Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has an LD_s0 (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu \mathrm{g}\) per 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 \(1.59 \times 10^{-21}\) g, what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD_so dosage for a person weighing 165 \(\mathrm{lb}\) ?

If we assume a 100 g sample of tetrodotoxin, we can easily find the amounts of each element using the percentages given: Carbon: \(41.38 \% \times 100 \ g = 41.38 \ g\) Nitrogen: \(13.16 \% \times 100 \ g = 13.16 \ g\) Hydrogen: \(5.37 \% \times 100 \ g = 5.37 \ g\) Oxygen: (100 - (41.38+13.16+5.37)) \% \( \times 100 \ g = 39.09 \ g\)

Calculate the moles for each element: Moles of Carbon: \( \frac{41.38 \ g}{12.01 \ g/mol} = 3.45 \ mol\) Moles of Nitrogen: \( \frac{13.16 \ g}{14.01 \ g/mol} = 0.940 \ mol\) Moles of Hydrogen: \( \frac{5.37 \ g}{1.01 \ g/mol} = 5.32 \ mol\) Moles of Oxygen: \( \frac{39.09 \ g}{16.00 \ g/mol} = 2.44 \ mol\)

Divide each mole amount by the smallest mole amount to find the mole ratios for each element: Mole ratio of Carbon: \(\frac{3.45}{0.940} = 3.67\approx4\) Mole ratio of Nitrogen: \(\frac{0.940}{0.940} = 1\) Mole ratio of Hydrogen: \(\frac{5.32}{0.940} = 5.66\approx6\) Mole ratio of Oxygen: \(\frac{2.44}{0.940} = 2.60\approx3\)

Using the mole ratios, we can now write the empirical formula for tetrodotoxin: C4N1H6O3 Now that we have the empirical formula, let's find the molecular formula.

Using the empirical formula, we can calculate the molar mass: C4N1H6O3: \((4 \times 12.01) + (1 \times 14.01) + (6 \times 1.01) + (3 \times 16.00) = 179.16 \ g/mol\)

Since the mass of three molecules is given, we can calculate the molar mass for the molecular formula: Molar mass of tetrodotoxin: \(\frac{1.59 \times 10^{-21} \ g \times 6.022 \times 10^{23} \ molecules/mol}{3 \ molecules} = 318.36 \ g/mol \)

Divide the molar mass of the molecular formula by the molar mass of the empirical formula to get the factor used to calculate the molecular formula: Factor: \(\frac{318.36 \ g/mol}{179.16 \ g/mol} = 1.78 \approx 2\) Multiply the empirical formula by the factor to obtain the molecular formula: (C4N1H6O3)2: C8N2H12O6

First, convert the person's weight to kilograms: Weight in kg: \(165 \ \mathrm{lb} \times \frac{1 \ \mathrm{kg}}{2.205 \ \mathrm{lb}} = 74.83 \ \mathrm{kg} \) Then, calculate the amount of tetrodotoxin required for LD50: LD50 dosage: \(74.83 \ \mathrm{kg} \times 10 \ \mu \mathrm{g/kg} = 748.3 \ \mu \mathrm{g} \) Now, convert the dosage to moles and then to molecules: Moles of tetrodotoxin: \(\frac{748.3 \ \mu \mathrm{g}}{318.36 \ \mathrm{g/mol} \times 10^6 \ \mu \mathrm{g/g}} = 2.35 \times 10^{-6} \ \mathrm{mol} \) Number of molecules: \(2.35 \times 10^{-6} \ \mathrm{mol} \times 6.022 \times 10^{23} \ \mathrm{molecules/mol} = 1.41 \times 10^{18} \ \mathrm{molecules} \) Thus, the LD50 dosage for a person weighing 165 lb would be about 1.41 x 10²¹ molecules of tetrodotoxin.