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Chapter 22: Problem 17

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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 \(1.03 \times 10^9 g \times \frac{38.00 g}{19.00 g} = 2.06 \times 10^9 g = 2.06 \times 10^6 kg\)
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bromine Extraction
Bromine is extracted from seawater through a process that takes advantage of its properties. Bromine exists in seawater as bromide ions (Br鈦�). 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鈦� + Cl鈧� 鈫� Br鈧� + 2Cl鈦� ).
  • Volatilization: Br鈧� 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鈦�) 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 km鲁:
First, convert the volume:
  • 1 km鲁 = 10鈦� m鲁 = 10鹿鈦� cm鲁 .
Next, use the density of seawater, given as 1.03 g/cm鲁:
  • Mass = Volume 脳 Density = 10鹿鈦� cm鲁 脳 1.03 g/cm鲁 = 1.03 脳 10鹿鈦� g = 1.03 脳 10鈦� 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 脳 10鈦� tons of seawater, 1.03 脳 10鈦� grams of fluoride can be extracted.
Conversions between volume and mass are a cornerstone in the extraction of chemical elements from seawater.

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Most popular questions from this chapter

In \(1988,\) G. J. Schrobilgen, professor of chemistry at McMaster University in Canada, reported the synthesis of an ionic compound, \([\mathrm{HCNKrF}]\left[\mathrm{AsF}_{6}\right],\) which consists of \(\mathrm{HCNKr} \mathrm{F}^{+}\) and \(\mathrm{AsF}_{6}^{-}\) ions. In the \(\mathrm{HCNKr} \mathrm{F}^{+}\) ion, the krypton is covalently bonded to both fluorine and nitrogen. Draw Lewis structures for these ions, and estimate the bond angles.

The heavier halogens (Cl, Br, and I) form compounds in which the central halogen atom, \(X\), is bonded directly to oxygen and to fluorine. Several examples are known, including those with formulas of the type \(\mathrm{FXO}_{2}, \mathrm{FXO}_{3, \text { and } \mathrm{F}_{3} \mathrm{XO} . \text { The structures of these }}\) molecules are all consistent with the VSEPR model. Draw Lewis structures and predict the geometries of (a) chloryl fluoride, \(\mathrm{FClO}_{2}\); (b) perchloryl fluoride, \(\mathrm{FClO}_{3} ;(\mathrm{c}) \mathrm{F}_{3} \mathrm{ClO}\).

Sodium nitrite can be made by passing oxygen and nitrogen monoxide gases into an aqueous solution of sodium carbonate. Write a balanced equation for this reaction.

Despite the fact that it has the higher molecular mass, XeO \(_{4}\) exists as a gas at \(298 \mathrm{K},\) whereas \(\mathrm{XeO}_{3}\) is a solid. Give a plausible explanation for this observation.

In the electrolysis of a sample of water \(22.83 \mathrm{mL}\) of \(\mathrm{O}_{2}(\mathrm{g})\) was collected at \(25.0^{\circ} \mathrm{C}\) at an oxygen partial pressure of \(736.7 \mathrm{mmHg} .\) Determine the mass of water that was decomposed.
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