times10^5$ cm is the radius of the Earth and $h=50.0\times10^5$ cm is the thickness of the crust.
However, since $h\ll R$, we can use the approximation $V\approx4\pi R^{2}h$.
$V = 4\pi(6371\times10^{5})^{2}(50.0\times10^{5})$
$V=4\pi\times6371^{2}\times10^{10}\times50.0\times10^{5}$
$V = 4\pi\times6371^{2}\times50.0\times10^{15}$
$V\approx4\times3.14\times(6.371\times10^{3})^{2}\times50.0\times10^{15}$
$V\approx4\times3.14\times40.59\times10^{6}\times50.0\times10^{15}$
$V\approx4\times3.14\times40.59\times50.0\times10^{21}$
$V\approx2.55\times10^{24}$ cm³.

Step2: Calculate the mass of the Earth's crust

Given the density of the crust $
ho = 2.6$ g/cm³, the mass of the crust $m_{crust}=
ho V$.
$m_{crust}=2.6\times2.55\times10^{24}$
$m_{crust}\approx6.63\times10^{24}$ g.

Step3: Calculate the mass of oxygen in the Earth's crust

Since the crust is 23.1% oxygen by mass, the mass of oxygen in the crust $m_{O - crust}=0.231\times m_{crust}$
$m_{O - crust}=0.231\times6.63\times10^{24}$
$m_{O - crust}\approx1.53\times10^{24}$ g.

Step4: Calculate the mass of the Earth's atmosphere

The atmosphere is modeled as a layer of thickness $h_{atm}=100\times10^{5}$ cm and average density $
ho_{atm}=0.99$ g/m³$ = 0.99\times10^{- 3}$ g/cm³. Using the approximation for the volume of the spherical - shell for the atmosphere $V_{atm}\approx4\pi R^{2}h_{atm}$, where $R = 6371\times10^{5}$ cm.
$V_{atm}\approx4\times3.14\times(6371\times10^{5})^{2}\times(100\times10^{5})$
$V_{atm}\approx4\times3.14\times40.59\times10^{16}\times100\times10^{5}$
$V_{atm}\approx5.1\times10^{24}$ cm³.
The mass of the atmosphere $m_{atm}=
ho_{atm}V_{atm}$
$m_{atm}=0.99\times10^{-3}\times5.1\times10^{24}$
$m_{atm}\approx5.05\times10^{21}$ g.

Assuming the atmosphere is 23.1% oxygen (a simplification, actual oxygen percentage in dry air is about 21%, but using the given 23.1% for consistency), the mass of oxygen in the atmosphere $m_{O - atm}=0.231\times m_{atm}$
$m_{O - atm}=0.231\times5.05\times10^{21}$
$m_{O - atm}\approx1.17\times10^{21}$ g.

Step5: Calculate the ratio of the mass of oxygen in the atmosphere to the mass of oxygen in the crust

The ratio $r=\frac{m_{O - atm}}{m_{O - crust}}$
$r=\frac{1.17\times10^{21}}{1.53\times10^{24}}$
$r\approx7.6\times10^{-4}$

Answer:

$7.6\times10^{-4}$