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Chapter 6: Problem 133

A student measured the relative rate of effusion of carbon dioxide and propane, \(\mathrm{C}_{3} \mathrm{H}_{8},\) and found that they effuse at exactly the same rate. Did the student make a mistake?

Short Answer

Expert verified
Answer: Using Graham's law of diffusion, we found that the ratio of the rates of effusion of carbon dioxide and propane is approximately 1.00, meaning that they effuse at almost the same rate. The student did not make a mistake; their observation of the effusion rates being the same is consistent with the molecular weights being very close. However, their rates of effusion are not exactly the same, but rather very close to each other.

Step by step solution

01

Find the molar mass of carbon dioxide (CO鈧�)

First, we need to find the molar mass of carbon dioxide. The molecular formula of carbon dioxide is CO鈧�, which means it is composed of one carbon atom and two oxygen atoms. Using their atomic masses, we can calculate the molar mass of CO鈧�: Molar mass of CO鈧� = 1 Carbon atom + 2 Oxygen atoms Molar mass of CO鈧� = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol
02

Find the molar mass of propane (C鈧僅鈧�)

Now, let's find the molar mass of propane. The molecular formula of propane is C鈧僅鈧�, which means it is composed of three carbon atoms and eight hydrogen atoms. Using their atomic masses, we can calculate the molar mass of propane: Molar mass of propane = 3 Carbon atoms + 8 Hydrogen atoms Molar mass of propane = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol
03

Apply Graham's Law of diffusion to compare rates of effusion

We can now use Graham's law of diffusion to compare the rates of effusion of carbon dioxide (CO鈧�) and propane (C鈧僅鈧�). In this case, we'll denote the rate of effusion of CO鈧� as \(R_1\) and the rate of effusion of propane as \(R_2\): $$ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{44.11}{44.01}} $$
04

Calculate the ratio of rates of effusion

Now, let's calculate the ratio of rates of effusion: $$ \frac{R_1}{R_2} = \sqrt{\frac{44.11}{44.01}} \approx 1.00 $$
05

Conclusion

According to Graham's law of diffusion, the ratio of the rates of effusion of carbon dioxide (CO鈧�) and propane (C鈧僅鈧�) is approximately 1.00, meaning that they effuse at almost the same rate. The student did not make a mistake; their observation of the effusion rates being the same is consistent with the molecular weights being very close. However, it's important to note that their rates of effusion are not exactly the same, but rather very close to each other.

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Key Concepts

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

Molar Mass Calculation
To understand the exercise, we first need to calculate the molar mass of the substances involved: carbon dioxide ( CO鈧�) and propane (C鈧僅鈧�). The molar mass of a compound is determined by adding up the atomic masses of all the atoms in a molecule. This is essentially the mass of one mole of that compound.

For carbon dioxide, the chemical formula is CO鈧�. It consists of one carbon atom and two oxygen atoms. Carbon has an atomic mass of 12.01 g/mol, and oxygen has an atomic mass of 16.00 g/mol. By adding these together, the molar mass of CO鈧� becomes:
  • Molar mass of carbon: 1 脳 12.01 g/mol = 12.01 g/mol
  • Molar mass of oxygen: 2 脳 16.00 g/mol = 32.00 g/mol
  • Total molar mass of CO鈧� = 12.01 + 32.00 = 44.01 g/mol
For propane (C鈧僅鈧�), the molecule consists of three carbon atoms and eight hydrogen atoms. Hydrogen has a smaller atomic mass of 1.01 g/mol, so the calculation is:
  • Molar mass of carbon: 3 脳 12.01 g/mol = 36.03 g/mol
  • Molar mass of hydrogen: 8 脳 1.01 g/mol = 8.08 g/mol
  • Total molar mass of propane = 36.03 + 8.08 = 44.11 g/mol
Understanding how to calculate molar mass is crucial in applying Graham's Law of Effusion, which deals with how gases escape through a tiny opening.
Carbon Dioxide
Carbon dioxide is a colorless gas with the chemical formula CO鈧�. It is an essential part of the Earth's atmosphere and is involved in the natural processes of respiration and photosynthesis. In the context of Graham's Law, carbon dioxide's properties are particularly suitable for demonstrating principles of effusion because it is a common and relatively simple diatomic molecule with predictable behavior.

By understanding CO鈧�'s chemical makeup, we can see why its molar mass is 44.01 g/mol. This is significant because effusion is influenced by the speed at which gas molecules can move. Lighter gases move more quickly, meaning the molar mass influences how it compares with other gases in terms of effusion rates.

For CO鈧�, being almost equal in molar mass to propane (44.11 g/mol) suggests that they would effuse at similar rates according to Graham's Law, which is counterintuitive if one would expect larger differences in behavior from looking at vastly different compounds like CO鈧� and C鈧僅鈧�.
Propane
Propane is a hydrocarbon with the chemical formula C鈧僅鈧�. It exists as a gas at standard temperature and pressure and is often used as a fuel. In exercises like the one described, understanding the composition of propane is key to applying chemical laws like Graham's Law of Effusion.

Propane's molecular structure contributes a molar mass of 44.11 g/mol, which is very close to that of carbon dioxide (44.01 g/mol). This proximity in mass means that their effusion rates are practically the same, which is a fascinating observation given their different chemical functions and uses.

Although propane is larger in terms of molecular structure compared to CO鈧�, it is surprisingly similar in effusion behavior. This shows off an interesting aspect of molecular chemistry where different substances can exhibit similar physical behaviors despite their structural differences.

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Most popular questions from this chapter

The pressure in an aerosol can is 1.5 atm at \(27^{\circ} \mathrm{C}\). The can will withstand a pressure of about 2 atm. Will it burst if heated in a campfire to \(450^{\circ} \mathrm{C} ?\)

What happens to the pressure of a gas under the following conditions? a. The absolute temperature is halved and the volume doubles. b. Both the absolute temperature and the volume double. c. The absolute temperature increases by \(75 \%,\) and the volume decreases by \(50 \%\).

Which noble gas is expected to deviate the most from ideal bchavior in a graph of \(P V / R T\) versus \(P ?\)

Using Wetlands to Treat Agricultural Waste Wetlands can play a significant role in removing fertilizer residues from rain runoff and groundwater; one way they do this is through denitrification, which converts nitrate ions to nitrogen gas: \(2 \mathrm{NO}_{3}^{-}(a q)+5 \mathrm{CO}(g)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)+5 \mathrm{CO}_{2}(g)\) Suppose \(200.0 \mathrm{g}\) of \(\mathrm{NO}_{3}^{-}\) flows into a swamp each day. What volume of \(\mathrm{N}_{2}\) would be produced at \(17^{\circ} \mathrm{C}\) and 1.00 atm if the denitrification process were complete? What volume of \(\mathrm{CO}_{2}\) would be produced? Suppose the gas mixture produced by the decomposition reaction is trapped in a container at \(17^{\circ} \mathrm{C} ;\) what is the density of the mixture if we assume that \(P_{\text {total }}=1.00\) atm?

The rate of effusion of an unknown gas is \(0.10 \mathrm{m} / \mathrm{s}\) and the rate of effusion of \(\mathrm{SO}_{3}(g)\) is \(0.052 \mathrm{m} / \mathrm{s}\) under identical experimental conditions. What is the molar mass of the unknown gas?
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