Question

Earth's surface area is \(5.10 \times 10^{8} \mathrm{~km}^{2} ;\) its crust has a mean thickness of \(35 \mathrm{~km}\) and a mean density of \(2.8 \mathrm{~g} / \mathrm{cm}^{3}\). The two most abundant elements in the crust are oxygen \(\left(4.55 \times 10^{5} \mathrm{~g} / \mathrm{t},\right.\) where \(\mathrm{t}\) stands for "metric ton"; \(1 \mathrm{t}=1000 \mathrm{~kg}\) ) and silicon ( \(2.72 \times 10^{5} \mathrm{~g} / \mathrm{t}\) ), and the two rarest nonradioactive elements are ruthenium and rhodium, each with an abundance of \(1 \times 10^{-4} \mathrm{~g} / \mathrm{L}\). What is the total mass of each of these elements in Earth's crust?

Short answer

Total mass: Oxygen - 2.275 × 10^{25} g, Silicon - 1.36 × 10^{25} g, Ruthenium and Rhodium - 1.785 × 10^{18} g.

Step-by-step solution

Calculate the Volume of the Crust

To find the volume of Earth's crust, use the formula for the volume of a cylinder \[ V = \text{Area} \times \text{Thickness} \] Given: - Earth's surface area \(A = 5.10 \times 10^{8} \, \text{km}^2\)- Mean thickness of the crust \(T = 35 \, \text{km}\)Thus, \[ V = 5.10 \times 10^{8} \times 35 = 1.785 \times 10^{10} \, \text{km}^3 \]Since 1 \( \text{km}^3 = 10^{15} \text{cm}^3\), then \[ V = 1.785 \times 10^{10} \times 10^{15} \, \text{cm}^3 = 1.785 \times 10^{25} \, \text{cm}^3 \]

Calculate the Mass of the Crust

Use the volume of the crust and the mean density to find its mass.Given:- Volume of the crust \(V = 1.785 \times 10^{25} \, \text{cm}^3\)- Mean density \(\rho = 2.8 \, \text{g/cm}^3\)The formula for mass is \[ M = \rho \times V \]Thus, \[ M = 2.8 \times 1.785 \times 10^{25} = 5.0 \times 10^{25} \, \text{g} \]

Calculate the Mass of Oxygen in the Crust

Given the abundance of oxygen: \[ 4.55 \times 10^{5} \, \text{g/t}\]Convert tons to grams:1 ton = 1000 kg = 1,000,000 g Thus, the fraction of the crust composed of oxygen is \[ \frac{4.55 \times 10^{5}}{1,000,000} = 0.455 \]Then,Mass of oxygen \[ M_{O} = 0.455 \times 5.0 \times 10^{25} = 2.275 \times 10^{25} \, \text{g} \]

Calculate the Mass of Silicon in the Crust

Given the abundance of silicon: \[ 2.72 \times 10^{5} \, \text{g/t}\]Convert tons to grams:1 ton = 1000 kg = 1,000,000 g Thus, the fraction of the crust composed of silicon is \[ \frac{2.72 \times 10^{5}}{1,000,000} = 0.272 \]Then,Mass of silicon \[ M_{Si} = 0.272 \times 5.0 \times 10^{25} = 1.36 \times 10^{25} \, \text{g} \]

Calculate the Mass of Ruthenium and Rhodium in the Crust

Given the abundance of ruthenium and rhodium: \[ 1 \times 10^{-4} \, \text{g/L} \]Convert \(cm^3\) to liters:1 liter = 1000 \(cm^3\)Thus, the fraction of the crust composed of ruthenium or rhodium is\[ \frac{1 \times 10^{-4}}{1000} = 1 \times 10^{-7} \]Then,Mass of ruthenium or rhodium \[ M_{Ru/Rh} = 1 \times 10^{-7} \times 1.785 \times 10^{25} = 1.785 \times 10^{18} \, \text{g} \]

Key concepts

Earth's Surface Area Calculation

To understand the Earth's crust composition, we first need to calculate its surface area. The surface area represents the total area covered by the Earth's crust. Imagine Earth as a big ball. The surface area is the outer layer that you would paint if you had to cover the entire ball.

