Question

The concentration of gold in seawater has been reported to be between \(5 \mathrm{ppt}\) (parts per trillion) and \(50 \mathrm{ppt}\). Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in \(1.0 \times 10^{3}\) gal of seawater.

Short answer

There are approximately \(4.918 \times 10^{-8}\) grams of gold in \(1.0 \times 10^{3}\) gallons of seawater.

Step-by-step solution

Conversion factors

We are given the volume of seawater in gallons, but we need the volume in liters for our calculation. The conversion factor is as follows: 1 gallon = 3.78541 liters

Convert the volume of seawater to liters

To convert the volume of seawater to liters, we need to multiply the given volume in gallons by the conversion factor: \(V_{seawater} (L) = V_{seawater} (gal) \times \frac{3.78541 L}{1 gal}\) Plug the given volume of seawater into the equation: \(V_{seawater}(L) = 1.0 \times 10^{3} gal \times \frac{3.78541 L}{1 gal}\) Calculate the volume in liters: \(V_{seawater}(L) = 3.78541 \times 10^{3} L\)

Calculate the grams of gold per liter of seawater

The concentration of gold in seawater is given as 13 ppt (parts per trillion). To calculate the grams of gold per liter of seawater, we will use the following conversion factors: 1 part per trillion (ppt) = \(1 \times 10^{-12}\) g/L We can compute the grams of gold per 1 liter of seawater: \(grams\_of\_gold\_per\_L = 13 ppt \times \frac{1 \times 10^{-12} g}{1 L}\) Calculate the grams of gold per liter: \(grams\_of\_gold\_per\_L = 13 \times 10^{-12} g/L\)

Calculate the total grams of gold in 1.0 x 10^3 gallons of seawater

To calculate the total grams of gold in the given volume of seawater, we use the following equation: Total grams of gold = grams of gold per liter of seawater × volume of seawater in liters Plug in the values calculated in the previous steps: Total grams of gold = \(13 \times 10^{-12} \frac{g}{L} \times 3.78541 \times 10^{3} L\) Calculate the total grams of gold: Total grams of gold = \(4.918 \times 10^{-8} g\) Thus, there are approximately \(4.918 \times 10^{-8}\) grams of gold in \(1.0 \times 10^{3}\) gallons of seawater.

Key concepts

Conversion Factors

Conversion factors are fundamental in transforming units from one measurement to another. They allow you to convert gallons of seawater into liters, as required in our exercise. This is crucial because scientific calculations often require measurements in metric units. In our context, the conversion factor used is:

• 1 gallon = 3.78541 liters

To utilize this factor, you simply multiply the number of gallons by the conversion factor to obtain the equivalent volume in liters. So, for our problem, with a volume of \(1.0 \times 10^{3}\) gallons, we compute:\[V_{seawater}(L) = 1.0 \times 10^{3} \text{ gal} \times \frac{3.78541 \text{ L}}{1 \text{ gal}} = 3.78541 \times 10^{3} \text{ L}\]
This simple multiplication shows how conversion factors act as bridges between different measurement systems, making complex calculations feasible and consistent.

Parts Per Trillion

Parts per trillion (ppt) is a unit of measurement for extremely low concentrations, notably used in environmental science and chemistry. It represents one part of a substance in a trillion parts of another, often used for measuring pollutants or trace elements like gold in seawater.
In our example, the concentration of gold in seawater is given as 13 ppt. This means there are 13 grams of gold in every \(1 \times 10^{12}\) grams of seawater:

• 1 ppt = \(1 \times 10^{-12} \) g/L

Using this knowledge, we can calculate the amount of gold per liter by performing: \[grams\_of\_gold\_per\_L = 13 \text{ ppt} \times \frac{1 \times 10^{-12} \text{ g}}{1 \text{ L}} = 13 \times 10^{-12} \text{ g/L}\]
This small number illustrates how ppt measurements effectively handle trace amounts, giving scientists a precise understanding of low-level concentrations in various environments.

Seawater Chemistry

Seawater chemistry involves understanding the composition of seawater, which is a complex mix of several dissolved elements and minerals. The oceans contain traces of virtually every natural element, including precious metals like gold.
The exercise highlights the method to determine gold concentration in a certain volume of seawater. Given that seawater has a uniform composition, understanding seawater chemistry allows estimation of the presence of various substances. Here, gold's concentration as 13 ppt allows for the calculation of its total mass in a large volume, such as:\[\text{Total grams of gold} = 13 \times 10^{-12} \frac{\text{g}}{\text{L}} \times 3.78541 \times 10^{3} \text{ L}\]
Thus, we find approximately \(4.918 \times 10^{-8}\) grams of gold. This shows the small but valuable quantities of metals present in our oceans, reflecting the vast but intricate nature of seawater chemistry, where even trace elements contribute to the mineral wealth of Earth's oceans.