Question

A glass vessel fitted with a stopcock valve has a mass of $337.428\mathrm{g}$ when evacuated. When filled with $\mathrm{Ar}$, it has a mass of $339.854\mathrm{g}$. When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of $339.076\mathrm{g}$. What is the mole percent of Ne in the gas mixture?

Short answer

The mole percent of Ne in the gas mixture is approximately 49.42%.

Step-by-step solution

Calculate the masses of Ar and Ar+Ne mixtures.

First, we need to find the mass of Ar and the mass of Ne+Ar mixture in the glass vessel when full. We can do this by subtracting the mass of the empty vessel from the mass of the vessel with Ar and the mass of the vessel with the Ne+Ar mixture. Mass of Ar in the glass vessel: $Mas{s}_{Ar}=Mas{s}_{Ar+Vessel}-Mas{s}_{Vessel}$ Mass of Ne+Ar mixture in the glass vessel: $Mas{s}_{Ne+Ar}=Mas{s}_{(Ne+Ar)+Vessel}-Mas{s}_{Vessel}$ Now inserting the given values: Mass of $Ar=339.854g-337.428g=2.426g$ Mass of $Ne+Ar=339.076g-337.428g=1.648g$

Calculate the number of moles of each element in the mixture.

Given the masses of each element, we can find the number of moles by dividing the masses by their respective molar masses. The molar mass of Ar is 39.95 g/mol and that of Ne is 20.18 g/mol. Number of moles of $Ar=\frac{Mas{s}_{Ar}}{Molarmas{s}_{Ar}}=\frac{2.426g}{39.95g/mol}=0.0607moles$ Number of moles of $Ne+Ar=\frac{Mas{s}_{Ne+Ar}}{Molarmas{s}_{Ne}+Molarmas{s}_{Ar}}=\frac{1.648g}{20.18g/mol+39.95g/mol}=0.0307moles$ Since the same volume of Ar has been replaced by the same volume of Ne+Ar mixture, at the same temperature and pressure, the number of moles of $Ar=0.0607moles$ in both cases.

Calculate the mole percent of Ne in the mixture.

Next, we need to find the difference in moles of Ar between the pure Ar case and the Ne+Ar mixture case. This difference corresponds to the number of moles of Ne in the mixture. Moles of Ne in the mixture: $Mole{s}_{Ne}=Mole{s}_{Ar}-Mole{s}_{(ArinNe+Ar)}=0.0607moles-0.0307moles=0.0300moles$ Now, we can calculate the mole percent of Ne in the mixture by dividing the moles of Ne by the total moles of Ne and Ar in the mixture and then multiplying it by 100. $MolePercen{t}_{Ne}=\frac{Mole{s}_{Ne}}{Mole{s}_{Ne}+Mole{s}_{(ArinNe+Ar)}}\times 100$ $MolePercen{t}_{Ne}=\frac{0.0300moles}{0.0300moles+0.0307moles}\times 100=49.42\mathrm{\%}$ The mole percent of Ne in the gas mixture is approximately 49.42%.

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that links the microscopic world of atoms to the macroscopic world we can measure. It represents the mass of one mole of a substance, which is defined as the mass in grams that is numerically equivalent to the atomic or molecular weight of the substance in atomic mass units (amu). For example, the molar mass of Argon (Ar) is 39.95 g/mol, which means one mole of Argon weighs 39.95 grams.

Understanding molar mass is crucial when dealing with stoichiometry, as it allows us to convert between the mass of a substance and the number of moles, thus bridging the gap between the mass of a compound and the number of its particles or entities, like atoms or molecules.

To calculate the molar mass, one must sum the atomic masses of all the atoms in a molecule. For elemental gases like Argon (Ar) and Neon (Ne), the molar mass is simply the atomic mass listed on the periodic table for that element. While dealing with gas mixtures, knowing the molar mass of each component allows you to determine how much of each gas is present in terms of moles.

Gas Mixture Composition

The composition of a gas mixture refers to the proportion of each component gas within the mixture. It can be expressed in various ways, such as by volume, by mass, or, as in our exercise, by mole percent. Mole percent or molar fraction is a way to express the concentration of a component in a mixture, calculated by taking the number of moles of that component and dividing it by the total number of moles of all components in the mixture.

In practical scenarios, such as the exercise provided, the gas mixture composition is determined by measuring the mass of the container with and without the gas and then converting these masses to moles using the molar mass of the respective gases. Mole percent is particularly valuable in stoichiometry because reactions often proceed according to mole ratios. Hence, knowing the mole percent can help predict the outcome of a reaction involving a gas mixture.

To enhance understanding, students can visualize the gas mixture as a collection of individual mole units of different gases, with the mole percent indicating the proportion of those units that belong to each type of gas. This approach can simplify the concept and make the calculations more accessible.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is the cornerstone of the quantitative analysis in chemistry, allowing chemists to predict the amounts of substances consumed and produced in a given reaction. The stoichiometric principles hinge on the conservation of mass and the concept of moles. It applies the molar ratios of compounds as they appear in a balanced chemical equation to relate the mass of one substance to the mass of another.

In the context of our exercise, stoichiometry is employed to convert between the mass of the gases in the mixture and the number of moles. This conversion is essential to determine the mole percent of each component in the mixture. Understanding stoichiometry requires a solid grasp of the mole concept, the molar mass of substances, and the ability to balance equations. It empowers students to understand not just the qualitative but also the quantitative aspects of reactions involving both pure compounds and mixtures, such as the gas mixture of Neon and Argon in the example provided.

A common challenge in stoichiometry is recognizing that the ratios of reactants and products are based on moles, and thus conversions to and from mass are an essential part of solving stoichiometric problems. Teachers and educational platforms can help students strengthen their stoichiometry skills by providing exercises that require these conversions and by emphasizing the importance of balanced chemical equations.