Problem 116 - Chemistry The Molecular Science

Question

Fritz Haber, a German chemist, proposed extracting gold from seawater as a way to pay off Germany's debt, \(\$ 28.8 \times 10^{6}\), after World War I. The value of gold at the time was \(\$ 21.25 /\) troy oz \((1\) troy \(o z=31.103 \mathrm{~g})\). The gold concentration in seawater is \(0.15 \mathrm{mg}\) gold/ton seawater \((1\) ton \(=2000 \mathrm{lb}\) ). Assume the density of seawater is \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) (a) Calculate the volume (in cubic kilometers) of seawater that would have had to be processed to obtain the required mass of gold. (b) By comparison, an Olympic-sized swimming pool is \(50 \mathrm{~m} \times 25 \mathrm{~m} \times 2.0 \mathrm{~m} .\) Calculate the number of Olympic-sized swimming pools required to hold the volume of seawater needed in part (a).

Step-by-Step Solution

Answer

(a) 247.047 km³; (b) 98,818,800 Olympic pools.

Step 1: Calculate Required Mass of Gold in Grams

First, we need to determine the total mass of gold required to pay off the debt in grams. We start by finding the total ounces of gold needed:\[\frac{28.8 \times 10^6 \text{ dollars}}{21.25 \text{ dollars/oz}} = \text{ounces of gold needed}\]By solving this, we find:\[\text{ounces of gold needed} = 1,353,882 \text{ ounces}\]Now, convert this mass from ounces to grams using the conversion factor:\[1,353,882 \text{ ounces} \times 31.103 \text{ g/oz} = 42,096,700 \text{ grams of gold}\]

Step 2: Calculate Volume of Seawater Needed

Calculate the volume of seawater needed to obtain 42,096,700 grams of gold. The concentration of gold in seawater is given as 0.15 mg/ton. First, convert this concentration to grams:\[0.15 \text{ mg/ton} = 0.00015 \text{ grams/ton}\]Then calculate the number of tons of seawater needed:\[\frac{42,096,700 \text{ grams of gold}}{0.00015 \text{ grams/ton}} = 280,644,666,667 \text{ tons of seawater}\]To find the cubic kilometers, convert tons to kilograms (since 1 ton = 907.185 kg) and use the density to find volume:\[280,644,666,667 \text{ tons} \times 907.185 \text{ kg/ton} = 2.54408 \times 10^{14} \text{ kg}\]Now using the density \(1.03 \text{ g/cm}^3\) or \(1030 \text{ kg/m}^3\), calculate the volume:\[\frac{2.54408 \times 10^{14} \text{ kg}}{1030 \text{ kg/m}^3} = 2.47047 \times 10^{11} \text{ m}^3\]Finally convert this volume to cubic kilometers:\[2.47047 \times 10^{11} \text{ m}^3 \times \frac{1 \text{ km}^3}{10^9 \text{ m}^3} = 247.047 \text{ km}^3\]

Step 3: Calculate the Number of Olympic-sized Swimming Pools

An Olympic-sized swimming pool has a volume of \(50 \text{ m} \times 25 \text{ m} \times 2.0 \text{ m} = 2500 \text{ m}^3\).To find the number of pools required:\[\frac{2.47047 \times 10^{11} \text{ m}^3}{2500 \text{ m}^3/pool} = 98,818,800 \text{ pools}\]

Key Concepts

Density and Volume Conversion

Density and volume conversion are fundamental principles in solving chemistry problems, especially when dealing with liquids. Density is defined as mass per unit volume and is generally measured in grams per cubic centimeter (g/cm³). For seawater, it is assumed to be 1.03 g/cm³, meaning each cubic centimeter of seawater weighs 1.03 grams. This piece of information is crucial when converting mass to volume or vice versa. In exercises like these, where you're asked to find the volume of a given mass, knowing the density allows you to use the formula:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]This formula lets you compute the seawater volume needed to obtain a specified mass of a solute (like gold in this case). Given the identity "1 m³ = 10^6 cm³", you can shift between different units of measurement, converting between meters, centimeters, and even kilometers for broader comprehension and comparison.

Concentration Calculations

Concentration calculations help determine how much of a substance is present in a given volume of solution. In our example, we deal with the concentration of gold in seawater, given as 0.15 milligrams per ton of seawater. This means every ton contains 0.15 mg of gold.To proceed with concentration calculations, you need to:

• Convert units of concentration to a consistent mass unit. Here, 0.15 mg is converted to 0.00015 grams per ton for simplicity.
• Determine the total mass of gold required (from previous calculations).
• Use the formula:\[ \text{Total tons of seawater} = \frac{\text{Total mass of gold required}}{\text{Gold concentration per ton}} \]

Breaking down each step ensures clarity and prevents mistakes while calculating how much seawater you'd need to process to extract the desired amount of gold.

Historical Scientific Proposals

The concept of extracting resources from natural sources is not new in scientific history. Fritz Haber's notion of mining gold from seawater post-World War I was not just ambitious but a reflection of the scientific curiosity and economic drive of that era. Haber, a Nobel laureate, was known for synthesizing ammonia from nitrogen and hydrogen gases, aiding agriculture massively. His idea to extract gold signifies the attempts by scientists to solve economic and technological challenges through chemistry. Although ultimately impractical (as calculations show vast volumes of seawater and tons of effort required against minute returns), this chemical endeavor illustrates the innovative spirit of historical scientific proposals. Such initiatives often faced limitations but provided a basis for future technological advancements and socio-economic discussions.