Nitroglycerin Decomposition Calculation How Many Grams Needed For 120 Grams Of Water

Nitroglycerin, an oily explosive liquid, can decompose rapidly upon shock or heating, producing a large amount of gas and heat. The balanced chemical equation for the decomposition of nitroglycerin is:

$4 C_3 H_5(NO_3)_3(l) \rightarrow 12 CO_2(g) + 6 N_2(g) + 10 H_2O(g) + O_2(g)$

This equation reveals the stoichiometry of the reaction, which is crucial for calculating the amount of nitroglycerin needed to produce a specific amount of water. In this comprehensive guide, we will walk through the process of calculating the mass of nitroglycerin required to produce 120 grams of water. This step-by-step approach will not only provide the answer but also enhance your understanding of stoichiometry and its applications in chemical reactions. Before we delve into the calculations, let's first understand the key concepts and definitions that underpin this process. This includes grasping the significance of molar mass, stoichiometric coefficients, and the mole concept, all of which play pivotal roles in chemical calculations. With a solid foundation in these principles, we can confidently tackle the problem at hand and ensure accurate results. Understanding these fundamental concepts is essential for anyone studying chemistry or working in related fields, as they form the bedrock of quantitative chemical analysis and synthesis. Let’s begin by clarifying these concepts before proceeding with the calculation.

Understanding Stoichiometry

Stoichiometry is the branch of chemistry that involves quantitative relationships between reactants and products in chemical reactions. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Stoichiometry allows us to predict the amounts of reactants and products involved in a chemical reaction.

Key Concepts in Stoichiometry

1. Molar Mass: The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you add up the atomic masses of all the atoms in the chemical formula. For instance, the molar mass of water ($H_2O$) is calculated by adding the atomic masses of two hydrogen atoms and one oxygen atom. The molar mass is a crucial conversion factor in stoichiometric calculations, allowing us to convert between mass and moles.

2. Mole Concept: The mole is the SI unit for the amount of substance. One mole contains exactly $6.02214076 imes 10^{23}$ elementary entities (Avogadro's number). The mole concept provides a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and kilograms. It simplifies calculations by allowing us to work with manageable numbers and relate them to real-world quantities. Using moles, we can easily determine the amount of reactants needed and products formed in a chemical reaction.

3. Stoichiometric Coefficients: Stoichiometric coefficients are the numbers placed in front of chemical formulas in a balanced chemical equation. These coefficients represent the relative number of moles of each reactant and product involved in the reaction. For example, in the balanced equation for nitroglycerin decomposition, the coefficient of nitroglycerin is 4, indicating that 4 moles of nitroglycerin are involved in the reaction. Stoichiometric coefficients are essential for determining the molar ratios between different substances in a reaction.

Significance of Balanced Chemical Equations

A balanced chemical equation is essential for stoichiometric calculations because it provides the mole ratios between reactants and products. The balanced equation ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the law of conservation of mass. Without a balanced equation, it is impossible to accurately determine the quantitative relationships between the substances involved in the reaction. Balancing chemical equations often involves adjusting coefficients to achieve atomic balance, and this balanced equation then serves as the foundation for all subsequent stoichiometric calculations. It is a critical first step in solving any quantitative chemistry problem.

Problem Statement and Given Information

In the context of the given reaction, we aim to determine the amount of nitroglycerin ($C_3H_5(NO_3)_3$) that will decompose to yield 120 grams of water ($H_2O$). The balanced chemical equation for the decomposition of nitroglycerin is:

$4 C_3 H_5(NO_3)_3(l) \rightarrow 12 CO_2(g) + 6 N_2(g) + 10 H_2O(g) + O_2(g)$

The key piece of information provided is that 120 grams of water are produced. Our task is to use this information, along with the stoichiometric relationships derived from the balanced equation, to calculate the mass of nitroglycerin required. This involves converting the mass of water to moles, using the mole ratio from the balanced equation to find the moles of nitroglycerin, and then converting moles of nitroglycerin back to grams. Accurate calculations require a clear understanding of these steps, and attention to detail in every conversion and calculation. Before we begin, it’s worth noting the significance of this type of calculation in practical applications. Understanding the quantitative relationships in chemical reactions is vital in various fields, including pharmaceuticals, manufacturing, and environmental science. Now, let’s proceed with the detailed calculations.

