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 about \(1.51 \times 10^{12} \mathrm{~L}\) of seawater to obtain $$\$1,000,000$$ worth of gold at \(50\%\) extraction efficiency.

Step-by-step solution

Convert gold concentration to a usable unit

We are given the gold concentration as \(0.05 \mathrm{~ppb}\) by mass. We need to convert this into a usable unit. One ppb is equal to \(10^{-9}\) or \(1 \mathrm{~g/g} \times 10^{-9}\). Thus, the concentration of gold in seawater is: \(0.05 \mathrm{~ppb} \times 10^{-9} \mathrm{~g/g} = 5 \times 10^{-11} \mathrm{~g/g}\)

Calculate the mass of gold in one liter of seawater

Now that we have the concentration of gold in seawater, we can calculate the mass of gold in one liter of seawater. We are given the density of seawater as \(1.03 \mathrm{~g/mL}\). Since there are \(1,000\) milliliters in a liter, there are \(1,000 \times 1.03 \mathrm{~g} = 1,030 \mathrm{~g}\) of seawater in one liter. The mass of gold in one liter of seawater, using the concentration we found in step 1, can be found by: \(m_{gold} = 5 \times 10^{-11} \mathrm{~g/g} \times 1,030 \mathrm{~g} = 5.15 \times 10^{-8} \mathrm{~g}\)

Calculate the troy ounces of gold for $$\$1,000,000$$

Since we are given the value of gold as $$\$800$$ per troy ounce, we can now calculate how many troy ounces of gold we would need to reach $$\$1,000,000$$. One troy ounce is equal to \(31.1 \mathrm{~g}\). Let's convert $$\$1,000,000$$ to troy ounces: \(troy \thinspace ounces = \frac{\$1,000,000}{\$800 \thinspace per \thinspace troy \thinspace ounce} = 1250 \thinspace troy \thinspace ounces\) Now we convert it to grams: \(mass = 1250 \thinspace troy \thinspace ounces \times 31.1 \mathrm{~g/troy \thinspace ounce} = 38,875 \mathrm{~g}\) Considering the gold recovery process is \(50\%\) efficient, we would need \(38,875 \thinspace g \times 2 = 77,750 \thinspace g\) of gold from the seawater.

Determine the liters of seawater needed for the desired amount of gold

Now that we know the mass of gold needed and the mass of gold in one liter of seawater, we can calculate the liters of seawater needed: \(Liters \thinspace of \thinspace seawater = \frac{77,750 \mathrm{~g}}{5.15 \times 10^{-8} \mathrm{~g/L}} = 1.51 \times 10^{12} \mathrm{~L}\) So, it would take about \(1.51 \times 10^{12} \mathrm{~L}\) of seawater to obtain $$\$1,000,000$$ worth of gold at \(50\%\) extraction efficiency.

Key concepts

Chemistry of Gold

Gold, known by its chemical symbol Au (from the Latin 'aurum'), is a noble metal which means it is resistant to corrosion and oxidation. This property, along with its rarity, has made it a sought-after metal for jewelry, currency, and industrial applications.

Being a 'soft' metal, gold is quite malleable and ductile, allowing it to be shaped into various forms. In its elemental form, gold is a bright yellow, dense, and lustrous material that can be found in rocks and, as mentioned in the exercise, dissolved in seawater.

The chemistry behind gold’s stability involves its electron configuration, which leads to less reactivity with other chemicals. This stability is precisely what makes gold extraction feasible even from seawater, where it exists only in trace amounts. The process of gold extraction from seawater would require overcoming challenges such as the extremely low concentration of the element, making it a technically demanding and costly endeavor.

Solution Concentration

Solution concentration is a key concept in chemistry, defining how much of a substance is dissolved in a specified volume of solvent. Common units of concentration include molarity, molality, and parts per billion (ppb), the latter being used in the exercise.

Parts per billion (ppb) indicates the number of parts of a substance in a billion parts of total mass, which is particularly useful for measuring extremely low concentrations. For instance, a concentration of 0.05 ppb means that for every billion parts of seawater, there are 0.05 parts of gold.

Understanding solution concentration is crucial when it comes to extracting substances from solutions, as in the case of gold extraction from seawater. Since the gold is present at such a minute level, sophisticated methods and vast volumes of seawater must be processed to obtain a meaningful amount of gold.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between reactants and products in a chemical reaction. It involves calculations that rely on the conservation of mass and the concept of mole ratios according to a balanced chemical equation.

In the context of extracting gold from seawater, as described in the exercise, stoichiometry is not employed in the traditional sense involving reactants and products, but rather to calculate the mass of gold that can be obtained from a certain volume of seawater. The exercise leverages stoichiometric calculations to determine how many liters of seawater would be needed to achieve a target mass of gold, incorporating factors like the solution concentration of gold and the efficiency of the recovery process.

These calculations can become complex when working with extremely small concentrations and large volumes, and they are essential to envision the feasibility of processes such as precious metal recovery from natural sources.