Question

A 0.732 g mixture of methane, \(\mathrm{CH}_{4}\), and ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) is burned, yielding \(2.064 \mathrm{g} \mathrm{CO}_{2} .\) What is the percent composition of this mixture (a) by mass; (b) on a mole basis?

Short answer

Using step 1 to 3, x (Moles of methane) and y (Moles of ethane) can be calculated. Once these are known, the composition by mass and mole basis can be calculated using formulas described in step 4.

Step-by-step solution

Reaction Stoichiometry

The combustion reactions for Methane and Ethane are:\[\mathrm{CH_{4 (g)} + 2O_{2 (g)} -> CO_{2 (g)} + 2H_{2O (g)}}\]and\[\mathrm{C_{2}H_{6 (g)} + 7/2O_{2 (g)} -> 2CO_{2 (g)} + 3H_{2O (g)}}\]which implies that 1 mole of Methane or 1 mole of Ethane will produce 1 or 2 moles of Carbon Dioxide respectively.

Formation of Equations

Now consider the initial moles of Methane as x and of Ethane as y. Since 1 mole of CH4 and 1 mole of C2H6 will produce 1 and 2 moles of CO2 respectively, the moles of CO2 produced are x + 2y. Also, as the mass of the gas mixture is known (0.732 g), can formulate the following equations:\[\mathrm{16.04x + 30.07y = 0.732} (equation 1)\]and\[\mathrm{x + 2y = mass of CO_{2} produced / 44.01} (equation 2)\]since 1 mole of CO2 has a mass of 44.01 g.

Solving the Equations

Substitute the given mass of CO2 produced (2.064 g) in equation 2 to get the moles of CO2 and then solve the 2 equations simultaneously to get the values of x and y. This will give the initial moles of CH4 and C2H6.

Calculation of % Composition

Once the x and y values are known, we can use the following formulas to determine the percent composition by mass and on a mole basis: \[\%composition (mass basis) = (mass component / mass total) * 100\]\[\%composition (mole basis) = (moles component / moles total) * 100\]

Key concepts

Stoichiometry

In chemical reactions, stoichiometry is about understanding the quantitative relationships between reactants and products. It's like a recipe for chemistry, where the ingredients (reactants) are combined in specific amounts to produce a certain number of products. Knowing these relationships helps us predict how much of each substance is needed or created in a reaction.

The combustion of methane and ethane showcases typical stoichiometric calculations. The balanced equations for the combustion reactions tell us that one mole of methane produces one mole of carbon dioxide, while one mole of ethane produces two moles of carbon dioxide. From this, we can deduce how much carbon dioxide will be produced based on the amount of methane or ethane burned. This ratio is vital for setting up equations that solve for unknown quantities in chemical mixtures.

These relationships are the backbone of calculating other elements in the reaction, such as masses, moles, and volume of gases. Analyzing the mass of reactants and resultant products based on stoichiometric ratios helps in determining the unknown parts of a chemical mixture like in the given problem with methane and ethane.

Mole Calculations

A mole is a universally-used unit in chemistry that represents a very large number of entities, such as atoms or molecules. It's the chemist's way of relating the mass of a substance to the number of particles it contains. In calculations, we prefer moles because they provide a more accurate reflection of a quantity's role in a reaction, unlike mass, which can be misleading.

During the exercise, we use mole calculations to understand how much carbon dioxide is produced from burning methane and ethane. By translating the mass of carbon dioxide formed (2.064 g) into moles using its molar mass (44.01 g/mol), we can directly correlate it with the number of moles of methane and ethane. This conversion is central to figuring out parts of the original gas mixture, through equations set up from the stoichiometric relationships we discussed earlier.

It's crucial for students to become familiar with mole conversions and practice moving fluidly between mass, molar mass, and moles, because it is a great tool for unlocking the chemical story behind a reaction.

Percent Composition

Percent composition is a way of expressing the proportion of each component in a mixture. It's like finding what percentage of a cake each ingredient takes up, allowing us to see the makeup of a substance by mass or number of moles.

In the given problem, once we determine how many moles of methane and ethane are present, we calculate the percent composition by mass by dividing the mass of each component by the total initial mass of the mixture. Similarly, on a mole basis, percent composition is calculated by dividing the number of moles of each component by the total number of moles in the mixture. These calculations allow us to understand not only the makeup of the original mixture but also to dive deeper into its chemical characteristics.

This understanding is essential in chemistry, especially when analyzing mixtures, because it provides a straightforward method for determining how much of each substance is present, thus helping with quality control, formulation, and adjustment of reactions to obtain desired results.