Calculating The Theoretical Yield Of Methanol A Stoichiometry Example

Introduction

In the realm of chemistry, understanding stoichiometry is paramount for predicting the outcome of chemical reactions. One crucial aspect of stoichiometry is determining the theoretical yield, which represents the maximum amount of product that can be formed from a given amount of reactants, assuming perfect reaction conditions and no loss of product. This article will delve into a step-by-step calculation of the theoretical yield of methanol ($CH_3OH$) when hydrogen ($H_2$) and carbon monoxide ($CO$) react, based on a specific scenario where 12.0 grams of $H_2$ is mixed with 74.5 grams of $CO$. This analysis will involve understanding the balanced chemical equation, calculating molar masses, identifying the limiting reactant, and finally, computing the theoretical yield in grams. This comprehensive guide aims to clarify the concepts involved and provide a clear methodology for solving similar stoichiometric problems.

Stoichiometry and Chemical Reactions

Stoichiometry, derived from the Greek words stoicheion (element) and metron (measure), is the quantitative relationship between reactants and products in chemical reactions. A balanced chemical equation provides the foundation for stoichiometric calculations, as it accurately represents the molar ratios of reactants and products. The balanced equation for the synthesis of methanol from carbon monoxide and hydrogen is:

$$CO + 2H_2 \rightarrow CH_3OH$$

This equation indicates that one mole of carbon monoxide reacts with two moles of hydrogen gas to produce one mole of methanol. This molar ratio is essential for determining how much product can be formed from given amounts of reactants. In real-world scenarios, reactions may not proceed to completion due to various factors such as equilibrium considerations, side reactions, or loss of product during purification. However, the theoretical yield provides an upper limit for the amount of product that can be obtained. To calculate the theoretical yield, several steps are involved, including converting reactant masses to moles, identifying the limiting reactant, and using stoichiometric ratios to determine the maximum amount of product that can be formed. This process ensures that chemists can accurately predict and optimize chemical reactions, making it a cornerstone of chemical synthesis and industrial processes.

Step-by-Step Calculation of Theoretical Yield

1. Write the Balanced Chemical Equation

The first step in any stoichiometry problem is to write the balanced chemical equation. This ensures that the molar ratios between reactants and products are correctly represented. The balanced equation for the synthesis of methanol from carbon monoxide and hydrogen is:

$$CO + 2H_2 \rightarrow CH_3OH$$

This equation tells us that one mole of carbon monoxide ($CO$) reacts with two moles of hydrogen gas ($H_2$) to produce one mole of methanol ($CH_3OH$). This molar ratio is crucial for all subsequent calculations.

2. Calculate the Molar Masses of Reactants and Products

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar masses, we sum the atomic masses of each element in the compound, which can be obtained from the periodic table.

• Molar mass of $H_2$: 2 × 1.008 g/mol = 2.016 g/mol
• Molar mass of $CO$: 12.01 g/mol (C) + 16.00 g/mol (O) = 28.01 g/mol
• Molar mass of $CH_3OH$: 12.01 g/mol (C) + 4 × 1.008 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol

These molar masses will be used to convert the given masses of reactants into moles, which is necessary for determining the stoichiometry of the reaction. Accurate molar masses are essential for precise stoichiometric calculations, as they form the basis for converting between mass and moles, allowing chemists to work with the balanced chemical equation effectively. This step is fundamental to ensuring the correct proportions of reactants are used and the expected amount of product is calculated.

3. Convert Grams of Reactants to Moles

To determine the number of moles of each reactant, we divide the given mass by its molar mass:

• Moles of $H_2$:

  $$\frac{12.0 \text{ g}}{2.016 \text{ g/mol}} = 5.95 \text{ mol}$$

• Moles of $CO$:

  $$\frac{74.5 \text{ g}}{28.01 \text{ g/mol}} = 2.66 \text{ mol}$$

Converting grams to moles allows us to work with the stoichiometric ratios in the balanced chemical equation. Moles are a fundamental unit in chemistry, as they directly relate to the number of molecules or atoms in a substance. This conversion is a critical step in stoichiometric calculations, as it bridges the gap between the macroscopic world (grams) and the microscopic world (moles), enabling accurate predictions of reaction outcomes. By calculating the moles of each reactant, we can determine which reactant is the limiting reactant, which ultimately dictates the maximum amount of product that can be formed.

