Question

The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given that (1) the amount of the reparations was \(\$ 28.8\) billion dollars, (2) the value of gold at the time was about \(\$ 21.25\) per troy ounce ( \(1 \text { troy ounce }=31.103 \mathrm{g}),\) and (3) gold occurs in seawater to the extent of \(4.67 \times 10^{17}\) atoms per ton of seawater \((1 \text { ton }=2000\) lb), how many cubic kilometers of seawater would have had to be processed to obtain the required amount of gold? Assume that the density of seawater is \(1.03 \mathrm{g} / \mathrm{cm}^{3}\) (Haber's scheme proved to be commercially infeasible, and the reparations were never fully paid.)

Short answer

The resulting calculation will give the total cubic kilometers of seawater required. This will be a large number, showing the infeasibility of Haber's idea to use seawater to pay for Germany's reparations after World War I.

Step-by-step solution

Determine the total weight of gold required

First, find out the total weight of gold required to pay off the reparations. This can be done by dividing the total reparations by the price of gold per troy ounce, then converting that to grams: \[ Weight = \frac{\$28.8 billion}{\$21.25 per troy ounce}\times 31.103g per troy ounce \]

Find the number of atoms of gold required

Next, use Avogadro's Law to convert the weight from grams to atoms. Avogadro's number (\(6.022 \times 10^{23}\)) represents the number of atoms in 1 mole, which weighs the atomic weight in grams: \[ Atoms = Weight(g) \times \frac{1 mole}{197.0g} \times \frac{6.022 \times 10^{23} atoms}{1 mole} = Weight \times \frac{6.022 \times 10^{23}}{197.0} \] We use \(197.0 g\) as it is the atomic weight of gold.

Calculate the total tons of seawater required

To find the total tons of seawater required, divide the total number of atoms required by the number of atoms per ton of seawater: \[ Tons = \frac{Atoms}{4.67 \times 10^{17} atoms per ton} \]

Convert tons of seawater to cubic kilometers

Finally, convert tons to kilograms, then kilograms to cubic centimeters (using the density of seawater), and finally cubic centimeters to cubic kilometers: \[ Cubic \, kilometers = \, Tons \times \frac{1000 \, kg}{1 ton} \times \frac{1 \, cm^3}{1.03 g} \times \frac{1 \, m^3}{10^6 \, cm^3} \times \frac{1 \, km^3}{10^9 \, m^3} \]

Key concepts

Avogadro's Law

One of the fascinating aspects of chemistry is understanding how seemingly minuscule particles, like atoms, are counted in such enormous quantities. Avogadro's Law is a critical part of this mystery. It says that equal volumes of gases, at the same temperature and pressure, contain an equal number of particles. This relationship is key to translating a macroscopic amount of a substance into the enormous number of atoms or molecules it contains.

In chemistry, Avogadro's number, which is approximately \(6.022 \times 10^{23}\), represents the number of atoms or molecules in a mole of a substance. This number is fundamental when you need to switch between measuring mass in the lab to counting atoms or molecules, as done in calculations involving gold and seawater. For example, if you know the atomic weight of gold, which is \(197.0 \text{ g/mol}\), you can use Avogadro's Law to determine how many atoms are in a given mass of gold by utilizing the following calculation:

\[ \text{Atoms} = \text{Weight (g)} \times \frac{6.022 \times 10^{23}}{197.0}\]

This conversion is crucial for determining the sheer amount of gold required for large tasks, such as extracting it from seawater! Understanding Avogadro's Law helps us link the visible world with the microscopic structures that compose it.

atomic weight

Atomic weight, also known as atomic mass, is the average mass of atoms of an element, calculated using the relative abundance of isotopes. It gives us the weight of a mole of atoms for a given element, indicating how heavy one atom is when compared to another element's atoms on average.

For gold, the atomic weight we typically use is \(197.0 \text{ g/mol}\). This means one mole of gold, a large number of \(6.022 \times 10^{23}\) atoms, weighs 197 grams. When dealing with chemical reactions or transformations, like calculating how much gold is present based on the number of atoms, the atomic weight becomes crucial. This mass allows chemists to convert easily between moles and grams, which is necessary when weighing how much of a substance you have or need.

This principle is central when converting the mass of gold extracted from seawater into an atom count or understanding the scale of industrial processes involving elements. By applying the atomic weight in calculations, chemists can precisely determine how much substance is involved.

density of seawater

The density of seawater is an interesting property that showcases its unique composition. Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm³). For seawater, this density is approximately \(1.03 \text{ g/cm}^3\).

This measure takes into account not just water, but the dissolved salts and other materials in it. This density becomes a crucial factor when converting between mass and volume in calculations, such as determining the volume of seawater needed to obtain a certain mass of gold.

To convert tons of seawater into volume, a chemist needs to know the density. Here's a basic conversion: multiply the tons by 1000 to get kilograms, then use the density to find the volume in cubic centimeters, and finally convert to cubic kilometers. This is represented in the calculation:

\[ \text{Cubic kilometers} = \text{Tons} \times \frac{1000 \text{ kg}}{1 \text{ ton}} \times \frac{1 \text{ cm}^3}{1.03 \text{ g}} \times \frac{1 \text{ m}^3}{10^6 \text{ cm}^3} \times \frac{1 \text{ km}^3}{10^9 \text{ m}^3} \]

This allows chemists and engineers to plan effectively when working on large-scale projects involving water resources, ensuring that the required quantities can be accurately measured and managed.