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Chapter 3: Problem 109

The explosive nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) has also been used as a drug to treat heart patients to relieve pain (angina pectoris). We now know that nitroglycerin produces nitric oxide (NO), which causes muscles to relax and allows the arteries to dilate. If each nitroglycerin molecule releases one NO per atom of \(\mathrm{N},\) calculate the mass percent of NO available from nitroglycerin.

Short Answer

Expert verified
The mass percent of Nitric Oxide (NO) available from Nitroglycerin (C3H5N3O9) is approximately 39.64%.

Step by step solution

01

Calculate the molar mass of Nitroglycerin and Nitric Oxide

First, we need to calculate the molar mass of Nitroglycerin (\(C_{3}H_{5}N_{3}O_{9}\)) and Nitric Oxide (NO). The atomic masses of Carbon (C), Hydrogen (H), Nitrogen (N), and Oxygen (O) are approximately 12.01 g/mol, 1.01 g/mol, 14.01 g/mol, and 16.00 g/mol respectively. Thus, the molar mass of Nitroglycerin can be calculated as follows: \(molar mass_{C3H5(NO3)3} = 3*12.01 g/mol + 5*1.01 g/mol + 3*14.01 g/mol + 9*16.00 g/mol = 227.09 g/mol \)For Nitric Oxide, the molar mass is calculated as follows:\(molar mass_{NO} = 1*14.01 g/mol + 1*16.00 g/mol = 30.01 g/mol \)
02

Calculate the mass of Nitric Oxide produced From One Molecule Of Nitroglycerin

Since each Nitroglycerin molecule releases one Nitric Oxide molecule per atom of Nitrogen, and Nitroglycerin has three Nitrogen atoms, we can conclude that each Nitroglycerin molecule will yield three Nitric Oxide molecules.The total mass of Nitric Oxide produced is calculated as follows:\(mass_{NO} = 3 * molar mass_{NO} = 3 * 30.01 g/mol = 90.03 g/mol\)
03

Calculate the mass percent of Nitric Oxide

With the total mass of Nitroglycerin and the Nitric Oxide produced, we can now calculate the mass percent of Nitric Oxide available from Nitroglycerin. Applying the formula for calculating percent composition:\(mass percent_{NO} = \frac{mass_{NO}}{molar mass_{C3H5(NO3)3}} * 100 = \frac{90.03 g/mol}{227.09 g/mol} * 100 = 39.64%\)

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