4. Determine the Limiting Reactant

The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed. To identify the limiting reactant, we compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.

From the balanced equation, the stoichiometric ratio of $CO$ to $H_2$ is 1:2. This means that for every mole of $CO$, we need two moles of $H_2$. Let's compare this to the actual mole ratio we have:

• Moles of $H_2$ required to react with 2.66 mol of $CO$: 2.66 mol $CO$ × (2 mol $H_2$ / 1 mol $CO$) = 5.32 mol $H_2$

We have 5.95 mol of $H_2$, which is more than the 5.32 mol required to react with all the $CO$. Therefore, $CO$ is the limiting reactant, and $H_2$ is the excess reactant. The limiting reactant dictates the maximum amount of product that can be formed because once it is consumed, the reaction stops. Identifying the limiting reactant is crucial for optimizing chemical reactions, as it allows chemists to ensure that the reactants are used efficiently and the maximum possible yield is achieved. This step is a cornerstone of stoichiometric calculations, providing a clear understanding of how much product can be realistically obtained from a given set of reactants.

5. Calculate the Theoretical Yield of $CH_3OH$ in Moles

Using the stoichiometry of the balanced equation, we can calculate the moles of $CH_3OH$ that can be produced from the limiting reactant ($CO$). The balanced equation shows that 1 mole of $CO$ produces 1 mole of $CH_3OH$. Therefore:

Moles of $CH_3OH$ = Moles of $CO$ = 2.66 mol

This calculation directly links the amount of limiting reactant to the amount of product formed, based on the stoichiometric coefficients in the balanced chemical equation. The theoretical yield in moles represents the maximum amount of product that can be obtained if the reaction proceeds to completion without any losses. This value serves as a benchmark for assessing the efficiency of a chemical reaction, providing a clear target for experimental outcomes. By calculating the theoretical yield in moles, we set the stage for converting this value into grams, which is a more practical unit for measuring product yield in the laboratory.

6. Convert Moles of $CH_3OH$ to Grams

To convert moles of $CH_3OH$ to grams, we multiply the number of moles by the molar mass of $CH_3OH$:

Theoretical yield of $CH_3OH$ = 2.66 mol × 32.04 g/mol = 85.22 g

Thus, the theoretical yield of methanol is 85.22 grams. This value represents the maximum amount of methanol that can be produced from the given amounts of reactants, assuming perfect reaction conditions and no losses during the process. Converting moles to grams provides a practical measure of the product yield, as mass is easily measurable in a laboratory setting. This final step completes the stoichiometric calculation, offering a clear prediction of the reaction outcome and serving as a crucial reference point for evaluating experimental results.

Conclusion

In summary, we have calculated the theoretical yield of methanol ($CH_3OH$) when 12.0 grams of hydrogen ($H_2$) is mixed with 74.5 grams of carbon monoxide ($CO$). By following a step-by-step approach, which included writing the balanced chemical equation, calculating molar masses, converting grams to moles, identifying the limiting reactant, and finally converting moles of product to grams, we determined the theoretical yield to be 85.22 grams. This calculation highlights the importance of stoichiometry in predicting the outcome of chemical reactions.

Understanding and calculating theoretical yield is crucial in various fields, including chemical synthesis, industrial chemistry, and research. It allows chemists to optimize reactions, minimize waste, and accurately predict the amount of product that can be obtained. While the theoretical yield represents an ideal scenario, it provides a valuable benchmark for assessing the efficiency of a reaction. In practical applications, the actual yield may be lower due to factors such as incomplete reactions, side reactions, and losses during product isolation and purification. Therefore, comparing the actual yield to the theoretical yield gives insight into the effectiveness of the reaction process.

The ability to perform stoichiometric calculations accurately is a fundamental skill for anyone working in chemistry. By mastering these concepts, chemists can design and execute experiments more effectively, leading to better outcomes and advancements in the field.

Answer

Therefore, based on our calculations, the correct answer to the question is not among the options provided. The theoretical yield of methanol is 85.22 grams, which differs from the given choices. This underscores the importance of performing accurate calculations in stoichiometry problems to arrive at the correct result. The options provided (A. 14.88 grams, B. 47.65 grams) do not match the calculated theoretical yield, highlighting the necessity of careful and precise calculations in chemistry.