Problem 17 - General Chemistry: Principles and Modern Applications

Question

The abundance of \(\mathrm{F}^{-}\) in seawater is \(1 \mathrm{g} \mathrm{F}^{-}\) per ton of seawater. Suppose that a commercially feasible method could be found to extract fluorine from seawater. (a) What mass of \(\mathrm{F}_{2}\) could be obtained from \(1 \mathrm{km}^{3}\) of seawater \(\left(d=1.03 \mathrm{g} \mathrm{cm}^{-3}\right) ?\) (b) Would the process resemble that for extracting bromine from seawater? Explain.

Step-by-Step Solution

Answer

From 1 kilometer cubed of seawater, \(2.06 \times 10^6 kg\) of fluorine could be obtained. The process wouldn't resemble bromine extraction because the extraction principles are based on different chemical properties.

Step 1: Converting volume to mass

Calculate the mass of seawater in 1 kilometer cubed. 1 cubic kilometer of water contains \(1km^3 = 1km \times 1km \times 1km = 10^3m \times 10^3m \times 10^3m = 10^9m^3 \times (10^2cm/m)^3 = 10^{15} cm^3 \).The mass of this seawater would be \(Volume \times Density = 10^{15}cm^3 \times 1.03g/cm^3 = 1.03 \times 10^{15}g = 1.03 \times 10^9 \) tons.

Step 2: Calculating fluorine amount

Calculate the amount of fluorine from this mass of seawater. Since every ton of seawater has 1 gram of F-, thus, the mass of F- would be \(1.03 \times 10^9 g \) (F- per ton of seawater). To find the mass of F2, which has a formula mass of approximately 38.00 g/mol remember F- indicates one fluorine atom and F2 indicates a diatomic molecule of fluorine (two fluorine atoms). Thus, the mass of F2 obtainable is \(1.03 \times 10^9 g \times \frac{38.00 g}{19.00 g} = 2.06 \times 10^9 g = 2.06 \times 10^6 kg\)

Step 3: Comparing to bromine extraction

The process of extracting fluorine from seawater would be different than that of extracting bromine. Bromine extraction is primarily a result of its solubility and volatility properties, where it can be released from water through aeration. Fluorine, on the other hand, is present in seawater as a fluoride ion and is not volatile. Consequently, a different extraction process would be required.

Key Concepts

Bromine Extraction

Bromine is extracted from seawater through a process that takes advantage of its properties. Bromine exists in seawater as bromide ions (Br⁻). These ions can be oxidized to bromine, which is a volatile liquid, using chlorine.
This process involves:

• Oxidation: Chlorine gas is used to oxidize bromide ions to bromine ( 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻ ).
• Volatilization: Br₂ is sparingly soluble and thus can be removed from the water easily.
• Collection: Bromine is collected by passing air through the seawater, which helps to volatilize and remove the bromine.

Bromine's relatively high volatility and solubility in non-aqueous solvents make this process efficient.

Seawater Chemistry

Seawater contains a complex mixture of elements and compounds, with chlorine and sodium ions making up a major part. Fluoride ions (F⁻) are present in very small amounts, usually 1 gram per ton of seawater.
These fluoride ions are stable and non-volatile, which differentiates them from elements like bromine. Several factors define seawater chemistry:

• Salinity: Mostly due to dissolved sodium and chloride ions.
• Minor Components: Elements like bromine and fluorine are trace elements, affecting extraction processes.
• Stable Compounds: Fluoride ions form stable compounds, often requiring specific chemical reactions for extraction.

Understanding seawater chemistry is crucial for developing efficient extraction methods.

Density and Volume Calculations

Density and volume calculations are essential when dealing with large volumes of seawater for extraction processes. To find the mass of seawater in 1 km³:
First, convert the volume:

• 1 km³ = 10⁹ m³ = 10¹⁵ cm³ .

Next, use the density of seawater, given as 1.03 g/cm³:

• Mass = Volume × Density = 10¹⁵ cm³ × 1.03 g/cm³ = 1.03 × 10¹⁵ g = 1.03 × 10⁹ tons.

Using these calculations:

• The mass of fluoride ions can be deduced. Fluoride ions are present as 1 gram per ton.
• Thus, in 1.03 × 10⁹ tons of seawater, 1.03 × 10⁹ grams of fluoride can be extracted.

Conversions between volume and mass are a cornerstone in the extraction of chemical elements from seawater.