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Someone claims that the mass and the mole fractions for a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\) gases are identical. Do you agree? Explain.

Short Answer

Expert verified
Answer: No, the mass and mole fractions for a mixture of CO2 and N2O gases are not identical.

Step by step solution

01

Calculate the molar mass of CO2 and N2O gases

First, we need to determine the molar mass of each gas. The molar mass of CO2 is calculated by adding the molar mass of one carbon atom (12.01 g/mol) and two oxygen atoms (15.999 g/mol each), and the molar mass of N2O is calculated by adding the molar mass of two nitrogen atoms (14.007 g/mol each) and one oxygen atom (15.999 g/mol). Molar mass of CO2 = 12.01 + (2 × 15.999) = 44.009 g/mol Molar mass of N2O = (2 × 14.007) + 15.999 = 44.013 g/mol
02

Assume a certain number of moles for each gas

Let's assume we have n1 moles of CO2 and n2 moles of N2O in the mixture. The total moles (nt) in the mixture will be: nt = n1 + n2
03

Calculate the mass of each gas

To calculate the mass (m) for each gas, we use the formula: m = n × Molar Mass Mass of CO2 = m1 = n1 × 44.009 g/mol Mass of N2O = m2 = n2 × 44.013 g/mol
04

Calculate the mass fraction of each gas

The mass fraction for each gas is calculated by dividing the mass of the individual gas by the total mass of the mixture. The total mass (mt) of the mixture would be: mt = m1 + m2 Mass fraction of CO2 = m1/mt = (n1 × 44.009) / ((n1 × 44.009) + (n2 × 44.013)) Mass fraction of N2O = m2/mt = (n2 × 44.013) / ((n1 × 44.009) + (n2 × 44.013))
05

Calculate the mole fraction of each gas

The mole fraction for each gas is calculated by dividing the moles of the individual gas by the total moles of the mixture. Mole fraction of CO2 = n1/nt = n1 / (n1 + n2) Mole fraction of N2O = n2/nt = n2 / (n1 + n2)
06

Compare the mass and mole fractions for each gas

Comparing the mass and mole fractions calculated in steps 4 and 5, we see that they are not identical for both gases: Mass fraction of CO2 ≠ Mole fraction of CO2 Mass fraction of N2O ≠ Mole fraction of N2O The mass and mole fractions for a mixture of CO2 and N2O gases are not identical. So, we do not agree with the claim.

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Key Concepts

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

Understanding Molar Mass
When studying chemistry, especially in exercises involving gas mixtures, the concept of molar mass is fundamental. Molar mass is defined as the mass of one mole of a substance. It is commonly expressed in grams per mole (g/mol). This property is critical because it links the mass of a substance to the amount of substance (number of moles).

To calculate the molar mass, one simply adds up the atomic masses of all the atoms in a molecule. For example, in carbon dioxide (CO2), you have one carbon atom and two oxygen atoms. Carbon's atomic mass is around 12.01 g/mol, and oxygen's is about 15.999 g/mol. The molar mass of CO2 is therefore the sum of these atomic masses: 44.009 g/mol.
Gas Mixture Composition Analysis
Gas mixtures, like the one in our exercise involving CO2 and N2O, have specific compositions that can be quantified in different ways. One way to characterize the composition of a gas mixture is by using mass fraction, which is the mass of a particular gas divided by the total mass of the mixture. Another way is by mole fraction, which is the amount (in moles) of a gas divided by the total moles in the mixture.

Understanding the difference between mass and mole fractions is pivotal when working with mixtures. Mole fractions are dimensionless quantities that give an idea of the 'share' each gas has in the total amount of substance. Mass fractions, although they might seem similar, take into account the different molar masses of each gas, resulting in a value dependent on both the number of moles and the type of molecules present.
The Role of Stoichiometry in Mixtures
Stoichiometry is the aspect of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. It helps to predict the amounts of substances consumed and produced in a reaction. However, stoichiometry is not limited to reactions; it also allows us to understand the relationships between different components of a mixture.

In our example with CO2 and N2O, stoichiometry helps us understand that even if the gases have almost identical molar masses, the individual amounts of each gas (in moles) will not result in identical mass fractions. The subtle differences in their molar masses, when multiplied by the respective number of moles of each gas, lead to this discrepancy. Thus, stoichiometry clarifies why the mass and mole fractions cannot be identical, which is a crucial point of understanding in this exercise.

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Most popular questions from this chapter

The mass diffusivity of ethanol \(\left(\rho=789 \mathrm{~kg} / \mathrm{m}^{3}\right.\) and \(M=46 \mathrm{~kg} / \mathrm{kmol}\) ) through air was determined in a Stefan tube. The tube has a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). Initially, the ethanol surface was \(10 \mathrm{~cm}\) from the top of the tube; and after 10 hours have elapsed, the ethanol surface was \(25 \mathrm{~cm}\) from the top of the tube, which corresponds to \(0.0445 \mathrm{~cm}^{3}\) of ethanol being evaporated. The ethanol vapor pressure is \(0.0684\) atm, and the concentration of ethanol is zero at the top of the tube. If the entire process was operated at \(24^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), determine the mass diffusivity of ethanol in air.

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was \(1800 \mathrm{~m}^{2}\). The skin of this balloon is \(2 \mathrm{~mm}\) thick and is made of a material whose helium diffusion coefficient is \(1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

A rubber object is in contact with nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(298 \mathrm{~K}\) and \(250 \mathrm{kPa}\). The solubility of nitrogen gas in rubber is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3}\).bar. The mass density of nitrogen at the interface is (a) \(0.049 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(0.064 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(0.077 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(0.092 \mathrm{~kg} / \mathrm{m}^{3}\) (e) \(0.109 \mathrm{~kg} / \mathrm{m}^{3}\)

Pure \(\mathrm{N}_{2}\) gas at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is flowing through a 10-m-long, 3-cm-inner diameter pipe made of 2 -mm-thick rubber. Determine the rate at which \(\mathrm{N}_{2}\) leaks out of the pipe if the medium surrounding the pipe is \((a)\) a vacuum and \((b)\) atmospheric air at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) with 21 percent \(\mathrm{O}_{2}\) and 79 percent \(\mathrm{N}_{2}\).

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