Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has an \(L D_{s 0}\) (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu \mathrm{g}\) per \(\mathrm{kg}\) of body mass. Tetrodotoxin is \(41.38 \%\) carbon by mass, \(13.16 \%\) nitrogen by mass, and \(5.37 \%\) hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \mathrm{~g}\), what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the \(L D_{50}\) dosage for a person weighing \(165 \mathrm{lb}\) ?

Assume a 100g sample of tetrodotoxin. In this sample, the mass percentages of each element can be expressed as: - Carbon: \(41.38g\) - Nitrogen: \(13.16g\) - Hydrogen: \(5.37g\) - Oxygen: \(100 - (41.38 + 13.16 + 5.37) = 39.09g\) Now, divide the mass of each element by their corresponding atomic masses to find the moles of each element: - Carbon: \(41.38g / 12.01g/mol \approx 3.44\) mol - Nitrogen: \(13.16g / 14.01g/mol \approx 0.94\) mol - Hydrogen: \(5.37g / 1.01g/mol \approx 5.31\) mol - Oxygen: \(39.09g / 16.00g/mol \approx 2.44\) mol Next, divide the moles of each element by the smallest number of moles (0.94 mol) to get the mole ratio: - Carbon: \(3.44/0.94 \approx 3.66 \approx 4\) - Nitrogen: \(0.94/0.94 \approx 1\) - Hydrogen: \(5.31/0.94 \approx 5.65 \approx 6\) - Oxygen: \(2.44/0.94 \approx 2.60 \approx 3\) The empirical formula of tetrodotoxin is \(C_4N_1H_6O_3 \), simplified to \(C_4NH_6O_3\).

First, find the molar mass of the empirical formula: - \(M_{C_4NH_6O_3} = 4 \times 12.01g/mol + 1 \times 14.01g/mol + 6 \times 1.01g/mol + 3 \times 16.00g/mol = 172.07g/mol\) Given that three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21}g\), the molar mass of tetrodotoxin can be calculated as: - \(M_{tetrodotoxin} = \frac{1.59 \times 10^{-21}g \times 6.022 \times 10^{23}molecules/mol}{3molecules} = 318.14g/mol\) To find the molecular formula, divide the molar mass of the molecular formula by the molar mass of the empirical formula: - \(\frac{318.14g/mol}{172.07g/mol} \approx 1.85\) Since the ratio is approximately equal to 2, the molecular formula of tetrodotoxin is \(C_8N_2H_{12}O_6\).

Convert 165 lb to kg: - \(165lb \times \frac{1kg}{2.205lb} \approx 74.83kg\) Calculate the mass of tetrodotoxin required for LD50 dosage (given the amount is 10µg per kg body mass): - Mass = \(74.83kg \times 10µg/kg = 748.3µg \) Convert the mass to grams: - \(748.3µg \times \frac{1g}{10^6µg} = 7.483 \times 10^{-4}g\) Finally, calculate the number of tetrodotoxin molecules required for the LD50 dosage: - Molecules of tetrodotoxin = \(\frac{7.483 \times 10^{-4}g}{318.14g/mol} \times 6.022 \times 10^{23}molecules/mol \approx 1.41 \times 10^{21} molecules\) The number of tetrodotoxin molecules in the \(LD_{50}\) dosage for a person weighing 165 lb is approximately \(1.41 \times 10^{21}\) molecules.