The nitrogen content of a .2476 g sample of a protein was determined by the Kjeldahl method. The NH3 produced with the whoosh was distilled into exactly 50.00mL of HCl in the receiving flask, which was in excess of the amount required to capture the ammonia gas. The excess HCl that was left over required 32.76 mL of .1234 M NaOH for neutralization.
In a preliminary lab, the HCl in the receiving flask had been previously standardized by neutralizing exactly 25.00mL of it with 21.15 mL of the same .1234 M NaOH.
What was the weight % nitrogen in the protein sample?
Just for the record what is a whoosh in a Kjeldahl determination.
Calculate the M of the HCl.Information to do that is in the second paragraph.
mols HCl initially = M x L = ?
mols excess HCl = M x L = ?
mols HCl used by the NH3 = difference
grams NH3 = mols NH3 x molar mass NH3
Then %NH3 = (g NH3/mass sample)*100 = ?
To find the weight % nitrogen in the protein sample, we need to calculate the number of moles of nitrogen in the sample and use it to determine the weight % nitrogen.
Let's break down the given information and the steps to solve the problem:
1. The nitrogen content of the protein sample was determined by the Kjeldahl method. This method involves converting the nitrogen in the protein to ammonia (NH3), which is captured in the receiving flask.
2. The NH3 produced in the reaction is distilled into a receiving flask containing excess HCl. This means that all the ammonia reacts with the HCl, leaving excess HCl behind.
3. The excess HCl in the receiving flask required 32.76 mL of 0.1234 M NaOH for neutralization. This information is useful for determining the amount of excess acid present.
4. In a preliminary lab, the HCl in the receiving flask was standardized by neutralizing exactly 25.00 mL of it with 21.15 mL of the same 0.1234 M NaOH. This information helps us calculate the concentration of HCl in the flask.
To solve the problem, we can follow these steps:
Step 1: Calculate the concentration of HCl in the flask:
Based on the preliminary lab data, we can use the neutralization reaction between HCl and NaOH to find the concentration of HCl in the receiving flask.
Given:
Volume of HCl in the flask (standardized) = 25.00 mL
Volume of NaOH used for neutralization = 21.15 mL
Concentration of NaOH = 0.1234 M
Using the reaction: HCl + NaOH → NaCl + H2O
We can set up a proportion:
(25.00 mL HCl) / (21.15 mL NaOH) = (x M HCl) / (0.1234 M NaOH)
Solving for x gives us the concentration of HCl in the receiving flask.
Step 2: Calculate the moles of NaOH used to neutralize the excess HCl:
From the given information, we know that 32.76 mL of 0.1234 M NaOH is used to neutralize the excess HCl in the receiving flask. This allows us to calculate the moles of NaOH used.
Step 3: Calculate the moles of excess HCl neutralized by NaOH:
Since the reaction between HCl and NaOH is 1:1, the moles of excess HCl neutralized will be equal to the moles of NaOH used.
Step 4: Calculate the moles of HCl in the receiving flask:
Using the concentration of HCl in the flask from Step 1 and the volume of NaOH used to neutralize the excess HCl, we can calculate the moles of HCl present in the receiving flask.
Step 5: Calculate the moles of NH3 produced:
Since the moles of HCl and NH3 produced in the Kjeldahl method are equal, we can use the moles of HCl in the receiving flask from Step 4 as the moles of NH3 produced.
Step 6: Calculate the moles of nitrogen in the protein sample:
Using the molar mass of nitrogen (N2), we can convert the moles of NH3 produced to moles of nitrogen.
Step 7: Calculate the weight % nitrogen in the protein sample:
Finally, we can calculate the weight % nitrogen in the protein sample by dividing the weight of nitrogen by the weight of the protein sample, and then multiplying by 100.
By following these steps and using the given information, you can calculate the weight % nitrogen in the protein sample.