Problem 50 - Chemistry and Chemical Reactivity

Question

Your body deals with excess nitrogen by excreting it in the form of urea, \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). The reaction producing it is the combination of arginine \(\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\right)\) with water to give urea and ornithine \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\) Arginine Urea Ornithine If you excrete 95 mg of urea, what mass of arginine must have been used? What mass of ornithine must have been produced?

Step-by-Step Solution

Answer

0.2755 g of arginine used; 0.209 g of ornithine produced.

Step 1: Find Molar Masses

Calculate the molar mass of each compound involved in the reaction. For urea, \( \mathrm{NH}_{2}\mathrm{CONH}_{2} \):- N: 14 g/mol, H: 1 g/mol, C: 12 g/mol, O: 16 g/mol- \( 2(14) + 4(1) + 12 + 16 = 60 \; \mathrm{g/mol} \)For arginine, \( \mathrm{C}_{6}\mathrm{H}_{14}\mathrm{N}_{4}\mathrm{O}_{2} \):- C: 12 g/mol, H: 1 g/mol, N: 14 g/mol, O: 16 g/mol- \( 6(12) + 14(1) + 4(14) + 2(16) = 174 \; \mathrm{g/mol} \)For ornithine, \( \mathrm{C}_{5}\mathrm{H}_{12}\mathrm{N}_{2}\mathrm{O}_{2} \):- \( 5(12) + 12(1) + 2(14) + 2(16) = 132 \; \mathrm{g/mol} \)

Step 2: Calculate Moles of Urea

Convert the mass of urea excreted into moles using its molar mass.Given mass of urea = 95 mg = 0.095 g.Moles of urea = \( \frac{\text{mass}}{\text{molar mass}} = \frac{0.095}{60} = 0.0015833 \; \mathrm{mol} \)

Step 3: Use Stoichiometry for Arginine

Use the reaction stoichiometry to find moles of arginine needed. The reaction ratio is 1:1. Therefore, moles of arginine = moles of urea = 0.0015833 mol.

Step 4: Calculate Mass of Arginine

Convert the moles of arginine to mass using its molar mass.Mass of arginine = moles \( \times \) molar mass = \( 0.0015833 \times 174 = 0.2755 \; \mathrm{g} \)

Step 5: Use Stoichiometry for Ornithine

Determine the moles of ornithine produced using the stoichiometry of the reaction. Since the equation is balanced, the moles of ornithine produced equal the moles of arginine used, which is 0.0015833 mol.

Step 6: Calculate Mass of Ornithine

Convert the moles of ornithine to mass using its molar mass.Mass of ornithine = moles \( \times \) molar mass = \( 0.0015833 \times 132 = 0.209 \; \mathrm{g} \)

Key Concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that helps us relate the mass of a substance to the amount of substance in moles. It's defined as the mass of one mole of a given element or compound, typically expressed in grams per mole (g/mol). Knowing the molar mass allows us to convert between the mass of a substance and the amount in moles, which is crucial for any stoichiometry calculations.

To calculate the molar mass of a compound, you add up the atomic masses of all the atoms in its formula. For instance:

• Urea (\( \text{NH}_2\text{CONH}_2 \)): Molar mass = \( 2(14) + 4(1) + 12 + 16 = 60 \; \text{g/mol} \)
• Arginine (\( \text{C}_6\text{H}_{14}\text{N}_4\text{O}_2 \)): Molar mass = \( 6(12) + 14(1) + 4(14) + 2(16) = 174 \; \text{g/mol} \)
• Ornithine (\( \text{C}_5\text{H}_{12}\text{N}_2\text{O}_2 \)): Molar mass = \( 5(12) + 12(1) + 2(14) + 2(16) = 132 \; \text{g/mol} \)

By understanding molar masses, we can easily transition from the amount of substance to its mass, which is vital for carrying out chemical reactions accurately in a lab.

Chemical Reactions

Chemical reactions are processes where substances, called reactants, are transformed into different substances known as products. This transformation involves rearranging the atoms of the reactants to form new products, often releasing or absorbing energy in the process.

In our example, the chemical reaction involves the conversion of arginine and water into urea and ornithine. The balanced chemical equation is:\[\text{C}_6\text{H}_{14}\text{N}_4\text{O}_2 + \text{H}_2\text{O} \rightarrow \text{NH}_2\text{CONH}_2 + \text{C}_5\text{H}_{12}\text{N}_2\text{O}_2\]This equation shows a 1:1 conversion where one mole of arginine reacts with one mole of water to produce one mole of urea and one mole of ornithine. These balanced equations are critical in ensuring that the stoichiometry - the calculation of reactants and products in chemical reactions - is accurate. Understanding the stoichiometry allows us to predict quantities of substances consumed and produced in a given reaction, which is essential for lab work and industrial processes.

Organic Chemistry

Organic chemistry is the branch of chemistry that deals with the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which include not just hydrocarbons but also compounds with any number of other elements, including hydrogen, nitrogen, oxygen, halogens, phosphorus, silicon, and sulfur.

In this exercise, we encounter substances such as arginine, urea, and ornithine, which are organic compounds. Each of these compounds contains carbon atoms, which is a primary feature of organic molecules. There are:

• Arginine: An amino acid that plays a significant role in nitrogen metabolism and is an essential component for the production of urea.
• Urea: A simple organic compound and primary nitrogenous waste product, soluble in water, crucial for excretion processes.
• Ornithine: A non-proteinogenic amino acid that is central to the urea cycle, responsible for converting ammonia into urea in the liver.

Understanding organic chemistry principles allows us to see how these molecules interact in biological systems and the significance of structures such as carbon chains, functional groups, and molecular geometry in their behavior and reactivity.