Question

A 9.780-g gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}\right) .\) Complete combustion to form carbon dioxide and water requires \(1.120\) mol oxygen. Calculate the mass percent of ethane in the original mixture.

Short answer

The mass percent of ethane in the original gaseous mixture is approximately 56.9%.

Step-by-step solution

Write the balanced combustion reactions for ethane and propane

First, let's write the balanced chemical equations for the combustion of ethane (C₂H₆) and propane (C₃H₈): Combustion of ethane: \[C_{2}H_{6} + \frac{7}{2} O_{2} \rightarrow 2 CO_{2} + 3 H_{2}O\] Combustion of propane: \[C_{3}H_{8} + 5O_{2} \rightarrow 3 CO_{2} + 4 H_{2}O\] Both reactions show the formation of carbon dioxide (CO₂) and water (H₂O) upon complete combustion.

Create a system of linear equations using moles

Let x be the number of moles of ethane (C₂H₆) and y be the number of moles of propane (C₃H₈) in the gaseous mixture. We can now create two linear equations based on the given information: (1) The total mass of the mixture = mass of ethane + mass of propane: \[9.780\ g = (x)(30.07\ g/mol) + (y)(44.10\ g/mol)\] (2) Total moles of oxygen (O₂) required for complete combustion = 1.120 mol: \[(\frac{7}{2}x) + (5y) = 1.120\ mol\]

Solve the system of linear equations for moles of ethane and propane

Now we need to isolate one of the variables, preferably in the simpler equation (2). It is simpler to isolate y and we have: \[y = \frac{1.120 - \frac{7}{2}x}{5}\] Substituting the expression for y from the second equation into the first equation: \[9.780 = 30.07x + 44.10\left(\frac{1.120 - \frac{7}{2}x}{5}\right)\] Solve for x: \[x \approx 0.185\ \text{mol}\] Now substitute x back into the expression for y: \[y \approx 0.099\ \text{mol}\]

Calculate the mass percentage of ethane

Now that we have the moles of ethane and propane, we can convert them to mass and then calculate the mass percentage of ethane in the mixture: Mass of ethane = (0.185 mol) × (30.07 g/mol) ≈ 5.563 g Mass of propane = (0.099 mol) × (44.10 g/mol) ≈ 4.365 g Mass percentage of ethane = (Mass of ethane / Total mass of the mixture) × 100 \[= \left(\frac{5.563}{9.780}\right) × 100 \approx 56.9\%\] The mass percent of ethane in the original gaseous mixture is approximately 56.9%.

Key concepts

Understanding Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They provide a concise way to express which reactants combine and what products they form. A well-balanced chemical equation obeys the law of conservation of mass, meaning the number of atoms of each element in the reactants side is equal to those in the products side. Take, for example, the combustion reactions provided in the exercise.

Through these equations, students can visualize how molecules like ethane (C2H6) and propane (C3H8) react with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). Balancing these equations requires an understanding of stoichiometry, ensuring that for every atom that reacts, there's a corresponding product atom. Here, it's also vital to interpret coefficients as mole ratios, which are used to determine the amount of each substance needed or produced.

The Role of Stoichiometry in Reactions

Stoichiometry is the section of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. For a combustion reaction, knowing the stoichiometry allows us to find out how much oxygen is needed to fully combust a certain amount of fuel, or conversely, how much fuel is needed for a given amount of oxygen.

In the given problem, stoichiometry is used to create a system of linear equations representing the mass and mole requirements of the reaction. The step-by-step calculations pivot around the concept of mole ratios derived from the balanced equations. Once you understand how to relate moles from the balanced equation to actual masses with the use of molar mass, determining the composition of a mixture like the one with ethane and propane becomes possible.

Calculating Mass Percent

Mass percent calculation is a method used to express the concentration of an element in a compound or a component in a mixture. It's calculated by dividing the mass of the element or component by the total mass of the compound or mixture and then multiplying by 100 to get a percentage.

When applied to the provided problem, mass percent allows us to determine the proportion of ethane in the original gaseous mixture. By discovering the moles of each gas through the system of linear equations and then converting these moles to mass using their respective molar masses, we arrive at individual masses for ethane and propane. Dividing the mass of ethane by the total mass and multiplying by 100 gives us the desired mass percent of ethane. This is not only crucial for homework problems but is also a fundamental concept in real-world applications, such as formulation of mixtures in the chemical industry and analyzing the composition of different substances.