Cellular Respiration Equation: Balancing And Mass Calculation

Hey guys! Let's dive into the fascinating world of cellular respiration, a fundamental process that fuels life as we know it. In this article, we'll tackle a specific chemical reaction within cellular respiration, focusing on balancing the equation and performing a stoichiometric calculation. Specifically, we're looking at the reaction of a simple hydrocarbon with oxygen. Let's break it down step by step.

Balancing the Chemical Equation for Cellular Respiration

Cellular respiration, at its core, is a combustion reaction. It involves the breakdown of a fuel molecule, like a carbohydrate or a hydrocarbon, in the presence of oxygen to release energy. The general form of the equation we're dealing with is:

$C_3H_8 + O_2 \rightarrow$

Our mission is to complete this equation and balance it, ensuring that the number of atoms of each element is the same on both sides of the arrow. This is crucial because it adheres to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. So, what exactly are the products of this reaction? Well, cellular respiration is like a controlled fire – it produces carbon dioxide ($CO_2$) and water ($H_2O$). Think of it like this: the hydrocarbon (our fuel) combines with oxygen, and the carbon and hydrogen atoms end up bonding with oxygen atoms to form these familiar products. Now we can write the unbalanced equation:

$C_3H_8 + O_2 \rightarrow CO_2 + H_2O$

But it's not balanced yet! See, the number of atoms on each side is different. So, let's put on our balancing hats and get this equation squared away.

Step-by-Step Balancing Act

1. Start with Carbon: We have 3 carbon atoms on the left ($C_3H_8$) and only 1 on the right ($CO_2$). To balance the carbon, we'll add a coefficient of 3 in front of the $CO_2$:

   $C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

2. Next Up: Hydrogen: There are 8 hydrogen atoms on the left ($C_3H_8$) and 2 on the right ($H_2O$). To balance the hydrogen, we'll add a coefficient of 4 in front of the $H_2O$:

   $C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

3. Oxygen, the Final Frontier: Now, let's count the oxygen atoms. On the right side, we have 3 $CO_2$ molecules, each with 2 oxygen atoms (3 * 2 = 6), and 4 $H_2O$ molecules, each with 1 oxygen atom (4 * 1 = 4). That's a total of 10 oxygen atoms on the right. On the left, we have $O_2$, which means we need 10 oxygen atoms on the left as well. To achieve this, we'll add a coefficient of 5 in front of the $O_2$:

   $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

4. The Grand Check: Let's double-check to make sure everything is balanced.

   • Carbon: 3 on the left, 3 on the right. Check!
   • Hydrogen: 8 on the left, 8 on the right. Check!
   • Oxygen: 10 on the left, 10 on the right. Check!

Voila! We have a balanced equation! The balanced chemical equation for the reaction of $C_3H_8$ with oxygen is:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Now that we've got the equation balanced, let's tackle the next part of the problem: stoichiometry.

Stoichiometry: Calculating the Mass of Oxygen Used

Stoichiometry, in simple terms, is like the recipe book for chemical reactions. It allows us to calculate the amounts of reactants and products involved in a chemical reaction. In this case, we want to find out how much oxygen ($O_2$) is needed to react with 45.5 grams of propane ($C_3H_8$). To do this, we'll need to use the balanced chemical equation and some good old-fashioned unit conversions.

The Stoichiometric Roadmap

Here's the plan of attack:

1. Grams to Moles (Propane): Convert the given mass of propane (45.5 grams) into moles using its molar mass.
2. Moles Propane to Moles Oxygen: Use the stoichiometric ratio from the balanced equation to determine the number of moles of oxygen required.
3. Moles to Grams (Oxygen): Convert the moles of oxygen back into grams using its molar mass.

Let's get to work!

Step 1: Grams to Moles of Propane ($C_3H_8$)

To convert grams to moles, we need the molar mass of propane. The molar mass is the mass of one mole of a substance, which is numerically equal to its atomic or molecular weight in grams per mole (g/mol). To calculate the molar mass of propane ($C_3H_8$), we add up the atomic masses of each element in the compound:

• Carbon (C): 3 atoms * 12.01 g/mol = 36.03 g/mol
• Hydrogen (H): 8 atoms * 1.01 g/mol = 8.08 g/mol

Molar mass of $C_3H_8$ = 36.03 g/mol + 8.08 g/mol = 44.11 g/mol

Now we can convert the given mass of propane (45.5 grams) to moles:

Moles of $C_3H_8$ = (45.5 grams) / (44.11 g/mol) = 1.0315 moles

Step 2: Moles of Propane to Moles of Oxygen ($O_2$)

This is where the balanced equation comes into play. The coefficients in the balanced equation tell us the molar ratios of the reactants and products. From the balanced equation:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

We see that 1 mole of $C_3H_8$ reacts with 5 moles of $O_2$. This gives us the stoichiometric ratio:

(5 moles $O_2$) / (1 mole $C_3H_8$)

Now we can use this ratio to find the moles of $O_2$ required:

Moles of $O_2$ = (1.0315 moles $C_3H_8$) * (5 moles $O_2$ / 1 mole $C_3H_8$) = 5.1575 moles $O_2$

Step 3: Moles to Grams of Oxygen ($O_2$)

Finally, we need to convert the moles of $O_2$ back into grams. To do this, we'll use the molar mass of $O_2$. The molar mass of $O_2$ is:

• Oxygen (O): 2 atoms * 16.00 g/mol = 32.00 g/mol

Now we can convert moles of $O_2$ to grams:

Mass of $O_2$ = (5.1575 moles) * (32.00 g/mol) = 165.04 grams

The Grand Finale: The Answer!

Therefore, 165.04 grams of $O_2$ are used when 45.5 grams of propane ($C_3H_8$) react. That's a lot of oxygen! It really highlights the power and energy packed within these chemical bonds. This detailed calculation really shows how stoichiometry works. It might seem like a lot of steps, but each step is logical and builds upon the previous one. Think of it like following a recipe – you need the right ingredients (molar masses), the right proportions (stoichiometric ratios), and the right steps to get the final product (the mass of oxygen).

In Conclusion

We've successfully balanced the chemical equation for the reaction of $C_3H_8$ with oxygen and calculated the mass of oxygen required for a given amount of propane. This exercise highlights the fundamental principles of chemistry, including balancing equations and stoichiometry, which are crucial for understanding chemical reactions. Remember, guys, chemistry is all about understanding how atoms interact and combine to form the world around us. Keep practicing, keep exploring, and keep asking questions! You've got this!