Problem 143 - Chemistry The Molecular Science

Question

The present average concentration (mass percent) of magnesium ions in seawater is \(0.13 \%\). A chemistry textbook estimates that if \(1.00 \times 10^{8}\) tons \(\mathrm{Mg}\) were taken out of the sea each year, it would take one million years for the Mg concentration to drop to \(0.12 \% .\) Do sufficient calculations to either verify or refute this statement. Assume that Earth is a sphere with a diameter of \(8000 \mathrm{mi}, 67 \%\) of which is covered by oceans to a depth of \(1 \mathrm{mi}\), and that no \(\mathrm{Mg}\) is washed back into the oceans at any time.

Step-by-Step Solution

Answer

The statement is verified; it would take about one million years.

Step 1: Calculate the Volume of the Oceans

First, we need to calculate the volume of the oceans. The Earth can be treated as a sphere with a diameter of 8000 miles. The formula for the volume of a sphere is \[ V = \frac{4}{3} \pi r^3 \], where \( r \) is the radius of the sphere. The radius of Earth is \( 4000 \text{ miles} \) (half of 8000 miles). The oceans cover 67% of the Earth and have a depth of 1 mile. So, we approximate the volume of the oceans as \( 0.67 \times \pi \times (4000^2) \times 1 \). Convert the volume from cubic miles to cubic feet since 1 mile = 5280 feet.

Step 2: Calculate the Total Mass of Magnesium in Oceans

To calculate the total mass of magnesium in the oceans, we need the density of seawater \( (\approx 63.9 \text{ lb/ft}^3) \). The equation for mass is \( \text{Density} \times \text{Volume} \). Given the magnesium concentration is 0.13%, we multiply the total mass of seawater by 0.013 to find the mass of magnesium.

Step 3: Determine Annual Decrease of Magnesium Concentration

Calculate how much magnesium is removed each year: \( 1.00 \times 10^8 \text{ tons} \). Convert tons to pounds (1 ton = 2000 pounds). Subtract this annual removal from the initial mass to find how much the mass of magnesium decreases in one year.

Step 4: Calculate Time for Desired Concentration Drop

We aim to reduce the magnesium concentration from 0.13% to 0.12%. Calculate the final desired mass of magnesium using the 0.12% concentration. Determine the annual decrease needed and calculate how many years it would take to achieve this decrease.

Key Concepts

Ocean Volume Calculation

To determine how much magnesium is present in the ocean, we need to first comprehend its vastness. One crucial step is the calculation of ocean volume. We proceed by considering the Earth as a sphere—a common simplification that allows us to use straightforward mathematical formulas. With a given diameter of 8000 miles, we obtain the Earth's radius as half of that, equalling 4000 miles. This radius is essential because it plugs into the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \]Since the oceans cover about 67% of Earth’s surface and average a depth of 1 mile, we only calculate this portion of the sphere’s volume. We simplify the volume calculation by multiplying 0.67 with the surface area of a spherical segment (base area times depth) as:\[ \text{Surface Area} \times \text{Depth} = \pi \times (4000^2) \times 1 \]Don't forget to convert cubic miles to cubic feet, as this matches with the density unit of seawater measurements. Remember, calculations of cumbersome spheres are simplified when considering only the relevant usable portions."

Magnesium Concentration

To tackle the problem of magnesium concentration in seawater, let's break it down. Magnesium ions constitute approximately 0.13% of seawater's mass. This percentage might seem tiny, but given the ocean's sheer volume, it accounts for a massive amount of magnesium.To find the total magnesium, you calculate the entire mass of seawater using its volume and density, then multiply by 0.13%. Density helps link volume to mass via the formula:\[ \text{Mass} = \text{Density} \times \text{Volume} \]Because we're given a density of roughly 63.9 lb/ft³, the calculation converts this gigantic water volume into a formidable weight, and 0.13% of that tells us how heavy the magnesium portion is. This conversion transforms a seemingly minuscule percentage into a remarkably significant figure. Understanding this helps clarify how such small concentrations can still lead to significant quantities.

Spherical Earth Model

The idea of using a spherical model for Earth simplifies our calculations, despite Earth's slight oblate shape. This model assumes symmetry and uniformity, making it easier to apply standard geometric formulas. Using this simplification, calculating surface and volume becomes predictable with basic math equations. The model necessitates assumptions like uniform depth for oceans covering a predefined surface area percentage (67% here). While these are simplifications, they help reach practical approximations for problems involving vast planetary scales. Keep in mind, the diameter considered is 8000 miles, setting the radius for spherical calculations at 4000 miles, which ties back into our earlier volume calculations.

Density of Seawater

Density is pivotal when relating volumes to masses. For seawater, this is essential due to its mixed components, which vary its density slightly from pure water. Here, we use an average seawater density of approximately 63.9 lb/ft³.This figure is critical when finding mass from volume, as shown by the equation:\[ \text{Mass} = \text{Density} \times \text{Volume} \]It's important to utilize this density to convert the volumetric data to a mass measurement, allowing further calculations about magnesium content. The knowledge of density allows you to see how a purely spatial measurement translates into a weight measurement, essential for understanding chemical concentrations across vast oceanic expanses. Understanding density also means grasping how it varies locally in oceans due to changes in salinity and temperature, offering a richer context to calculations based on average figures.