Percent Yield Calculation: Aluminum Oxide Formation

Hey guys! Let's dive into a common chemistry problem: calculating percent yield. This is super important because it tells us how efficient a chemical reaction really is. We're going to break down an example involving aluminum and oxygen reacting to form aluminum oxide. So, grab your calculators, and let's get started!

Understanding the Problem

Before we jump into the calculations, let's clearly define what percent yield actually means. In simple terms, it's the ratio of the actual yield (the amount of product you actually get from a reaction) to the theoretical yield (the maximum amount of product you could get based on the stoichiometry of the reaction), expressed as a percentage. Think of it like this: if you're baking cookies and the recipe should give you 24 cookies (theoretical yield), but you only end up with 20 (actual yield), your percent yield tells you how close you were to perfection.

The problem we're tackling today states that 2.00 g of aluminum reacts with 0.750 g of molecular oxygen to produce 0.350 g of aluminum oxide. We are given the balanced chemical equation: $Al(s) + O_2(g) ightarrow Al_2O_3(s)$. The key here is to figure out what the theoretical yield of aluminum oxide is, so we can compare it to the actual yield of 0.350 g and calculate the percent yield. To determine the theoretical yield, we need to identify the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thereby determining the maximum amount of product that can be formed. The other reactant is known as the excess reactant, which means there will be some leftover after the reaction is complete. Identifying the limiting reactant is crucial because it dictates the theoretical yield of the product.

Why is Percent Yield Important?

Knowing the percent yield of a reaction is vital in chemistry for several reasons. Firstly, it provides a measure of the reaction's efficiency. A low percent yield can indicate that the reaction conditions weren't optimal, some product was lost during the process, or that there were side reactions occurring. Secondly, percent yield helps in cost analysis in industrial processes. If a reaction has a consistently low yield, it might be too expensive to use on a large scale. Thirdly, it assists in troubleshooting experimental procedures. If the actual yield is significantly lower than the theoretical yield, it prompts chemists to investigate potential sources of error or loss in the experiment. Finally, percent yield is crucial for assessing the reliability and reproducibility of a chemical reaction. High percent yields, obtained consistently, suggest that the reaction is reliable and the procedure is well-controlled.

Step 1: Balancing the Chemical Equation

The very first thing we always need to do in stoichiometry problems is to make sure our chemical equation is balanced. This ensures that we're working with the correct mole ratios. Our unbalanced equation is:

$Al(s) + O_2(g) ightarrow Al_2O_3(s)$

To balance it, we need to make sure we have the same number of each type of atom on both sides of the equation. Let's start with oxygen. We have 2 oxygen atoms on the left ($O_2$) and 3 on the right ($Al_2O_3$). To balance this, we can use a common multiple, which is 6. Place a coefficient of 3 in front of $O_2$ and a coefficient of 2 in front of $Al_2O_3$:

$Al(s) + 3O_2(g) ightarrow 2Al_2O_3(s)$

Now, let's balance aluminum. We have 1 aluminum atom on the left and 4 on the right (2 molecules of $Al_2O_3$, each containing 2 Al atoms). So, we place a coefficient of 4 in front of $Al$:

$4Al(s) + 3O_2(g) ightarrow 2Al_2O_3(s)$

Now our equation is balanced! This balanced equation tells us that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. This mole ratio is the foundation for our calculations.

The Importance of a Balanced Equation

Balancing the chemical equation is a cornerstone of stoichiometry and crucial for accurate calculations in chemistry. An unbalanced equation violates the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, the number of atoms of each element must be the same on both sides of the equation. The balanced equation provides the correct mole ratios between reactants and products, which are essential for determining theoretical yields, identifying limiting reactants, and calculating percent yields. Without a balanced equation, the stoichiometric calculations would be based on incorrect proportions, leading to inaccurate results. This foundational step ensures that chemical reactions are represented accurately, allowing for reliable quantitative analysis and predictions.

Step 2: Converting Grams to Moles

Next, we need to convert the masses of our reactants (aluminum and oxygen) from grams to moles. Why moles? Because the balanced chemical equation gives us the mole ratios, not mass ratios. To do this, we'll use the molar mass of each substance. The molar mass is the mass of one mole of a substance, and we can find these values on the periodic table.

• Molar mass of Aluminum (Al) = 26.98 g/mol
• Molar mass of Oxygen ($O_2$) = 32.00 g/mol (remember, oxygen exists as a diatomic molecule, $O_2$)

Now, let's convert:

• Moles of Aluminum = (2.00 g Al) / (26.98 g/mol) = 0.0741 mol Al
• Moles of Oxygen = (0.750 g $O_2$) / (32.00 g/mol) = 0.0234 mol $O_2$

We now know the number of moles of each reactant we started with. This is a crucial piece of information for figuring out which reactant is going to run out first – the limiting reactant.

The Significance of Moles in Chemical Calculations

The concept of the mole is central to stoichiometry and chemical calculations because it provides a bridge between the macroscopic world of grams and the microscopic world of atoms and molecules. A mole is defined as the amount of a substance that contains as many entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12, which is approximately $6.022 imes 10^{23}$ entities (Avogadro's number). By converting grams to moles, we can work with the stoichiometric ratios defined by the balanced chemical equation, which express the relationships between reactants and products at the molecular level. Moles allow chemists to perform quantitative analysis, predict product yields, determine reaction efficiencies, and understand the fundamental proportions involved in chemical transformations. This conversion is vital for accurate and meaningful interpretations of experimental results and for designing chemical processes.

