Chapter 3, Problem 3.138P

(a)

Interpretation Introduction

Interpretation:

The mass percentage of nitrogen in urea, in arginine, and in ornithine is to be calculated.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture multiplied by 100.

The formula to calculate the mass percent is as follows:

  $\text{Mass}\text{percent}=\left(\frac{\text{mass}\text{of}\text{solute}}{\text{mass}\text{of}\text{solution}}\right)100$

The formula to calculate mass of a substance is as follows:

  $\text{Mass of a substance}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

(b)

Interpretation Introduction

Interpretation:

The mass of nitrogen that can be excreted as urea when $135.2\text{ g}$ of ornithine is produced is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\frac{\text{given mass}\text{of compound}\left(\text{g}\right)}{\text{molecular mass of compound}\left(\text{g/mol}\right)}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.