Your body deals with excess nitrogen by excreting it in the form of urea, NH2CONH2. The reaction producing it is the combination of arginine (C6H14N4O2) with water to give urea and ornithine (C5H12N2O2).
C6H14N4O2 + H2O NH2CONH2 + C5H12N2O2
[Molar masses: 174.2 18.02 60.06 132.2]
If you excrete 95 mg of urea, what quantity of arginine must have been used? What quantity of ornithine must have been produced?
i got .276g of arginine and .209g of ornithine
You can use this simple stoichiometry example I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
O OK
.0276
To determine the quantity of arginine that must have been used to produce 95 mg of urea, we can set up a proportion based on the balanced chemical equation and the molar masses given.
Let's calculate the moles of urea produced first:
Mass of urea = 95 mg = 0.095 g
Molar mass of urea = 60.06 g/mol
Moles of urea = Mass of urea / Molar mass of urea
Moles of urea = 0.095 g / 60.06 g/mol
Next, we can set up a mole ratio between arginine and urea using the coefficients from the balanced chemical equation:
1 mole of C6H14N4O2 produces 1 mole of NH2CONH2
Since the mole ratio is 1:1, the moles of arginine used must be equal to the moles of urea produced.
Therefore, the moles of arginine used = Moles of urea = 0.095 g / 60.06 g/mol
To calculate the mass of arginine used, we can use the molar mass of arginine (174.2 g/mol):
Mass of arginine used = Moles of arginine used x Molar mass of arginine
Mass of arginine used = (0.095 g / 60.06 g/mol) x 174.2 g/mol
Now we can plug in the values and calculate:
Mass of arginine used = (0.095 g / 60.06 g/mol) x 174.2 g/mol
Mass of arginine used = 0.275 g
Therefore, the quantity of arginine that must have been used is 0.275 g.
Similarly, we can calculate the quantity of ornithine produced using the same method.
Moles of ornithine produced = Moles of urea (since the mole ratio is 1:1)
To calculate the mass of ornithine produced, we can use the molar mass of ornithine (132.2 g/mol):
Mass of ornithine produced = Moles of ornithine produced x Molar mass of ornithine
Mass of ornithine produced = (0.095 g / 60.06 g/mol) x 132.2 g/mol
Mass of ornithine produced = 0.211 g
Therefore, the quantity of ornithine that must have been produced is 0.211 g.