Step-by-Step Calculation

Step 1: Calculate the Molar Mass of Water ($H_2O$)

The molar mass of water is the sum of the atomic masses of its constituent elements: hydrogen (H) and oxygen (O). The atomic mass of hydrogen is approximately 1.008 g/mol, and the atomic mass of oxygen is approximately 16.00 g/mol. Water ($H_2O$) consists of two hydrogen atoms and one oxygen atom.

$Molar ext{ }Mass ext{ }of ext{ }H_2O = (2 imes 1.008 ext{ }g/mol) + (1 imes 16.00 ext{ }g/mol)$

$Molar ext{ }Mass ext{ }of ext{ }H_2O = 2.016 ext{ }g/mol + 16.00 ext{ }g/mol$

$Molar ext{ }Mass ext{ }of ext{ }H_2O = 18.016 ext{ }g/mol$

So, the molar mass of water is approximately 18.016 g/mol. This value will be used to convert the given mass of water into moles. Knowing the molar mass is essential for translating between the macroscopic world of grams, which we can measure, and the microscopic world of moles, which relates directly to the number of molecules. This conversion is a fundamental step in stoichiometric calculations. Once we have the number of moles of water, we can then use the balanced chemical equation to determine the corresponding number of moles of nitroglycerin.

Step 2: Convert Grams of Water to Moles

To convert the given mass of water (120 grams) to moles, we use the molar mass of water calculated in Step 1 (18.016 g/mol). The conversion is done using the formula:

$Moles = \frac{Mass}{Molar ext{ }Mass}$

$Moles ext{ }of ext{ }H_2O = \frac{120 ext{ }g}{18.016 ext{ }g/mol}$

$Moles ext{ }of ext{ }H_2O ext{ }\approx ext{ }6.66 ext{ }moles$

Therefore, 120 grams of water is approximately equal to 6.66 moles. This conversion is a critical step because it allows us to work with the mole ratios defined in the balanced chemical equation. Working in moles is essential for comparing the amounts of different substances in a chemical reaction. Now that we know the moles of water produced, we can use the stoichiometry of the reaction to determine the moles of nitroglycerin that reacted to produce this amount of water. This will bring us one step closer to finding the mass of nitroglycerin.

Step 3: Use the Stoichiometric Ratio to Find Moles of Nitroglycerin

From the balanced chemical equation, the stoichiometric ratio between nitroglycerin ($C_3H_5(NO_3)_3$) and water ($H_2O$) is:

$4 ext{ }moles ext{ }of ext{ }C_3H_5(NO_3)_3 : 10 ext{ }moles ext{ }of ext{ }H_2O$

This ratio can be simplified to:

$\frac{4 ext{ }moles ext{ }C_3H_5(NO_3)_3}{10 ext{ }moles ext{ }H_2O}$

Using this ratio, we can find the number of moles of nitroglycerin that decomposed to produce 6.66 moles of water:

$Moles ext{ }of ext{ }C_3H_5(NO_3)_3 = 6.66 ext{ }moles ext{ }H_2O imes \frac{4 ext{ }moles ext{ }C_3H_5(NO_3)_3}{10 ext{ }moles ext{ }H_2O}$

$Moles ext{ }of ext{ }C_3H_5(NO_3)_3 = 6.66 imes 0.4 ext{ }moles$

$Moles ext{ }of ext{ }C_3H_5(NO_3)_3 ext{ }\approx ext{ }2.664 ext{ }moles$

Therefore, approximately 2.664 moles of nitroglycerin decomposed to produce 120 grams (6.66 moles) of water. The stoichiometric ratio is a cornerstone of stoichiometric calculations, allowing us to relate the amounts of different substances in a chemical reaction accurately. With the moles of nitroglycerin calculated, the final step is to convert this value back to grams, providing us with the answer to our problem.

