What Is the Empirical Formula for Tetrodotoxin?

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Discussion Overview

The discussion revolves around determining the empirical formula for tetrodotoxin based on combustion analysis data and calculating the lethal dose (LD50) for a person. It includes both theoretical and practical aspects of chemical analysis and calculations related to toxicology.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the combustion analysis results and seeks to determine the empirical formula for tetrodotoxin, initially arriving at C11H17N5O6 instead of the expected C11H17N3O8.
  • Another participant suggests that showing the combustion reaction could help identify potential mistakes in the calculations.
  • Some participants agree with the initial calculation of N5, indicating a possible misinterpretation of the data.
  • A later reply indicates that the teacher made an error in one of the masses, confirming that the correct empirical formula is C11H17N3O8 after recalculation.
  • Discussion shifts to calculating the mass of tetrodotoxin corresponding to the LD50 for a 70 kg person, with one participant questioning the role of Avogadro's number in this calculation.
  • Another participant confirms that Avogadro's number applies universally to moles of any substance.
  • There is a humorous exchange regarding the small amount of toxin required to be lethal, emphasizing the potency of tetrodotoxin.

Areas of Agreement / Disagreement

Participants generally agree on the empirical formula being C11H17N3O8 after addressing the teacher's error, but there is some initial disagreement regarding the calculations leading to the empirical formula. The discussion about the LD50 calculation remains unresolved, with differing levels of confidence in the answers provided.

Contextual Notes

Some calculations depend on the accuracy of the combustion analysis data and the interpretation of the results. There are unresolved steps in the calculation of the LD50 dosage based on body weight.

Who May Find This Useful

Students studying chemistry, particularly in the context of organic compounds and toxicology, as well as those interested in empirical formula determination and related calculations.

Astronomer1
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1. The Question: The pufferfish is an Asian delicacy, but also contains a deadly toxin, tetrodotoxin. The LD50 (lethal dose to kill 50% of the population) is 10 micrograms/kg. An analysis of 18.34 mg of tetrodoxin produced 27.82 mg of carbon dioxide, 13.2 mg of nitrogen dioxide, and 8.83 mg water in a combustion reaction.

a) Determine the empirical formula for tetrodotoxin. b) What mass of tetrodotoxin is the LD50 for a 70kg person?


2. The answer is C11H17N3O8, but I'm not getting it.

After converting to mass, and then back to moles, then dividing by the smallest # of moles, and then multiplying by the factor ("common denominator") that gives all integer values...I'm not getting what I'm supposed to be getting. I've also checked 5 times to make sure I'm not punching something wrong in the calculator.


3. I'm not going to type out the WHOLE thing of how I arrived to my answer, cause it took about one page of writing to arrive at the answer. But I'm not getting C11H17N3O8 for the empirical formula, I keep getting C11H17N5O6 which is WRONG. Not sure what I'm doing wrong, but if someone could help me, that would be great. Thanks! (I've attempted this question 5-6 times already, and keep getting the same WRONG answer.)
 
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You could show the reaction which results from the combustion of TD. There may be a mistake there. If you don't show something, you can't expect help.
 
Your calculation seems to be right. Maybe your teacher misread a 5 for a 3 when setting up the exercise.
 
I got N5 as well.
 
Thanks everyone. It turns out that the teacher made an error in 1 of the masses. So the empirical formula (re-calculated) IS still C11H17N3O8. But now I'm stuck on Part b. I calculated the molar mass of tetrodotoxin which is 319g/mol. Does Avogradro's # play a role here in the calculation? I'm clueless again.
 
There's 1 Avogadro's number of molecules in a mole, if that's what you are asking. That works for any substance.

For part b, you are given the LD50 dosage (10 micrograms/kg body weight) How much toxin is required for the LD50 dosage if a person has a mass of 70 kg (This can be solved by looking at the units)
 
700 micrograms? I don't know...
 
Fine. Give him a whole gram. He'll be deader than last week's news.

The point about deadly toxins is, it doesn't take a whole lot of it to kill someone.
 
700 μg it is.
 

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