Your body deals with excess nitrogen by excreting it in the...

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?

Key Concepts

Chemical Reaction Balancing

Balancing chemical reactions ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the conservation of mass. This balance provides the mole ratios needed to carry out stoichiometric calculations accurately.

Molar Mass Calculation

Molar mass calculation is the process of determining the mass of one mole of a chemical compound by summing the atomic masses of its constituent elements. This conversion factor is crucial for translating between moles and grams in stoichiometric problems.

Mole Concept

The mole concept serves as a bridge between the microscopic scale of atoms and molecules and the macroscopic scale of grams and liters. It allows the conversion between mass and the number of particles, which is essential for performing quantitative chemical calculations.

Stoichiometry

Stoichiometry involves using the mole ratios from a balanced chemical equation to relate the quantities of reactants and products. It enables one to determine how much of a reactant is needed to produce a given amount of product, or vice versa.

Transcript

00:01 Okay.

00:03 This is about, ooh, i erase something there.

00:07 I'm not even sure how to pronounce a couple of these.

00:09 Urea, arginine and ornithine.

00:13 We have these substances.

00:16 Our body deals with excess nitrogen by excreting the nitrogen in the form of urea, which is listed in the first product in this equation.

00:31 It also gives the ornithine, and i'm not even sure what ornithine is, and the argonine reacts with water.

00:45 So we're given the following information.

00:48 If we excrete 95 milligrams of urea, what mass of argonine must have been used, and what mass of ornithine must have been produced? i'm going to go look at the top of this and see if i have more questions up on the top of that page.

01:10 I do not.

01:11 So the mass of argonine, this will be easy.

01:14 We've got to remember to change our grams, our milligrams to grams.

01:20 And fortunately, when i plugged these chemical formulas into the molar mass calculator, they all came up.

01:28 So i didn't have to figure out any of these.

01:34 We're going to be doing two mass mass calculations.

01:39 And the mass calculation, 95 milligrams, move the decimal 3 to the left.

01:54 These are what i'm going to call these three substances.

02:05 The molar mass of a is 60 .06 grams per mole.

02:11 That's the rg.

02:12 Wait a minute here.

02:14 Oh, that's urea.