Question

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Short answer

(a) 7.04 g of oxygen can be obtained. (b) Percent yield is 93.0%.

Step-by-step solution

Calculate Molar Mass of Nitroglycerin

The molar mass of nitroglycerin \( \left( \text{C}_3 \text{H}_5 \text{N}_3 \text{O}_9 \right) \) is calculated by summing the atomic masses of its constituent atoms: \( (3 \times 12.01) + (5 \times 1.01) + (3 \times 14.01) + (9 \times 16.00) = 227.09 \text{ g/mol} \).

Calculate Moles of Nitroglycerin

Using the molar mass, calculate the moles of nitroglycerin in \( 200 \text{ g} \): \( \frac{200 \text{ g}}{227.09 \text{ g/mol}} = 0.880 \text{ mol} \).

Determine Moles of Oxygen Produced

From the balanced equation, 4 moles of nitroglycerin yield 1 mole of \( \text{O}_2 \). Therefore, \( 0.880 \text{ mol} \) of nitroglycerin will produce \( \frac{0.880}{4} = 0.220 \text{ mol} \) of \( \text{O}_2 \).

Convert Moles of Oxygen to Grams

Calculate the mass of \( \text{O}_2 \) produced: \( 0.220 \text{ mol} \times 32.00 \text{ g/mol} = 7.04 \text{ g} \).

Calculate Percent Yield

The percent yield is calculated as \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \). The actual yield is given as \( 6.55 \text{ g} \), and the theoretical yield is \( 7.04 \text{ g} \). Thus, the percent yield is \( \frac{6.55}{7.04} \times 100\% = 93.0\% \).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, are transformed into different substances, known as products. In the decomposition of nitroglycerin, the chemical reaction is represented by the equation: \[ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}_{3} \mathrm{O}_{9} \rightarrow 6 \mathrm{N}_{2} + 12 \mathrm{CO}_{2} + 10 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \] This equation shows that from four molecules of nitroglycerin, you obtain six molecules of nitrogen gas, twelve molecules of carbon dioxide, ten molecules of water, and one molecule of oxygen gas. The balanced chemical equation illustrates the principle of the conservation of matter, which states that atoms are neither created nor destroyed in a chemical reaction.

• The coefficients (e.g., 4, 6, 12) denote the number of moles of each substance involved in the reaction.
• Balancing chemical equations ensures that the mass and charge are balanced, meaning what you start with should be what you end with.

Molar Mass

Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). Understanding molar mass helps in converting between the mass of a substance and the amount of substance in moles, which is essential in stoichiometric calculations. For nitroglycerin \((\text{C}_3 \text{H}_5 \text{N}_3 \text{O}_9)\), calculating the molar mass involves adding up the atomic masses of all the atoms in one molecule:

• Carbon (C): 12.01 g/mol \, \times 3 = 36.03 g/mol
• Hydrogen (H): 1.01 g/mol \, \times 5 = 5.05 g/mol
• Nitrogen (N): 14.01 g/mol \, \times 3 = 42.03 g/mol
• Oxygen (O): 16.00 g/mol \, \times 9 = 144.00 g/mol

Adding these gives a total molar mass of 227.09 g/mol for nitroglycerin. Knowing the molar mass allows us to convert the weight of a substance into moles, which is vital for calculating how much of each product will be formed during a reaction.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, expressed as the ratio of the actual yield to the theoretical yield, multiplied by 100. Theoretical yield is the maximum amount of product that could be formed from the given amounts of reactants, calculated using stoichiometry. However, in practical scenarios, the actual yield is often less due to various factors like incomplete reactions, side reactions, or loss of product during recovery. In the nitroglycerin decomposition exercise, the theoretical yield of oxygen was calculated to be 7.04 grams. Given that the actual yield obtained from the experiment was 6.55 grams, the percent yield is given by the formula: \[ \text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\% = \left(\frac{6.55}{7.04}\right) \times 100\% = 93.0\% \] Some possible reasons for a percent yield less than 100% include:

• Incomplete reactions where not all reactants converted to products.
• Loss of product during separation and purification processes.
• Side reactions that produce other unwanted products.

Calculating the percent yield allows chemists to understand the practicality and efficiency of their reactions.

Gases in Chemical Reactions

In many chemical reactions, especially in the context of explosives like nitroglycerin, the production of gases is an important feature. Gaseous products in reactions possess significant volume expansion, often leading to increased pressure and hence, dramatic effects such as explosions. For the decomposition of nitroglycerin, a large volume of gases such as nitrogen (\(\text{N}_2\)), carbon dioxide (\(\text{CO}_2\)), and oxygen (\(\text{O}_2\)) is produced. These gases are initially formed under high pressure and temperature, causing them to expand rapidly. This expansion is what translates the chemical reaction into a mechanical explosion.

• Gases have high kinetic energy and tend to spread out quickly, filling any available space.
• The rapid increase in volume from gas formation contributes to the explosive power of many reactions.
• Balancing chemical equations is crucial to predicting the amount of gas produced in a reaction, which in turn allows for accurate predictions about the reaction's outcome and safety measures needed.