Sucrose Combustion: Calculating Moles And Reactants

Hey everyone! Today, we're diving into the fascinating world of chemistry, specifically looking at the combustion of sucrose, that sweet stuff we know as table sugar. We'll be breaking down the chemical equation, figuring out how much of each reactant we have, and ultimately calculating the moles of sucrose involved in the reaction. It's a fun and practical application of stoichiometry, and by the end, you'll have a solid grasp of how to analyze chemical reactions like this one. So, grab your lab coats (just kidding... unless?) and let's get started!

The Combustion of Sucrose: A Closer Look

Alright, let's start with the basics: the chemical equation for the combustion of sucrose. This equation describes what happens when sucrose ($C_{12}H_{22}O_{11}$) reacts with oxygen ($O_2$): $C_{12}H_{22}O_{11} + 12O_2 ightarrow 12CO_2 + 11H_2O$. This equation tells us a whole lot! It shows that one molecule of sucrose reacts with twelve molecules of oxygen to produce twelve molecules of carbon dioxide and eleven molecules of water. But, because chemistry is all about quantities, we're not just interested in the molecules. We want to know how much of each substance we're dealing with in terms of moles and grams.

Before we can start crunching numbers, let's establish a clear understanding of the equation. This is a balanced chemical equation, which means that the number of atoms of each element is the same on both sides of the arrow. This is super important because it follows the law of conservation of mass. You can't just create or destroy atoms in a chemical reaction! So, the coefficients (the numbers in front of each chemical formula) are the heart of the equation. They tell us the mole ratio, or the ratio of moles of reactants and products involved in the reaction. For example, the equation tells us that 1 mole of sucrose reacts with 12 moles of oxygen. Understanding the balanced equation is the key to solving pretty much any stoichiometry problem, so take a moment to appreciate its importance. The balanced equation gives us the recipe for the chemical reaction, and it allows us to know what's going on, quantitatively.

Identifying Reactants and Products

Let's get even more specific about our cast of characters. In this reaction, the reactants are sucrose ($C_{12}H_{22}O_{11}$) and oxygen ($O_2$. These are the ingredients that go in. Sucrose, as mentioned, is the sugar we eat, and oxygen is the stuff we breathe. When these guys get together in the presence of enough heat (like a match or a flame), they undergo a chemical change. This change produces the products, which are carbon dioxide ($CO_2$) and water ($H_2O$. Carbon dioxide is a gas, and water is the familiar liquid. It is worth noting that water is a product of the reaction. We all know that you can't burn water, but water can be produced as a result of the burning of sucrose. We often observe this when we cook: the steam or vapor that comes from the burning sugar is actually water, a product of the combustion. The transformation of reactants into products is a sign that a chemical reaction has taken place, and in the case of combustion, it releases a lot of energy in the form of heat and light. So, when sugar burns, it's not just disappearing; it's undergoing a transformation into new substances. This is a fundamental concept in chemistry – matter is neither created nor destroyed, it just changes form.

Calculating Moles of Sucrose and Oxygen

Now, let's get down to the actual math! The question is: if we have 10.0 grams of sucrose and 8.0 grams of oxygen, how many moles of sucrose are available for this reaction? To solve this, we need to convert the mass of each reactant into moles. The calculation involves using the molar mass of each substance. The molar mass is the mass of one mole of a substance. It's expressed in grams per mole (g/mol) and can be found on the periodic table by adding up the atomic masses of all the atoms in a molecule.

To find the moles of sucrose:

1. Find the molar mass of sucrose ($C_{12}H_{22}O_{11}$):
   • Carbon (C): 12 atoms × 12.01 g/mol = 144.12 g/mol
   • Hydrogen (H): 22 atoms × 1.01 g/mol = 22.22 g/mol
   • Oxygen (O): 11 atoms × 16.00 g/mol = 176.00 g/mol
   • Total molar mass = 144.12 + 22.22 + 176.00 = 342.34 g/mol
2. Convert grams of sucrose to moles:
   • Moles of sucrose = (mass of sucrose) / (molar mass of sucrose)
   • Moles of sucrose = (10.0 g) / (342.34 g/mol) = 0.0292 moles

To find the moles of oxygen:

1. Find the molar mass of oxygen ($O_2$):
   • Oxygen (O): 2 atoms × 16.00 g/mol = 32.00 g/mol
2. Convert grams of oxygen to moles:
   • Moles of oxygen = (mass of oxygen) / (molar mass of oxygen)
   • Moles of oxygen = (8.0 g) / (32.00 g/mol) = 0.250 moles

So, we've got approximately 0.0292 moles of sucrose and 0.250 moles of oxygen available for the reaction. With this information, we can determine the limiting reactant.

Determining the Limiting Reactant

The limiting reactant is the one that gets used up first in a chemical reaction. It determines the maximum amount of product that can be formed. To determine the limiting reactant, we can use the mole ratio from the balanced equation ($C_{12}H_{22}O_{11} + 12O_2 ightarrow 12CO_2 + 11H_2O$).

• For sucrose: Based on the balanced equation, 1 mole of sucrose requires 12 moles of oxygen. We have 0.0292 moles of sucrose. To completely react, we would need 0.0292 moles sucrose × 12 moles O2 / 1 mole sucrose = 0.3504 moles of oxygen. We can compare the required moles of oxygen for full consumption of sucrose, with the actual moles of oxygen available.
• For oxygen: According to the balanced equation, 12 moles of oxygen react with 1 mole of sucrose. We have 0.250 moles of oxygen. To completely react, we would need 0.250 moles O2 * 1 mole sucrose / 12 moles O2 = 0.0208 moles sucrose. Now we can compare the amount of sucrose that would be required for full consumption of oxygen, with the amount of sucrose we actually have.

Now we can compare the required and available amounts. The limiting reactant is the one that will be completely used up first. In our case, oxygen will run out before sucrose does, and therefore, oxygen is the limiting reactant. We can use the information to then calculate the moles of carbon dioxide and water produced. However, the original question only asked for the moles of sucrose available, and we have determined that amount to be 0.0292 moles.

Conclusion: Wrapping Up the Sucrose Combustion

So, there you have it, folks! We've successfully analyzed the combustion of sucrose, a core example of stoichiometry. We went over the balanced chemical equation, calculated the number of moles of each reactant, and now understand the concept of a limiting reactant. This knowledge can be applied to all sorts of chemical reactions and is super useful when you're working in a lab or just curious about how things work. Remember that the balanced equation is the key to solving these types of problems, and always make sure your equation is balanced before starting any calculations. Keep practicing, and you'll be a stoichiometry whiz in no time!

This process is crucial for understanding chemical reactions, predicting product yields, and making sure reactions proceed as planned. Keep an eye out for more chemistry discussions, and happy experimenting! Remember, understanding chemical reactions isn't just about memorizing equations; it's about seeing how the world works at a molecular level. Keep asking questions, keep learning, and most importantly, keep that curiosity burning! Until next time, stay curious!