Problem 64 - Chemistry The Molecular Science

Question

Nitroglycerin decomposes violently according to the equation $$4 \mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(\ell) \longrightarrow$$ \(12 \mathrm{CO}_{2}(\mathrm{~g})+10 \mathrm{H}_{2} \mathrm{O}(\ell)+6 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) Calculate the mass (in grams) of each gaseous product produced from \(1.00 \mathrm{~g}\) nitroglycerin.

Step-by-Step Solution

Answer

Masses: CO2 = 0.581 g, N2 = 0.185 g, O2 = 0.0352 g.

Step 1: Find the Molar Mass of Nitroglycerin

The chemical formula for nitroglycerin is \( \mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{NO}_{3})_{3} \). To find its molar mass, sum the atomic masses: Carbon (C) has an atomic mass of about 12.01, Hydrogen (H) is about 1.01, Nitrogen (N) is about 14.01, and Oxygen (O) is about 16.00. Calculating, we get: \(3(12.01) + 5(1.01) + 3(14.01 + 3 \times 16) = 227.09 \, \text{g/mol} \).

Step 2: Determine Moles of Nitroglycerin

Use the molar mass to find the number of moles of nitroglycerin in 1.00 g. Using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \), we get: \[ \frac{1.00 \, \text{g}}{227.09 \, \text{g/mol}} = 0.00440 \, \text{mol}. \]

Step 3: Apply Stoichiometry to Find Moles of Gaseous Products

According to the balanced equation, 4 moles of nitroglycerin produce 12 moles of \(\mathrm{CO}_{2}\), 6 moles of \(\mathrm{N}_{2}\), and 1 mole of \(\mathrm{O}_{2}\). Calculate moles of each gas: - \(\mathrm{CO}_{2}\): \(0.00440 \, \text{mol} \times \frac{12}{4} = 0.0132 \, \text{mol}\)- \(\mathrm{N}_{2}\): \(0.00440 \, \text{mol} \times \frac{6}{4} = 0.00660 \, \text{mol}\)- \(\mathrm{O}_{2}\): \(0.00440 \, \text{mol} \times \frac{1}{4} = 0.00110 \, \text{mol}\)

Step 4: Calculate Mass of Each Gaseous Product

Use the moles of each gas and their molar masses to find the mass:- \(\mathrm{CO}_{2}:\) \(0.0132 \, \text{mol} \times 44.01 \, \text{g/mol} = 0.581 \, \text{g}\)- \(\mathrm{N}_{2}:\) \(0.00660 \, \text{mol} \times 28.02 \, \text{g/mol} = 0.185 \, \text{g}\)- \(\mathrm{O}_{2}:\) \(0.00110 \, \text{mol} \times 32.00 \, \text{g/mol} = 0.0352 \, \text{g}\)

Key Concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They display the reactants and products, showing the transformation of substances.
This helps in understanding the stoichiometry of the reactions, which involves balancing the number of atoms for each element on both sides. Here, for nitroglycerin's decomposition, the balanced equation is crucial.
It ensures that we accurately relate moles of reactants to moles of products.

• The balance of the equation reveals quantitative relationships.
• Comes in handy for determining the proportions of substances involved.

For nitroglycerin, the coefficients indicate that 4 molecules of nitroglycerin yield 12 molecules of carbon dioxide, along with water, nitrogen, and oxygen.
It informs us that, to fully decompose nitroglycerin, specific amounts of products are consistently formed. This helps in stoichiometric calculations, like those shown in the exercise.

Molar Mass Calculation

Calculating the molar mass is a vital step in stoichiometry. With nitroglycerin as an example, we must calculate its molar mass to determine how many moles are in a given mass.
Use the periodic table to find the atomic masses:

• Carbon: 12.01 g/mol
• Hydrogen: 1.01 g/mol
• Nitrogen: 14.01 g/mol
• Oxygen: 16.00 g/mol

For nitroglycerin, the molar mass calculation combines these values according to its formula, \( \mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{NO}_{3})_{3} \).
By adding the atomic masses: \(3(12.01) + 5(1.01) + 3(14.01 + 3 \times 16.00) = 227.09 \, \text{g/mol} \). Molar mass allows us to convert mass to moles, which is a stepping stone for further stoichiometric calculations.

Nitroglycerin Decomposition

Nitroglycerin is known for its explosive decomposition reaction, which involves a rapid breakdown into simpler substances.
This reaction releases gases such as carbon dioxide, nitrogen, and oxygen. The decomposition is represented in the balanced equation:

• Combustion of nitroglycerin also produces water as a byproduct.
• Understanding this process involves examining each gas product's amount generated.
• It involves calculating moles and converting these to mass.

Through stoichiometry, we find that decomposing 1.00 g of nitroglycerin produces measurable amounts of each gas.
For instance, it produces 0.0132 mol of \(\mathrm{CO}_2\), which converts to a mass of 0.581 g using the previous steps.
Knowing these specifics is vital for safety and handling, as well as for theoretical and practical applications of nitroglycerin.