Question

Gold is found in seawater at very low levels, about \(0.05 \mathrm{Ppb}\) by mass. Assuming that gold is worth about \(\$ 1300\) per troy ounce, how many liters of seawater would you have to process to obtain \(\$ 1,000,000\) worth of gold? Assume the density of seawater is \(1.03 \mathrm{~g} / \mathrm{mL}\) and that your gold recovery process is \(50 \%\) efficient.

Short answer

To obtain $1,000,000 worth of gold with a 50% efficient recovery process, you would need to process approximately \(9.292 \times 10^{11}\) liters of seawater.

Step-by-step solution

Convert gold concentration from ppb to mass ratio

The gold concentration in seawater is given as \(0.05 \mathrm{Ppb}\). To convert this to a mass ratio (grams of gold per gram of seawater), we use the conversion: 1 ppb = \(1 \times 10^{-9}\) mass ratio So, the gold concentration in the seawater is \(0.05 \times (1 \times 10^{-9}) = 5 \times 10^{-11} \frac{\text{g of gold}}{\text{g of seawater}}\).

Convert the value of gold per troy ounce to the value of gold per gram

We are given the value of gold as \(\$1300\) per troy ounce. There are approximately 31.1035 grams in a troy ounce. Therefore, the value of gold per gram will be \(\frac{\$1300}{31.1035 \text{ g}} \approx \$41.797 \frac{\text{USD}}{\text{g}}\).

Calculate the required gold mass to reach $1,000,000 worth

To find the mass of gold needed to reach $1,000,000 worth, we can use the value of gold per gram: \(\frac{\$1,000,000}{\$ 41.797 \frac{\text{USD}}{\text{g}}} \approx 23923.24 \text{ grams}\).

Account for the efficiency of the gold recovery process

The gold recovery process is 50% efficient, meaning that we would only recover half of the gold extracted from the seawater. Thus, we need to double the amount of gold that needs to be extracted: \(23923.24 \text{ grams} \times 2 = 47846.48 \text{ grams}\).

Calculate the volume of seawater needed to process

We know the gold concentration in seawater, the mass of gold that needs to be extracted, and the density of seawater. Now, we can calculate the mass of seawater needed using the mass ratio (gold concentration): \(\frac{47846.48 \text{ grams of gold}}{5 \times 10^{-11} \frac{\text{g of gold}}{\text{g of seawater}}} \approx 9.569 \times 10^{14} \text{ grams of seawater}\). Now we will convert the mass of seawater to volume using the density of seawater, \(1.03 \frac{\text{g}}{\mathrm{mL}}\): \(\frac{9.569 \times 10^{14} \text{ grams}}{1.03 \frac{\text{g}}{\mathrm{mL}}} \approx 9.292 \times 10^{14} \mathrm{mL}\). Finally, we will convert milliliters to liters: \(9.292 \times 10^{14} \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 9.292 \times 10^{11} \mathrm{L}\). Therefore, you would need to process approximately \(9.292 \times 10^{11}\) liters of seawater to obtain $\$1,000,000 worth of gold with a 50% efficient recovery process.

Key concepts

Concentration Conversion

Understanding concentration conversion is crucial when dealing with substances diluted in large quantities, like gold in seawater. Concentration is usually denoted in parts per billion (ppb), indicating how many units of a substance are contained in a billion units of a mixture. To solve problems involving extracting substances at such low concentrations, converting ppb to a mass ratio is a necessary step. This means calculating how many grams of the substance (in this case, gold) are present in one gram of the mixture (seawater).

For gold, the concentration in seawater is given as 0.05 ppb, which implies that for every billion grams of seawater, there are 0.05 grams of gold. The conversion to a mass ratio is given by multiplying 0.05 by the factor of 10^-9 (since 1 ppb equals 10^-9). Therefore, the mass ratio for gold in seawater is calculated to be a mere 5 x 10^-11, underscoring the minuscule amount of gold dissolved in the ocean's vastness.

Value of Gold Per Gram

The economic feasibility of extracting gold from seawater largely hinges on the current market value of gold. The price of gold can vary significantly, influencing whether gold recovery is worthwhile. In exercises like this, we're often given the value of gold per troy ounce, a common unit of measure for precious metals. However, to calculate recovery amounts and needed volumes, we need to know the value per gram.

By dividing the value per troy ounce by the number of grams in a troy ounce (approximately 31.1035 grams), we can find the corresponding value per gram. For instance, with gold valued at \(1300 per troy ounce, this translates to roughly \)41.797 per gram, allowing us to determine how much gold needs to be extracted to reach a certain monetary goal.

Gold Recovery Process Efficiency

Gold recovery process efficiency is a measure of how well a process can extract gold from a material, in this case, from seawater. It is represented as a percentage and affects the total amount of gold that needs to be extracted to achieve a certain yield. If the efficiency is 50%, this implies that half of the gold that goes through the recovery process will be lost due to imperfections in the method.

This loss must be factored into our calculations. Therefore, if the goal is to obtain a particular value of gold, we need to account for this inefficiency by targeting to extract double the calculated amount of gold. This way, even after losing half, you would still end up with the desired quantity.

Seawater Density and Volume Calculation

Calculating the volume of seawater needed for gold extraction is the final step, involving two key parameters: seawater density and the amount of seawater required (which in turn depends on the gold concentration and the total amount of gold we aim to extract). Seawater density is commonly measured in grams per milliliter (g/mL), with an average density of about 1.03 g/mL. This means that for every milliliter of seawater, there are 1.03 grams of seawater.

We use this density value to convert the total mass of seawater necessary for the gold extraction into volume. Once the mass is calculated using the mass ratio from the concentration of gold, the volume can be found by dividing the mass by the density, resulting in a value expressed in milliliters. This value is then converted to liters to obtain a more manageable number, essential for practical considerations in large-scale operations like extracting gold from seawater.