Problem 81 - 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 \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\right)\). $$\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}$$ If you excrete \(95 \mathrm{mg}\) of urea, what mass of arginine must have been used? What mass of ornithine must have been produced?

Step-by-Step Solution

Answer

0.2754 g of arginine was used; 0.2089 g of ornithine was produced.

Step 1: Calculate Molar Mass of Urea

First, we calculate the molar mass of urea, \( \mathrm{NH}_2 \mathrm{CONH}_2 \). - Nitrogen (N): 14.01 g/mol \times 2 = 28.02 g/mol- Carbon (C): 12.01 g/mol \times 1 = 12.01 g/mol- Oxygen (O): 16.00 g/mol \times 1 = 16.00 g/mol- Hydrogen (H): 1.01 g/mol \times 4 = 4.04 g/molThus, the molar mass of \( \mathrm{NH}_2 \mathrm{CONH}_2 \) is:\[ 28.02 + 12.01 + 16.00 + 4.04 = 60.07 \text{ g/mol} \]

Step 2: Calculate Moles of Urea

Convert the mass of excreted urea (95 mg) to grams and then find the number of moles using its molar mass.\[ 95 \text{ mg} = 0.095 \text{ g} \]Now, calculate the moles:\[ \text{Moles of urea} = \frac{0.095 \text{ g}}{60.07 \text{ g/mol}} \approx 0.00158 \text{ mol} \]

Step 3: Moles Conversion to Arginine

To find the moles of arginine needed, use the balanced chemical equation where 1 mol of arginine produces 1 mol of urea. Therefore, \( 0.00158 \) mol of arginine is needed.

Step 4: Calculate Mass of Arginine Used

Calculate the molar mass of arginine, \( \mathrm{C}_6 \mathrm{H}_{14} \mathrm{N}_4 \mathrm{O}_2 \).- Carbon (C): 12.01 g/mol \times 6 = 72.06 g/mol- Hydrogen (H): 1.01 g/mol \times 14 = 14.14 g/mol- Nitrogen (N): 14.01 g/mol \times 4 = 56.04 g/mol- Oxygen (O): 16.00 g/mol \times 2 = 32.00 g/molThus, the molar mass of arginine is:\[ 72.06 + 14.14 + 56.04 + 32.00 = 174.24 \text{ g/mol} \]Calculate the mass of arginine used:\[ \text{Mass} = 0.00158 \text{ mol} \times 174.24 \text{ g/mol} \approx 0.2754 \text{ g} \]

Step 5: Moles Conversion to Ornithine

From the balanced equation, the same amount of moles of ornithine is produced as the urea or arginine. Therefore, \( 0.00158 \) mol of ornithine is produced.

Step 6: Calculate Mass of Ornithine Produced

Calculate the molar mass of ornithine, \( \mathrm{C}_5 \mathrm{H}_{12} \mathrm{N}_2 \mathrm{O}_2 \).- Carbon (C): 12.01 g/mol \times 5 = 60.05 g/mol- Hydrogen (H): 1.01 g/mol \times 12 = 12.12 g/mol- Nitrogen (N): 14.01 g/mol \times 2 = 28.02 g/mol- Oxygen (O): 16.00 g/mol \times 2 = 32.00 g/molThus, the molar mass of ornithine is:\[ 60.05 + 12.12 + 28.02 + 32.00 = 132.19 \text{ g/mol} \]Calculate the mass of ornithine produced:\[ \text{Mass} = 0.00158 \text{ mol} \times 132.19 \text{ g/mol} \approx 0.2089 \text{ g} \]

Key Concepts

Chemical Reaction

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. In the given reaction, arginine reacts with water to form two new compounds: urea and ornithine. This type of reaction is called a hydrolysis reaction, where water is used to break down a compound. The overall balanced chemical equation is:

• Reactants: Arginine (\(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\)) and water
• Products: Urea (\(\mathrm{NH}_{2} \mathrm{CONH}_{2}\)) and ornithine (\(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\))

Each element's atoms are conserved during the reaction, illustrating the principle of mass conservation. This means that the number of atoms for each element is the same on both sides of the equation.

Molar Mass

Molar mass is a crucial concept in stoichiometry, as it allows us to convert between the mass of a substance and the amount of substance in moles. The molar mass of a compound is calculated by summing the atomic masses of all the atoms present in one molecule of the compound. It is expressed in grams per mole (g/mol).
For example, to find the molar mass of urea \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\), one must consider the atomic masses of nitrogen, carbon, oxygen, and hydrogen:

• Nitrogen (N): \(14.01 \text{ g/mol} \times 2 = 28.02 \text{ g/mol}\)
• Carbon (C): \(12.01 \text{ g/mol} \)
• Oxygen (O): \(16.00 \text{ g/mol}\)
• Hydrogen (H): \(1.01 \text{ g/mol} \times 4 = 4.04 \text{ g/mol}\)

This takes us to a total of \(60.07 \, \text{g/mol}\) for urea. Calculating the molar mass allows us to understand how much of a substance is involved in a chemical reaction.

Arginine

Arginine is an amino acid, symbolized as \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\), and plays a key role in the body's metabolism. In the given chemical reaction, a mole of arginine reacts with water to form urea and ornithine. To determine the amount of arginine used or needed in the reaction, you need to first calculate its molar mass:

• Carbon (C): \(12.01 \text{ g/mol} \times 6 = 72.06 \text{ g/mol}\)
• Hydrogen (H): \(1.01 \text{ g/mol} \times 14 = 14.14 \text{ g/mol}\)
• Nitrogen (N): \(14.01 \text{ g/mol} \times 4 = 56.04 \text{ g/mol}\)
• Oxygen (O): \(16.00 \text{ g/mol} \times 2 = 32.00 \text{ g/mol}\)

The total molar mass of arginine is \(174.24 \, \text{g/mol}\). This information helps us calculate how much arginine is consumed when a specific amount of urea is produced.

Ornithine

Ornithine, represented as \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\), is another product in the reaction and it is formed in a 1:1 molar ratio with urea. To calculate the mass of ornithine produced, we first need its molar mass, determined by adding the masses of its constituent atoms:

• Carbon (C): \(12.01 \text{ g/mol} \times 5 = 60.05 \text{ g/mol}\)
• Hydrogen (H): \(1.01 \text{ g/mol} \times 12 = 12.12 \text{ g/mol}\)
• Nitrogen (N): \(14.01 \text{ g/mol} \times 2 = 28.02 \text{ g/mol}\)
• Oxygen (O): \(16.00 \text{ g/mol} \times 2 = 32.00 \text{ g/mol}\)

The molar mass totals \(132.19 \, \text{g/mol}\). This calculation helps us understand the amount of ornithine produced in relation to the amount of urea generated, providing insights into the conservation and transformation of matter in this chemical process.