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Chapter 5: Problem 141

Sulfur hexafluoride, \(\mathrm{SF}_{6},\) is an extremely dense gas. How does its density compare with the density of air? Use a molar mass for air of \(29.0 \mathrm{~g} / \mathrm{mol}\).

Short Answer

Expert verified
The density of SF6 is approximately 5.04 times greater than the density of air.

Step by step solution

01

Determine the Molar Mass of SF6

To calculate the molar mass of sulfur hexafluoride (SF6), we have to sum the atomic masses of sulfur and fluorine. Sulfur has an atomic mass of approximately 32.07 g/mol, and fluorine has an atomic mass of approximately 19.00 g/mol. Since there are six fluorine atoms, the calculation is:\[\text{Molar mass of } \text{SF}_6 = 32.07 \text{ g/mol (S)} + 6 \times 19.00 \text{ g/mol (F)} = 32.07 \text{ g/mol} + 114.00 \text{ g/mol} = 146.07 \text{ g/mol}.\]
02

Compare Densities Using Molar Masses

The density of a gas is proportional to its molar mass. Thus, we can compare the densities of SF6 and air by comparing their molar masses. The molar mass of air is given as 29.0 g/mol. The ratio of the densities is therefore equal to the ratio of the molar masses:\[\text{Density ratio} = \frac{\text{Molar mass of } \text{SF}_6}{\text{Molar mass of air}} = \frac{146.07 \text{ g/mol}}{29.0 \text{ g/mol}} \approx 5.04\]This indicates that SF6 is approximately 5.04 times denser than air.

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Key Concepts

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

Molar Mass Calculation
The molar mass calculation is a fundamental step in understanding properties such as density and reactivity of a compound. To calculate the molar mass of a compound, you sum the atomic masses of all the atoms in a molecule. In the case of sulfur hexafluoride (\(\text{SF}_6\), it consists of one sulfur atom and six fluorine atoms.
  • Sulfur (S) has an atomic mass of approximately 32.07 g/mol.
  • Fluorine (F) has an atomic mass of approximately 19.00 g/mol.
To find the molar mass, combine these values:
- Calculate the total mass of the fluorine atoms: \(6 \times 19.00 = 114.00\) g/mol.
- Add this to the molar mass of sulfur: \(32.07 + 114.00 = 146.07\) g/mol.
This means that the molar mass of \(\text{SF}_6\) is 146.07 g/mol, a crucial figure for further comparisons.
Sulfur Hexafluoride
Sulfur hexafluoride (\(\text{SF}_6\)) is an unusual gas with several interesting properties. One of its most significant characteristics is its high density, which results from its relatively large molar mass compared to other common gases.
  • \(\text{SF}_6\) is composed of sulfur and fluorine, giving it a higher molar mass than many simple molecules.
  • This factor contributes to its density鈥攖he essence of the exercise.
Furthermore, \(\text{SF}_6\) is known for being inert and non-toxic, often used in the electrical industry for insulation and in other applications for its ability to displace less dense gases.
Density Comparison
Density comparison is a useful method in chemistry to understand how heavy or light a gas is compared to another. The density of a gas directly correlates with its molar mass when conditions like temperature and pressure remain constant.
The concept becomes simple with the formula:
- Density (\(d\)) is proportional to Molar Mass (\(M\)):\[d \propto M\]For sulfur hexafluoride (\(\text{SF}_6\)), its significant molar mass of 146.07 g/mol makes it much denser than air, which has a molar mass of 29.0 g/mol.
  • Formula for density ratio: \[\text{Density ratio} = \frac{\text{Molar mass of } \text{SF}_6}{\text{Molar mass of air}}\]
  • Plug in the values: \(\frac{146.07}{29.0} \approx 5.04\)
Thus, \(\text{SF}_6\) is approximately 5.04 times denser than air, highlighting its physical heaviness in a simple, mathematical form.

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Most popular questions from this chapter

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

A vessel containing \(39.5 \mathrm{~cm}^{3}\) of helium gas at \(25^{\circ} \mathrm{C}\) and \(106 \mathrm{kPa}\) was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required \(7.5 \mathrm{~cm}^{3}\) of ethanol, what was the final temperature of the helium?

At what temperature does the rms speed of \(\mathrm{O}_{2}\) molecules equal \(475 . \mathrm{m} / \mathrm{s}\) ?

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5.118 Raoul Pictet, the Swiss physicist who first liquefied oxygen, attempted to liquefy hydrogen. He heated potassium formate, \(\mathrm{KCHO}_{2}\), with \(\mathrm{KOH}\) in a closed 2.50-Lvessel. \(\mathrm{KCHO}_{2}(s)+\mathrm{KOH}(s) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2}(g)\) If \(75.0 \mathrm{~g}\) of potassium formate reacts in a \(2.50-\mathrm{L}\) vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.
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