Hydrocarbon Molecular Formula: Burning 8 Liters, What's The Formula?

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Hey guys! Today, we're diving into a fascinating chemistry problem involving the combustion of hydrocarbons. We'll break down how to determine the molecular formula of a hydrocarbon gas given the volumes of reactants and products in a combustion reaction. Let's get started!

Understanding Hydrocarbon Combustion

Hydrocarbon combustion is a fundamental chemical process where a hydrocarbon reacts with oxygen to produce carbon dioxide and water. The general equation for this reaction is:

CxHy+O2→CO2+H2O\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

Where x and y represent the number of carbon and hydrogen atoms in the hydrocarbon molecule, respectively. To figure out the molecular formula (CxHy\text{C}_x\text{H}_y), we need to find these values of x and y. This is where stoichiometry, the study of the quantitative relationships between reactants and products in chemical reactions, comes into play. To really grasp this, think of it like baking a cake – you need the right amounts of ingredients to get the recipe right! In our case, we need to figure out the 'recipe' for the hydrocarbon.

Stoichiometry and Gas Volumes

The key principle we'll use here is Avogadro's Law. This law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. In simpler terms, for gases reacting under the same conditions, the volume ratio is equal to the mole ratio. Mole ratio is like the secret code that unlocks the relationship between different substances in a chemical reaction. It tells us exactly how many molecules or moles of one substance are needed to react with or produce a certain amount of another substance.

This means if we have the volumes of the gases involved in a reaction, we can directly use those volumes as ratios to determine the stoichiometric coefficients in the balanced chemical equation. Balancing chemical equations is like balancing a seesaw – you need the same number of atoms on both sides to make it stable and accurate. Stoichiometric coefficients are the numbers in front of each chemical formula in a balanced equation, and they tell us the proportion of each substance involved in the reaction.

Problem Setup: Cracking the Hydrocarbon Code

Now, let’s apply this to our problem. We're given that 8 liters of a hydrocarbon gas require 24 liters of oxygen for complete combustion, and the reaction produces 16 liters of carbon dioxide. Let's write down what we know:

  • Volume of hydrocarbon gas: 8 liters
  • Volume of oxygen ($ ext{O}_2$): 24 liters
  • Volume of carbon dioxide ($ ext{CO}_2$): 16 liters

Our goal is to determine the molecular formula of the hydrocarbon, $ ext{C}_x\text{H}_y$. Remember, we're essentially trying to find the subscripts x and y.

Setting Up the Balanced Equation

First, we'll write a general balanced equation for the combustion reaction:

CxHy+aO2→bCO2+cH2O\text{C}_x\text{H}_y + a\text{O}_2 \rightarrow b\text{CO}_2 + c\text{H}_2\text{O}

Here, a, b, and c are the stoichiometric coefficients we need to determine. Think of these coefficients as the balancing weights on our chemical seesaw. They ensure that the number of atoms of each element is the same on both the reactant and product sides of the equation.

Solving the Problem: Unveiling the Formula

Using Volume Ratios as Mole Ratios

Since the volumes of the gases are given, we can use them directly as mole ratios. Let's simplify the volume ratios:

  • Hydrocarbon : $ extO}_2$ $ ext{CO_2$ = 8 : 24 : 16

Divide each volume by the smallest volume (8 liters) to get the simplest ratio:

  • Hydrocarbon : $ extO}_2$ $ ext{CO_2$ = 1 : 3 : 2

This gives us the values for the coefficients in our balanced equation, at least for the hydrocarbon and carbon dioxide. So, we now know that for every 1 mole (or volume) of hydrocarbon, we get 2 moles (or volumes) of carbon dioxide.

Determining the Value of x

The coefficient in front of $ ext{CO}_2$ (which is 2) directly tells us the value of x in the hydrocarbon formula. This is because all the carbon atoms in the hydrocarbon end up in the carbon dioxide molecules. If we get 2 CO2 molecules for every hydrocarbon molecule, there must be 2 carbon atoms in each hydrocarbon molecule. So:

  • x = 2

Balancing Carbon Atoms

Now we know that x = 2. This means our hydrocarbon is $ ext{C}_2\text{H}_y$. Let’s plug this into our equation:

C2Hy+3O2→2CO2+cH2O\text{C}_2\text{H}_y + 3\text{O}_2 \rightarrow 2\text{CO}_2 + c\text{H}_2\text{O}

See how we've used the ratio we found earlier to fill in the coefficients for the hydrocarbon (1), oxygen (3), and carbon dioxide (2)? This is like fitting puzzle pieces together! Now, our next mission is to figure out 'y', the number of hydrogen atoms.

