Ethane & Propane Pyrolysis: Molar Composition Calculation

Hey guys! Let's dive into a cool chemistry problem involving the pyrolysis of an ethane and propane mixture. We're going to figure out the molar percentage composition and the molar mass of the original mixture. This might sound complex, but we'll break it down step by step to make it super clear. Think of it like baking a cake – we just need the right ingredients and the right recipe (or in this case, the balanced chemical equations!).

Understanding Pyrolysis

Before we jump into the calculations, let's quickly recap what pyrolysis actually is. Essentially, it's a thermal decomposition process. This means we're using heat to break down molecules into smaller ones, without the presence of oxygen. In our scenario, we're heating up a mixture of ethane (C₂H₆) and propane (C₃H₈), and they're cracking into simpler hydrocarbons. The problem tells us that both ethane and propane are completely transformed, which is a crucial piece of information. It simplifies things because we know all of the starting materials react fully.

Now, why is pyrolysis so important? Well, it's a key process in the petrochemical industry. It's used to produce a variety of valuable chemicals, like ethene (C₂H₄), propene (C₃H₆), and methane (CH₄) – the very compounds mentioned in our problem! These products are then used to make plastics, fuels, and a whole host of other things we use every day. So, understanding the chemistry behind pyrolysis is not just an academic exercise; it has real-world applications.

To solve this problem effectively, we need to write out the balanced chemical equations for the pyrolysis of both ethane and propane. This will give us the stoichiometric relationships between the reactants and the products, which are essential for calculating the molar compositions. Remember, stoichiometry is like the secret language of chemistry – it tells us exactly how much of each substance is involved in a reaction. Once we have these equations, we can use the given molar ratio of the products (CH₄:C₂H₄:C₃H₆ = 1:2:2) to work backwards and figure out the composition of the initial ethane and propane mixture.

Setting Up the Equations

Okay, let's get down to the nitty-gritty and write those balanced chemical equations. This is where we see the magic of chemistry in action! We'll start with ethane (C₂H₆) pyrolysis. Ethane can decompose in several ways, but for this problem, we'll focus on the reactions that produce the products mentioned in the final mixture: methane (CH₄), ethene (C₂H₄), and propene (C₃H₆) – although propene isn't directly produced from ethane, it gives us a clue about the overall reaction pathways involved. A common pyrolysis reaction for ethane is:

C₂H₆ → C₂H₄ + H₂

This equation shows ethane cracking into ethene and hydrogen. While hydrogen isn't mentioned in the final molar ratio, it's an important product to keep in mind. Now, let's look at propane (C₃H₈) pyrolysis. Propane can also break down in multiple ways, but again, we're interested in reactions that produce methane, ethene, and propene. Some key reactions for propane pyrolysis include:

C₃H₈ → C₂H₄ + CH₄ C₃H₈ → C₃H₆ + H₂

The first reaction produces ethene and methane, while the second produces propene and hydrogen. Notice how these reactions connect the dots – they show us how both ethane and propane can contribute to the final mixture of CH₄, C₂H₄, and C₃H₆. Now, here's the crucial part: we need to balance these equations to ensure that the number of atoms of each element is the same on both sides of the equation. This is the law of conservation of mass in action! The equations above are already balanced, which makes our lives easier.

With these balanced equations in hand, we're ready to start setting up a system of equations to solve for the unknowns – the molar amounts of ethane and propane in the initial mixture. This is where the algebraic puzzle begins, and it's super satisfying when all the pieces fall into place!

Solving for Molar Composition

Alright, let's put on our algebra hats and get to solving! This is where we'll use the information given in the problem – the molar ratio of the products (CH₄:C₂H₄:C₃H₆ = 1:2:2) – and our balanced chemical equations to figure out the initial composition of the ethane and propane mixture. Let's start by defining some variables. Let's say:

• x = the number of moles of ethane (C₂H₆) in the initial mixture
• y = the number of moles of propane (C₃H₈) in the initial mixture

Now, we need to relate these variables to the amounts of products formed. This is where our balanced equations come into play. For every mole of ethane that reacts, one mole of ethene is produced (according to the first equation we wrote). So, x moles of ethane will produce x moles of ethene. For propane, things are a little more interesting. From the second propane pyrolysis reaction, one mole of propane produces one mole of ethene and one mole of methane. From the third reaction, one mole of propane produces one mole of propene. So, y moles of propane will contribute to both ethene and propene production.

Now, let's express the moles of each product in terms of x and y:

• Moles of CH₄ = y (from the reaction C₃H₈ → C₂H₄ + CH₄)
• Moles of C₂H₄ = x + y (x from ethane pyrolysis, y from the reaction C₃H₈ → C₂H₄ + CH₄)
• Moles of C₃H₆ = y (from the reaction C₃H₈ → C₃H₆ + H₂)

We know the molar ratio of CH₄:C₂H₄:C₃H₆ is 1:2:2. This gives us two equations:

• (x + y) / y = 2/1 (Ratio of C₂H₄ to CH₄)
• y / y = 2/2 (Ratio of C₃H₆ to CH₄ or C₂H₄, which simplifies to 1=1 and doesn't give new information)

Simplifying the first equation, we get: x + y = 2y x = y

This is a key result! It tells us that the number of moles of ethane in the initial mixture is equal to the number of moles of propane. Now we can easily calculate the molar percentage composition.

Calculating Molar Percentage and Molar Mass

Now that we know x = y, we're in the home stretch! We can easily calculate the molar percentage composition of the initial mixture. Since the number of moles of ethane and propane are equal, the molar fraction of each is 0.5 (because each constitutes half of the total moles). To express this as a percentage, we multiply by 100%:

• Molar percentage of ethane = (0.5) * 100% = 50%
• Molar percentage of propane = (0.5) * 100% = 50%

So, the initial mixture was 50% ethane and 50% propane. Pretty cool, huh? Now, let's calculate the molar mass of the mixture. To do this, we need the molar masses of ethane and propane:

• Molar mass of ethane (C₂H₆) = (2 * 12.01) + (6 * 1.01) = 30.08 g/mol
• Molar mass of propane (C₃H₈) = (3 * 12.01) + (8 * 1.01) = 44.11 g/mol

The molar mass of the mixture is the weighted average of the molar masses of its components, where the weights are the mole fractions:

Molar mass of mixture = (0.5 * 30.08 g/mol) + (0.5 * 44.11 g/mol) = 15.04 g/mol + 22.055 g/mol = 37.095 g/mol

Therefore, the molar mass of the initial mixture is approximately 37.10 g/mol.

Wrapping Up

So, there you have it! We've successfully calculated the molar percentage composition and the molar mass of the ethane and propane mixture undergoing pyrolysis. We found that the initial mixture was 50% ethane and 50% propane, and its molar mass was approximately 37.10 g/mol. This problem nicely illustrates how we can use stoichiometry and balanced chemical equations to understand and predict the outcomes of chemical reactions. Chemistry can be awesome, especially when we can solve puzzles like this!

Remember, the key to tackling problems like this is to break them down into smaller, manageable steps. Write out the balanced equations, define your variables, and then use the given information to set up a system of equations. And don't be afraid to ask for help if you get stuck! Keep practicing, and you'll become a chemistry whiz in no time!