Question

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 \(\mathrm{lb}\) ?

Short answer

The empirical formula of tetrodotoxin is C4N1H6O3, and its molecular formula is C8N2H12O6. The LD50 dosage for a person weighing 165 lb would be approximately 1.41 x 10²¹ molecules of tetrodotoxin.

Step-by-step solution

Find the amount of each element in 100 g of tetrodotoxin

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)) \% \( \times 100 \ g = 39.09 \ g\)

Convert the masses into moles

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: \( \frac{39.09 \ g}{16.00 \ g/mol} = 2.44 \ mol\)

Calculate the mole ratios

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: \(\frac{2.44}{0.940} = 2.60\approx3\)

Determine the empirical formula

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.

Calculate the molar mass of the empirical formula

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

Calculate the molar mass from the given mass of three molecules

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

Determine the molecular formula

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

Calculate the LD50 dosage for a person weighing 165 lb

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.

Key concepts

Chemical Composition

Understanding the chemical composition of a substance is essential to define its empirical and molecular formulas. In the case of tetrodotoxin, the chemical composition is given in terms of percentage mass for different elements. These are elements like carbon, nitrogen, hydrogen, and oxygen. For example, tetrodotoxin is composed of:

• 41.38% carbon
• 13.16% nitrogen
• 5.37% hydrogen
• 39.09% oxygen

Knowing these percentages allows us to calculate the mass of each element in a hypothetical 100-gram sample, serving as a basis for further calculations. This streamlined method helps in determining the empirical formula, which provides a simplified representation of the actual chemical structure. It shows the basic ratio of elements in the compound without detailing the exact count of atoms.

Mole Calculations

Mole calculations are critical in converting mass information into a mole ratio, which helps find the empirical formula. A mole is a fundamental unit in chemistry, representing Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules or atoms. To translate the mass of each element into moles, we divide the given mass of an element by its molar mass. For example:

• Moles of Carbon: \( \frac{41.38 \text{ g}}{12.01 \text{ g/mol}} = 3.45 \text{ mol} \)
• Moles of Nitrogen: \( \frac{13.16 \text{ g}}{14.01 \text{ g/mol}} = 0.940 \text{ mol} \)
• Moles of Hydrogen: \( \frac{5.37 \text{ g}}{1.01 \text{ g/mol}} = 5.32 \text{ mol} \)
• Moles of Oxygen: \( \frac{39.09 \text{ g}}{16.00 \text{ g/mol}} = 2.44 \text{ mol} \)

These calculations give a mole amount for each component, which are then used to determine the empirical formula by dividing by the smallest mole amount to achieve a whole number ratio.

LD50 Dosage

LD50 refers to the dosage of a substance that is lethal to 50% of a tested population. In this context, it is especially significant when dealing with toxic chemicals like tetrodotoxin. The given LD50 for tetrodotoxin is 10 micrograms per kilogram of body mass. Calculating the LD50 dosage involves considering the body mass of the individual in question.
For instance, converting a person's weight from pounds to kilograms and applying the LD50 value provides the critical dose needed to reach a potentially lethal level. The calculation follows:

• Convert mass from pounds to kilograms: \(165 \text{ lb} \times \frac{1 \text{ kg}}{2.205 \text{ lb}} = 74.83 \text{ kg}\)
• Calculate LD50 dosage: \(74.83 \text{ kg} \times 10 \text{ \(\mu\)g/kg} = 748.3 \text{ \(\mu\)g}\)

This amount is then translated into moles and the number of molecules, providing a clearer picture of the potential impact at a molecular level.

Molar Mass Calculation

Calculating molar mass is key in converting between grams and moles, ultimately giving insight into the substance's molecular formula. The empirical formula provides the most basic ratio, but it may not represent the actual molecule size or atoms count. To find the molecular formula, it's necessary to determine the molar mass based on the given molecular weight data.
In tetrodotoxin's case, the molecular formula is determined by multiplying the empirical formula by a factor. The molar mass of the empirical formula (C4N1H6O3) is calculated first:

• Molar Mass of Empirical Formula: \( (4 \times 12.01) + (1 \times 14.01) + (6 \times 1.01) + (3 \times 16.00) = 179.16 \text{ g/mol} \)

Using the known mass of three molecules, we calculate the molecular formula mass, checking how many times it fits into the experimental molar mass:

• Molecular Mass Calculation: \( \frac{1.59 \times 10^{-21} \text{ g} \times 6.022 \times 10^{23} \text{ molecules/mol}}{3 \text{ molecules}} = 318.36 \text{ g/mol} \)

Finally, determining the factor (close to 2) allows us to multiply the subscripts in the empirical formula by this factor, giving the final molecular formula: C8N2H12O6.