Problem 42 - Chemistry The Central Science

Question

Gold is found in seawater at very low levels, about 0.05 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 to prosity of seawater is 1.03 \(\mathrm{g} / \mathrm{mL}\) and that your gold recovery process is 50\(\%\) efficient.

Step-by-Step Solution

Answer

You would need to process approximately 934,579,439 liters of seawater to obtain $1,000,000 worth of gold, considering the given conditions of 0.05 ppb gold concentration, $1300 per troy ounce gold worth, 1.03 g/mL seawater prosity, and 50% gold recovery process efficiency.

Step 1: Convert ppb to mass fraction

First, we need to convert the gold concentration in seawater from parts per billion (ppb) to mass fraction. 1 ppb means 1 µg gold per 1g of seawater, so 0.05 ppb = 0.05 µg of gold per 1g of seawater. Hence the mass fraction would be: \(0.05 \times 10^{-6}\)

Step 2: Calculate the mass of gold in 1 liter of seawater

Next, we need to find the mass of gold in 1 liter of seawater. We know that the prosity of seawater is 1.03 g/mL, meaning 1 liter (1000 mL) of seawater has a mass of 1000 mL × 1.03 g/mL = 1030 g. Multiplying the mass of seawater by the mass fraction of gold: \(1030\,\text{g}\, \times 0.05 \times 10^{-6}\, = 5.15 \times 10^{-5}\,\text{g}\)

Step 3: Convert grams to troy ounces

Since the worth of gold is given in USD per troy ounce, we have to convert the mass of gold from grams to troy ounces.1 troy ounce is approximately equal to 31.1 grams. So, we convert the gold mass as follows: \((5.15 \times 10^{-5} \text{ g}) \times \frac{1\, \text{troy oz}}{31.1\,\text{g}} = 1.65 \times 10^{-6}\,\text{troy oz}\)

Step 4: Calculate the worth of gold in 1 liter of seawater

Now, we can find the worth of gold in 1 liter of seawater by multiplying the mass of gold in troy ounces by the worth of gold per troy ounce (1300 USD/troy oz): \(1.65 \times 10^{-6} \,\text{troy oz}\, \times 1300\,\text{USD/troy oz} = 0.00214\,\text{USD}\)

Step 5: Adjust for gold recovery process efficiency

Since the gold recovery process is only 50% efficient, we have to adjust the worth of gold that we can obtain from 1 liter of seawater: \(0.00214\,\text{USD} \times 50\% = 0.00107\,\text{USD}\)

Step 6: Calculate the number of liters to process

Finally, we can find the number of liters of seawater we need to process to obtain $1,000,000 worth of gold by dividing the required worth by the worth of gold in 1 liter of seawater, adjusted for recovery process efficiency: \(\frac{1,000,000\,\text{USD}}{0.00107\,\text{USD}} = 934,579,439\,\text{liters}\) So, you would need to process approximately 934,579,439 liters of seawater to obtain $1,000,000 worth of gold, considering the given conditions.

Key Concepts

Mass Fraction

Mass fraction is an essential concept when dealing with very tiny concentrations, like the amount of gold in seawater. It represents the ratio of the mass of a particular substance to the total mass of the mixture.
In simpler terms, it's a way to express how much of a substance is contained in a mixture relative to the whole.

• For example, if gold in seawater is stated to be 0.05 ppb (parts per billion), it means there are 0.05 micrograms (µg) of gold per gram of seawater.
• Converting ppb to mass fraction involves multiplying the concentration by a factor of \(10^{-6}\) to handle the units of micrograms to grams. Therefore, 0.05 ppb is expressed as \(0.05 \times 10^{-6}\).

This small mass fraction, once determined, can be used to calculate the total mass of gold in larger quantities of seawater, which is crucial for extraction calculations.

Gold Recovery Efficiency

Gold recovery efficiency is a vital factor in any gold extraction process. It indicates the percentage of gold that can be successfully extracted from a particular source.
In many practical applications, not all the gold present in a mixture can be retrieved, owing to various technical limitations.

• For example, a 50% gold recovery efficiency means that only half of the gold present in the seawater can be recovered during the extraction process.
• To factor in this efficiency, you need to adjust the expected yield by multiplying the theoretical amount of recovered gold by 0.50.

This reduced yield directly impacts calculations for determining the volume of seawater required to extract a desired amount of gold.

Seawater Density

The density of seawater is a crucial parameter when determining the amount of a substance within a liquid.
In this exercise, the prosity or density of seawater is given as 1.03 g/mL, meaning that each milliliter of seawater weighs 1.03 grams.

• Knowing the density is essential because it allows conversions between volume and mass, such as calculating how many grams of seawater are in a liter (1000 mL), which is 1030 grams.
• This mass is then used along with the mass fraction of gold to figure out how much gold is present in that volume of seawater.

Understanding seawater density helps perform accurate calculations when estimating how much of a metal can be extracted from a given volume of seawater.