Volume Calculation

After finding the surface area, we calculate the volume of the Earth's crust. To do this, we use the formula for the volume of a cylinder because the Earth's crust is like a thin shell around the planet. The formula is:
\[ V = \text{Area} \times \text{Thickness} \]
Given the Earth's surface area as \[5.10 \times 10^8 \text{km}^2 \] and the crust thickness as 35 km, we find:
\[ V = 5.10 \times 10^8 \text{km}^2 \times 35 \text{km} = 1.785 \times 10^{10} \text{km}^3 \]
Energy conversion helps us to change km³ into cm³ for easier handling:
\[ 1 \text{km}^3 = 10^{15} \text{cm}^3 \]
Thus, the volume becomes:
\[ V = 1.785 \times 10^{10} \text{km}^3 \times 10^{15} \text{cm}^3 = 1.785 \times 10^{25} \text{cm}^3 \]
We now have the volume of the Earth's crust in cm³.

Mass Calculation

With the volume known, we can calculate the mass of the Earth's crust. Using the formula:
\[ M = \rho \times V \]
where \[ \rho \] is the mean density and \[ V \] is the volume, we substitute the values given:
Given, \[ \rho = 2.8 \text{g/cm}^3 \ and V = 1.785 \times 10^{25} \text{cm}^3 \]
Thus,
\[ M = 2.8 \times 1.785 \times 10^{25} = 5.0 \times 10^{25} \text{g} \]
We now know the total mass of the Earth's crust.

Elemental Abundance

The next step is to understand which elements are abundant in the Earth's crust. The two most abundant elements are oxygen and silicon. The given abundances of oxygen and silicon in a metric ton of the crust are:

• Oxygen: \[ 4.55 \times 10^{5} \text{g/t} \]
• Silicon: \[ 2.72 \times 10^{5} \text{g/t} \]
  To find the fraction of these elements, we convert tons to grams. As 1 ton equals 1,000,000 grams, the fraction of oxygen and silicon in the crust is:
  
  • Oxygen: \[ \frac{4.55 \times 10^{5}}{1,000,000} = 0.455 \]
  • Silicon: \[ \frac{2.72 \times 10^5}{1,000,000} = 0.272 \]
  Next, calculate their masses:
  • Mass of oxygen:
    \[ M_{O} = 0.455 \times 5.0 \times 10^{25} = 2.275 \times 10^{25} \text{g} \]
  • Mass of silicon:
    \[ M_{Si} = 0.272 \times 5.0 \times 10^{25} = 1.36 \times 10^{25} \text{g} \]
  This gives us the mass of oxygen and silicon within the Earth's crust.

Density

Finally, let's understand the role of density in these calculations. Density helps us determine how much mass is in a given volume. For the Earth's crust, the given mean density is:
\[ \rho = 2.8 \text{g/cm}^3 \]
This means that each cubic centimeter of the Earth's crust weighs 2.8 grams. Knowing the density makes it easier to convert between volume and mass, essential for calculating the mass of entire regions such as the Earth's crust. For rarer elements like ruthenium and rhodium, understanding their density helps calculate their total mass using their given abundance:

• Mass fraction:
  \[ \frac{1 \times 10^{-4} \text{g/L}}{1000} = 1 \times 10^{-7} \text{g/cm}^3 \]
• Mass of ruthenium or rhodium:
  \[ M_{Ru/Rh} = 1 \times 10^{-7} \times 1.785 \times 10^{25} = 1.785 \times 10^{18} \text{g} \]

This concludes our understanding of Earth's crust composition through surface area, volume, mass, elemental abundance, and density calculations.