Question

Gold is found in seawater at very low levels, about 0.05 ppb by mass. Assuming that gold is worth about \(\$ 800\) 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

It would take approximately \(1,509,533,428\) liters of seawater to obtain \(\$ 1,000,000\) worth of gold with a 50% efficient recovery process.

Step-by-step solution

Mass of gold in seawater

Given that the gold concentration in seawater is 0.05 ppb (parts per billion). Let's consider a sample of 1 liter (1000 mL) of seawater. Using the density of seawater, we can find the mass of this sample: mass_seawater = volume * density mass_seawater = 1000 mL * 1.03 g/mL = 1030 g Now, we can calculate the mass of gold in this 1-liter seawater sample: mass_gold = gold_concentration * mass_seawater mass_gold = 0.05 ppb * 1030 g = 0.00000005 * 1030 g = 0.0000515 g

Convert gold mass to troy ounces

To find the value of gold, we need to convert the mass from grams to troy ounces (1 troy ounce ≈ 31.1035 g): mass_gold_troy_ounce = mass_gold / 31.1035 mass_gold_troy_ounce ≈ 0.0000515 g / 31.1035 ≈ 0.000001656 troy ounces

Calculate the worth of gold in a 1-liter sample

Now, calculate the worth of 0.000001656 troy ounces of gold at a rate of \(\$ 800\) per troy ounce: gold_worth = mass_gold_troy_ounce * gold_rate gold_worth ≈ 0.000001656 * 800 ≈ \$ 0.0013248

Calculate the worth of gold recovered considering process efficiency

Considering a gold recovery process with 50% efficiency, the worth of gold recovered from the 1-liter sample would be: gold_recovered_worth = gold_worth * efficiency gold_recovered_worth ≈ \$ 0.0013248 * 0.5 ≈ \$ 0.0006624

Estimate the volume of seawater required to obtain \(1,000,000 worth of gold

Finally, divide the target of \)\$1,000,000$ by the gold recovered worth of a 1-liter seawater sample to find the required number of liters: volume_required = target_worth / gold_recovered_worth volume_required ≈ 1,000,000 / 0.0006624 ≈ 1,509,533,428 liters It would take approximately 1,509,533,428 liters of seawater to obtain \(\$ 1,000,000\) worth of gold with a 50% efficient recovery process.

Key concepts

Gold Concentration Calculations

Understanding gold concentration calculations is vital in the process of gold extraction from seawater. In our exercise, the concentration of gold in seawater was given as 0.05 parts per billion (ppb) by mass. This seemingly small number requires context to grasp its significance in gold extraction endeavors.

To visualize what 0.05 ppb means, we consider that a billion parts of something equates to 1,000,000,000 units. Hence, 0.05 ppb corresponds to 0.05 units of gold in every 1,000,000,000 units of seawater. This proportion translates to an incredibly minute quantity when compared to the mass of seawater we interact with on a daily basis. To find the direct mass of gold in a liter of seawater, we multiply the concentration by the mass of seawater (1030 grams for 1 liter). This level of precision in calculation is essential when contemplating the feasibility or profitability of extracting gold from such a large volume of seawater.

Seawater Density

The density of a substance is a measure of its mass per unit volume. For seawater, the density can have minor variations based on temperature and salinity, but a standard value often used is 1.03 grams per milliliter (g/mL), as stated in the exercise.

This density is crucial when determining the mass of seawater we are working with, which in turn is needed to calculate the mass of gold it contains. Due to the density of seawater being higher than that of pure water, primarily because of the salts dissolved in it, every liter (1,000 milliliters) weighs slightly more than a kilogram – specifically, 1,030 grams in our exercise.

Conversion to Troy Ounces

Gold is traditionally weighed in troy ounces, and understanding how to convert from grams to troy ounces is essential in the context of gold trade and evaluation. One troy ounce is equivalent to approximately 31.1035 grams.

The given mass of gold retrieved from the liter of seawater, once calculated in grams, needs to be converted into troy ounces to assess its market value. This conversion is done by dividing the mass in grams by 31.1035. As we compute this in our exercise, we see how the minute mass of gold in seawater is even smaller when expressed in troy ounces, underscoring the challenge in economically justifying the extraction of gold from seawater.

Efficiency of Gold Recovery Process

No gold recovery process is 100% efficient, and accounting for this efficiency is critical when estimating the actual yield of gold from any extraction operation. In our exercise, we assumed a 50% efficiency rate, which means that only half of the gold present in the seawater can be recovered and sold.

The efficiency greatly impacts the final calculations, doubling the amount of seawater that needs to be processed to achieve a specific monetary goal. By incorporating the efficiency rate into our value calculation, we can estimate more realistic figures for the amount of raw material (seawater) needed and set practical expectations for the profitability of the gold extraction process.