Problem 45 - General Chemistry: Principles and Modern Applications

Question

Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a \(0.312 \mathrm{g}\) sample yields \(0.458 \mathrm{g} \mathrm{CO}_{2}\) and \(0.374 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The nitrogen content of a \(0.486 \mathrm{g}\) sample is converted to \(0.226 \mathrm{g} \mathrm{N}_{2} .\) What is the empirical formula of dimethylhydrazine?

Step-by-Step Solution

Answer

The empirical formula of Dimethylhydrazine is \( \mathrm{CH}_{4}\mathrm{N}_{2} \).

Step 1: Find moles of Carbon

The carbon that formed in the compound is now in the form of \( \mathrm{CO}_{2} \). One mole of \( \mathrm{CO}_{2} \) contains one mole of carbon. So, using the given mass of the \( \mathrm{CO}_{2} \), the moles of carbon can be found by dividing the mass by the molar mass of \( \mathrm{CO}_{2} \). In this case, that would be \(0.458 / 44.01 = 0.0104 \) moles.

Step 2: Find moles of Hydrogen

The hydrogen that was present in the compound is now in the form of \( \mathrm{H}_{2}\mathrm{O} \). Each mole of \( \mathrm{H}_{2}\mathrm{O} \) contains two moles of hydrogen. So, the moles of hydrogen can be found by first finding the moles of \( \mathrm{H}_{2}\mathrm{O} \) (0.374 / 18.015) and then multiplying by 2 to get the number of moles of hydrogen. This equation comes to about \(0.0414 \) moles.

Step 3: Find moles of Nitrogen

The nitrogen content is present as \( \mathrm{N}_{2} \), one mole of which contains two moles of nitrogen. So, in order to find the moles of nitrogen, the weight of the \( \mathrm{N}_{2} \) is divided by the molar mass of nitrogen and then multiplied by 2. Here, that is \( (0.226 / 28.02) * 2 = 0.0161 \) moles.

Step 4: Determine the Empirical Formula

Normalize the obtained moles to get simplest whole number ratio, by dividing each value by the smallest value among the three. Here, that would mean dividing \(0.0104, 0.0414 \), and \(0.0161 \) by \(0.0104 \). This results in a mole ratio of \( C:1, H:4, N:2 \) denoting the empirical formula to be \( \mathrm{CH}_{4}\mathrm{N}_{2} \)

Step 5: Confirm the Answer

The empirical formula determines the simplest whole-number ratio of atoms in a compound. In this case, \( \mathrm{CH}_{4}\mathrm{N}_{2} \) reflects a compound that is composed of 1 atom of carbon, 4 atoms of hydrogen, and 2 atoms of nitrogen. Therefore, the empirical formula of Dimethylhydrazine is \( \mathrm{CH}_{4}\mathrm{N}_{2} \).

Key Concepts

Dimethylhydrazine

Dimethylhydrazine is an interesting compound primarily used as a rocket fuel. It consists of carbon, hydrogen, and nitrogen. In the context of chemistry, understanding its empirical formula helps us determine the simplest ratio of these atoms in the compound. By isolating and analyzing each element in reactions, scientists can unravel complex structures and verify the proportions of each type of atom present.
Additionally, its use as a rocket fuel underscores the importance of chemical reactions and stoichiometric calculations to ensure precision in fuel mixtures, potentially impacting the success of missions.

• Used in rocket propellants
• Comprises carbon, hydrogen, and nitrogen
• Empirical formula reveals basic atomic ratios

This fundamental understanding balances function with safety and efficiency, showcasing chemistry's role in real-world applications.

Molecular Composition

Molecular composition refers to the types and quantities of atoms present in a compound. In the case of dimethylhydrazine, we are interested in understanding how its atoms are arranged in a formula. By figuring out its empirical formula, experiments such as combustion in the presence of excess oxygen reveal these atomic counts by converting them into measurable by-products like carbon dioxide and water.
The collected data is then used to calculate the amount of each element, helping us to map the entire composition.

• This exercise involves calculating moles of carbon, hydrogen, and nitrogen
• From the mass of these compounds, we derive atomic counts

Results lead to determining the simplest empirical formula, providing a more straightforward view of its molecular identity.

Chemical Reactions

Chemical reactions transform substances. The combustion of dimethylhydrazine in oxygen is a clear example. Here, carbon forms carbon dioxide, hydrogen forms water, and nitrogen forms nitrogen gas. Each transformation helps us capture and quantify elemental moles through accurate measurement.
This process is vital because it helps translate complex molecular compositions into simpler forms that are easier to analyze. Through these transformations, the true nature of dimethylhydrazine becomes evident, aiding in calculating its empirical formula systematically.

• Combustion transforms original substances into gases
• It allows for calculating elemental moles

These reactions also reveal stability and potential uses in different applications, crucial for advancing material sciences and related fields.

Stoichiometry

Stoichiometry is the framework that chemists use to measure relative quantities in reactions. It provides the math behind balancing reactions to determine the amount of reactants and products formed. In our example involving dimethylhydrazine, stoichiometry allows us to calculate the quantities of each atom in the compound by analyzing combustion products.
Using the ratios derived from stoichiometric calculations ensures precise chemical equations, confirming the understanding of dimethylhydrazine’s empirical formula. This concept is not just academic; it has significant real-world applications in industries where precise chemical ratios are vital.

• Involves balancing chemical equations
• Helps calculate reactant and product quantities
• Ensures formula precision

Understanding stoichiometry deepens our grasp of chemistry’s foundational principles, leading to accurate exploration and practical synthesis of compounds.