Step 4: Calculate the Molar Mass of Nitroglycerin ($C_3H_5(NO_3)_3$)

The molar mass of nitroglycerin is the sum of the atomic masses of its constituent elements: carbon (C), hydrogen (H), nitrogen (N), and oxygen (O). The atomic masses are approximately:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.008 g/mol
• Nitrogen (N): 14.01 g/mol
• Oxygen (O): 16.00 g/mol

Nitroglycerin ($C_3H_5(NO_3)_3$) consists of 3 carbon atoms, 5 hydrogen atoms, 3 nitrogen atoms, and 9 oxygen atoms (3 nitrate groups, each with 3 oxygen atoms).

$Molar ext{ }Mass ext{ }of ext{ }C_3H_5(NO_3)_3 = (3 imes 12.01 ext{ }g/mol) + (5 imes 1.008 ext{ }g/mol) + (3 imes 14.01 ext{ }g/mol) + (9 imes 16.00 ext{ }g/mol)$

$Molar ext{ }Mass ext{ }of ext{ }C_3H_5(NO_3)_3 = 36.03 ext{ }g/mol + 5.04 ext{ }g/mol + 42.03 ext{ }g/mol + 144.00 ext{ }g/mol$

$Molar ext{ }Mass ext{ }of ext{ }C_3H_5(NO_3)_3 = 227.10 ext{ }g/mol$

So, the molar mass of nitroglycerin is approximately 227.10 g/mol. This value is essential for converting the moles of nitroglycerin to grams. A correct molar mass is crucial for accurate stoichiometric calculations, as it directly impacts the final result. With the molar mass of nitroglycerin now calculated, we can proceed to the final step of finding the mass of nitroglycerin required.

Step 5: Convert Moles of Nitroglycerin to Grams

To convert moles of nitroglycerin (2.664 moles) to grams, we use the molar mass of nitroglycerin calculated in Step 4 (227.10 g/mol). The conversion is done using the formula:

$Mass = Moles imes Molar ext{ }Mass$

$Mass ext{ }of ext{ }C_3H_5(NO_3)_3 = 2.664 ext{ }moles imes 227.10 ext{ }g/mol$

$Mass ext{ }of ext{ }C_3H_5(NO_3)_3 ext{ }\approx ext{ }605.09 ext{ }g$

Therefore, approximately 605.09 grams of nitroglycerin will decompose to give 120 grams of water. This final step completes the calculation, providing the answer to the problem. By converting the moles of nitroglycerin back to grams, we obtain a tangible mass that can be measured in a laboratory setting. This result highlights the practical application of stoichiometry in determining the quantities of reactants and products in chemical reactions.

Final Answer

In conclusion, approximately 605.09 grams of nitroglycerin ($C_3H_5(NO_3)_3$) will decompose to produce 120 grams of water. This result is obtained through a series of stoichiometric calculations, including converting mass to moles, using stoichiometric ratios from the balanced chemical equation, and converting moles back to mass. Understanding and applying these concepts are crucial for solving quantitative problems in chemistry.

Practice Problems

To reinforce your understanding of stoichiometry, here are some practice problems:

1. If 200 grams of nitroglycerin decompose, how many grams of carbon dioxide ($CO_2$) are produced?
2. How many moles of nitrogen gas ($N_2$) are formed when 500 grams of nitroglycerin decompose?
3. If 150 grams of water are produced, what mass of nitroglycerin was initially present?

Working through these problems will help solidify your grasp of stoichiometric principles and enhance your problem-solving skills in chemistry. Remember to follow the same step-by-step approach we used in this guide: balance the equation, convert grams to moles, use mole ratios, and convert back to grams if necessary.

Conclusion

Stoichiometry is a fundamental concept in chemistry that allows us to understand and predict the quantitative relationships in chemical reactions. By mastering the principles of stoichiometry, one can accurately calculate the amounts of reactants and products involved in a chemical reaction. This guide provided a detailed, step-by-step approach to calculating the mass of nitroglycerin required to produce a specific amount of water, illustrating the practical application of stoichiometry. Accurate calculations in chemistry are essential for various applications, including pharmaceutical development, chemical manufacturing, and environmental science. The ability to perform these calculations accurately ensures the correct use of chemicals, the safe execution of experiments, and the efficient production of chemical substances. The problem we addressed in this guide is a prime example of how stoichiometry helps in understanding the quantitative aspects of chemical reactions, contributing significantly to the broader field of chemistry. We hope this guide has helped you better understand the process of stoichiometric calculations and their importance in chemistry.