Step 3: Identifying the Limiting Reactant

Okay, this is where things get a little more involved, but don't worry, we'll break it down. The limiting reactant is the reactant that is completely used up in a reaction, thus determining the maximum amount of product that can be formed. To find it, we'll compare the mole ratio of the reactants we have to the mole ratio from the balanced equation.

Our balanced equation is:

$4Al(s) + 3O_2(g) ightarrow 2Al_2O_3(s)$

This tells us that 4 moles of Al react with 3 moles of $O_2$. Let's figure out how much $O_2$ we need to react completely with the Al we have (0.0741 mol):

(0. 0741 mol Al) * (3 mol $O_2$ / 4 mol Al) = 0.0556 mol $O_2$

This calculation tells us that we need 0.0556 moles of $O_2$ to react completely with our 0.0741 moles of Al. But we only have 0.0234 moles of $O_2$. Since we don't have enough $O_2$, it's the limiting reactant. Aluminum is the excess reactant in this case.

Why Limiting Reactant Matters

Identifying the limiting reactant is critical in stoichiometry because it dictates the maximum amount of product that can be formed in a chemical reaction. The limiting reactant is the reactant that is completely consumed, thereby halting the reaction when it runs out. The other reactants, present in excess, will have some amount remaining after the reaction ceases. Therefore, the amount of product formed is directly determined by the stoichiometry of the limiting reactant, not the excess reactants. Accurate determination of the limiting reactant is essential for predicting theoretical yields, optimizing reaction conditions, and ensuring efficient use of resources in chemical processes. In industrial applications, using the correct proportions of reactants based on the limiting reactant concept can significantly reduce costs and waste, making the process more economical and environmentally sustainable.

Step 4: Calculating the Theoretical Yield

Now that we know oxygen is the limiting reactant, we can calculate the theoretical yield of aluminum oxide ($Al_2O_3$). This is the maximum amount of product we could form if all the limiting reactant reacted. We'll use the mole ratio from the balanced equation again:

$4Al(s) + 3O_2(g) ightarrow 2Al_2O_3(s)$

This tells us that 3 moles of $O_2$ produce 2 moles of $Al_2O_3$. Let's convert our moles of $O_2$ (0.0234 mol) to moles of $Al_2O_3$:

(0. 0234 mol $O_2$) * (2 mol $Al_2O_3$ / 3 mol $O_2$) = 0.0156 mol $Al_2O_3$

Now, we need to convert moles of $Al_2O_3$ to grams. We'll use the molar mass of $Al_2O_3$:

• Molar mass of $Al_2O_3$ = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol

Theoretical yield of $Al_2O_3$ = (0.0156 mol) * (101.96 g/mol) = 1.59 g $Al_2O_3$

So, the theoretical yield of aluminum oxide is 1.59 grams. This is the maximum amount we could possibly make in this reaction.

Theoretical Yield as an Ideal Outcome

The theoretical yield represents the ideal or maximum amount of product that can be obtained from a chemical reaction, assuming that all of the limiting reactant is converted into product and no product is lost during the process. It is a crucial benchmark in chemistry because it provides a standard against which the actual outcome of a reaction can be compared. The calculation of the theoretical yield involves using stoichiometry and the balanced chemical equation to determine the mole ratio between the limiting reactant and the desired product. By understanding the theoretical yield, chemists can evaluate the efficiency of a reaction, identify potential sources of error or loss, and optimize reaction conditions to maximize product formation. It serves as a key metric in assessing the viability and scalability of chemical processes, particularly in industrial settings where yield optimization is critical for economic and sustainable production.

Step 5: Calculating the Percent Yield

Finally, we can calculate the percent yield! We know:

• Actual yield = 0.350 g $Al_2O_3$ (given in the problem)
• Theoretical yield = 1.59 g $Al_2O_3$ (calculated in the previous step)

The formula for percent yield is:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

Percent Yield = (0.350 g / 1.59 g) * 100% = 22.0%

So, the percent yield of this reaction is 22.0%. This means that only 22.0% of the maximum possible amount of aluminum oxide was actually produced.

The Significance of Percent Yield

The percent yield is a crucial metric in chemistry for evaluating the efficiency of a chemical reaction. It quantifies the ratio of the actual yield (the amount of product obtained experimentally) to the theoretical yield (the maximum amount of product that could be formed based on stoichiometry), expressed as a percentage. A high percent yield indicates that the reaction was conducted efficiently with minimal loss of product, whereas a low percent yield suggests that the reaction was less efficient, possibly due to incomplete reactions, side reactions, or loss of product during isolation and purification steps. The percent yield helps chemists assess the effectiveness of reaction conditions, identify potential sources of error, and optimize procedures to maximize product formation. In industrial chemistry, achieving high percent yields is essential for economic viability and sustainability, as it minimizes waste and ensures efficient utilization of resources. The percent yield serves as a key performance indicator in chemical synthesis, driving improvements in reaction design and execution.

Conclusion

There you have it! We've calculated the percent yield of the reaction between aluminum and oxygen. It might seem like a lot of steps, but once you get the hang of it, it's pretty straightforward. Remember to always balance your equation, convert to moles, find the limiting reactant, calculate the theoretical yield, and then finally calculate the percent yield. Understanding percent yield is essential for anyone studying chemistry, as it helps us evaluate the efficiency of chemical reactions in the lab and in industrial processes. Keep practicing, and you'll become a pro at these calculations in no time! Keep rocking, guys!