Let's solve the problem step by step: --- **Step 1: Calculate the mass of oxygen in the Earth's crust** - The crust is made of sanidine (\(KAlSi_3O_8\)), density = \(2.5 \, \text{g/cm}^3 = 2500 \, \text{kg/m}^3\) - Crust thickness = \(50.0 \, \text{km} = 50,000 \, \text{m}\) - Earth's radius = \(6371 \, \text{km} = 6,371,000 \, \text{m}\) - Volume of crust \(V_{\text{crust}} = \frac{4}{3}\pi (R_o^3 - R_i^3)\) - \(R_o = 6,371,000\,\text{m}\) (outer radius) - \(R_i = 6,371,000 - 50,000 = 6,321,000\,\text{m}\) (inner radius) \[ V_{\text{crust}} = \frac{4}{3} \pi [(6,371,000)^3 - (6,321,000)^3] \, \text{m}^3 \] \[ V_{\text{crust}} \approx \frac{4}{3} \pi [2.59 \times 10^{20} - 2.53 \times 10^{20}] \] \[ V_{\text{crust}} \approx \frac{4}{3} \pi (0.06 \times 10^{20}) \approx 0.251 \times 10^{20} \, \text{m}^3 \] - Mass of crust \(= \text{density} \times \text{volume} = 2500 \times 0.251 \times 10^{20} = 0.6275 \times 10^{23} \, \text{kg}\) - Sanidine (\(KAlSi_3O_8\)): Molar mass = 39 (K) + 27 (Al) + 3×28 (Si) + 8×16 (O) = 39 + 27 + 84 + 128 = **278 g/mol** - Mass fraction of O: \( \frac{128}{278} \approx 0.460 \) - Mass of oxygen in crust: \[ m_{\text{O,crust}} = 0.460 \times 0.6275 \times 10^{23} = 0.288 \times 10^{23} \, \text{kg} \] --- **Step 2: Calculate the mass of oxygen in the Earth's atmosphere** - Atmosphere thickness = \(100 \, \text{km} = 100,000 \, \text{m}\) - Density = \(0.99 \, \text{kg/m}^3\) - Outer radius for atmosphere: \(6,371,000 + 100,000 = 6,471,000\,\text{m}\) - Volume of atmosphere: \[ V_{\text{atm}} = \frac{4}{3}\pi \left[(6,471,000)^3 - (6,371,000)^3\right] \] \[ (6,471,000)^3 \approx 2.71 \times 10^{20}, \quad (6,371,000)^3 \approx 2.59 \times 10^{20} \] \[ \Delta = 0.12 \times 10^{20} \] \[ V_{\text{atm}} = \frac{4}{3}\pi (0.12 \times 10^{20}) \approx 0.503 \times 10^{20} \, \text{m}^3 \] - Mass of atmosphere \(= 0.99 \times 0.503 \times 10^{20} = 0.498 \times 10^{20} \, \text{kg}\) - Oxygen = \(23.1\%\) by mass \[ m_{\text{O,atm}} = 0.231 \times 0.498 \times 10^{20} = 0.115 \times 10^{20} \, \text{kg} \] --- **Step 3: Calculate the ratio** \[ \text{Ratio} = \frac{m_{\text{O,atm}}}{m_{\text{O,crust}}} = \frac{0.115 \times 10^{20}}{0.288 \times 10^{23}} \] \[ = \frac{0.115}{0.288} \times 10^{-3} \] \[ \approx 0.399 \times 10^{-3} \] \[ = 3.99 \times 10^{-4} \] **Final Answer (rounded to two significant digits):** \[ \boxed{4.0 \times 10^{-4}} \] --- **Summary of steps:** 1. Found the mass of oxygen in the crust using density, volume, and stoichiometry. 2. Found the mass of oxygen in the atmosphere using density, volume, and given percentage. 3. Divided the two masses to get the final ratio.

The problem involves calculating and comparing the amounts of oxygen stored in Earth's crust and atmosphere by first determining their respective volumes based on geometric modeling of Earth's layers, then converting these volumes to masses using density values, and finally applying chemical composition data to find the oxygen content. This process uses concepts of volume calculation for spherical shells, density-mass relationships, and proportional reasoning based on chemical composition to arrive at the ratio of oxygen mass in the atmosphere to that in the crust.