Determining the Value of y

To find y, we need to balance the hydrogen atoms. We'll use the fact that the number of hydrogen atoms on both sides of the equation must be equal. Let's look at the products side of the equation. We have c molecules of $ ext{H}_2 ext{O}$, and each water molecule has 2 hydrogen atoms. So, the total number of hydrogen atoms on the products side is 2c. On the reactants side, we have y hydrogen atoms in the hydrocarbon molecule $ ext{C}_2\text{H}_y$. Therefore:

  • y = 2c

Balancing Oxygen Atoms

Now, let's balance the oxygen atoms. On the reactants side, we have 3 molecules of $ ext{O}_2$, which means we have 3 * 2 = 6 oxygen atoms. On the products side, we have 2 molecules of $ ext{CO}_2$ (2 * 2 = 4 oxygen atoms) and c molecules of $ ext{H}_2 ext{O* (c oxygen atoms). So, the total number of oxygen atoms on the products side is 4 + c. Equating the number of oxygen atoms on both sides:

  • 6 = 4 + c

Solving for c:

  • c = 2

Calculating y

Now that we know c = 2, we can find y:

  • y = 2c = 2 * 2 = 4

So, we've discovered that the value of y, which represents the number of hydrogen atoms in our hydrocarbon, is 4. We're getting closer to unveiling the mystery formula!

Finalizing the Balanced Equation

Let's plug c = 2 back into our balanced equation:

C2Hy+3O2→2CO2+2H2O\text{C}_2\text{H}_y + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}

The Grand Finale: The Molecular Formula

We've determined that x = 2 and y = 4. Therefore, the molecular formula of the hydrocarbon is C2H4\text{C}_2\text{H}_4.

The Answer

The molecular formula of the hydrocarbon is C2H4\text{C}_2\text{H}_4. This corresponds to option (A). It's like solving a puzzle, isn't it? Each piece of information we gather helps us fit the bigger picture together and find the answer.

Key Takeaways

Let's recap the key steps we took to solve this problem:

  1. Write the General Balanced Equation: CxHy+aO2→bCO2+cH2O\text{C}_x\text{H}_y + a\text{O}_2 \rightarrow b\text{CO}_2 + c\text{H}_2\text{O}. This sets the stage for our calculation and gives us a framework to work with.
  2. Use Volume Ratios as Mole Ratios: This is a crucial step. Avogadro's Law allows us to use the given volumes as mole ratios, simplifying our calculations.
  3. Determine x from the $ ext{CO}_2$ Volume: The volume of carbon dioxide produced directly relates to the number of carbon atoms in the hydrocarbon.
  4. Balance Hydrogen and Oxygen Atoms: Balancing the equation involves making sure we have the same number of atoms of each element on both sides. This helps us find the coefficients and, ultimately, the values of y.
  5. Write the Molecular Formula $ ext{C}_x\text{H}_y$: Once we have x and y, we have our molecular formula!

Tips and Tricks

  • Always start with a balanced equation: This is the foundation of any stoichiometry problem. A balanced equation ensures that you're accounting for every atom involved in the reaction.
  • Use the simplest ratios: Simplifying the volume ratios makes the calculations easier.
  • Check your work: Make sure the number of atoms of each element is the same on both sides of the final equation. This is like proofreading your work to catch any mistakes.

Conclusion: Mastering Hydrocarbon Combustion

So, there you have it! We've successfully determined the molecular formula of a hydrocarbon using the principles of stoichiometry and Avogadro's Law. These kinds of problems might seem daunting at first, but with a clear understanding of the concepts and a step-by-step approach, you can crack them every time. Keep practicing, and you'll become a hydrocarbon combustion master! Chemistry is like a detective story – you gather the clues, piece them together, and solve the mystery. Happy solving, guys!