Question

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has a \(\mathrm{LD}_{50}\) (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 LD \(_{50}\) dosage for a person weighing \(165 \mathrm{lb}\) ?

Short answer

The empirical formula of tetrodotoxin is \(C_{4}H_{6}N_{1}O_{3}\) and the molecular formula is \(C_{8}H_{12}N_{2}O_{6}\). The lethal dosage of tetrodotoxin for a person weighing 165 lbs is approximately \(1.36 \times 10^{18}\) molecules.

Step-by-step solution

Convert mass percentages to moles

To find the ratios of carbon, nitrogen, and hydrogen in the empirical formula, we need to first convert the mass percentages to moles. Let's assume we have 100 g of the substance. This means that we have 41.38 g of carbon, 13.16 g of nitrogen, and 5.37 g of hydrogen. Using the molar masses of the elements (C: 12.01 g/mol, N: 14.01 g/mol, H: 1.01 g/mol), we can convert the amounts of each element to moles: Carbon: \(\frac{41.38 \, \mathrm{g}}{12.01 \, \mathrm{g/mol}} = 3.44 \, \mathrm{mol}\) Nitrogen: \(\frac{13.16 \, \mathrm{g}}{14.01 \, \mathrm{g/mol}} = 0.938 \, \mathrm{mol}\) Hydrogen: \(\frac{5.37 \, \mathrm{g}}{1.01 \, \mathrm{g/mol}} = 5.31 \, \mathrm{mol}\)

Determine the empirical formula

Now that we have the amount of each element in moles, we can find the simplest whole number ratios by dividing them by the smallest value. In this case, the smallest value is 0.938 (for nitrogen): Carbon: \(\frac{3.44}{0.938} \approx 3.67\) Nitrogen: \(\frac{0.938}{0.938} = 1\) Hydrogen: \(\frac{5.31}{0.938} \approx 5.66\) These values are approximately 3.67, 1, and 5.66. Since we need whole numbers, we can round to 4, 1, and 6. The remaining mass is from oxygen, which we can calculate by subtracting the mass percentages of other elements from 100: Oxygen: \(100 - (41.38 + 13.16 + 5.37) = 40.09 \% \) Now calculate the number of moles: Oxygen: \(\frac{40.09 \, \mathrm{g}}{16.00 \, \mathrm{g/mol}} = 2.51 \, \mathrm{mol}\) Oxygen: \(\frac{2.51}{0.938} \approx 2.68 \approx 3\) So, the empirical formula of tetrodotoxin is \(C_{4}H_{6}N_{1}O_{3}\).

Calculate the molecular formula

We are given that three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \, \mathrm{g}\). To find the molecular formula, we need to determine the molar mass of one molecule using the empirical formula and compare it to the given mass. First, find the molar mass of the empirical formula: Molar mass: \(4 \times 12.01 + 6 \times 1.01 + 1 \times 14.01 + 3 \times 16.00 = 140.0 \, \mathrm{g/mol}\) Now, find the mass of one molecule in the empirical formula: Mass of one molecule: \(\frac{140.0 \, \mathrm{g/mol}}{6.022 \times 10^{23} \, \mathrm{mol^{-1}}} \approx 2.32 \times 10^{-22} \, \mathrm{g}\) Since we have the mass of one molecule in the empirical formula and the mass of one molecule in the molecular formula, we can determine the ratio by dividing the given mass by the empirical mass: \(\frac{1.59 \times 10^{-21} \, \mathrm{g}}{3 \times 2.32 \times 10^{-22} \, \mathrm{g}} \approx 2.28 \approx 2\) This ratio indicates that the molecular formula is 2 times the empirical formula: \(C_{8}H_{12}N_{2}O_{6}\).

Calculate the number of molecules for a lethal dosage

We need to calculate the number of molecules of tetrodotoxin that would be lethal to a person weighing 165 lbs. First, convert the weight to kilograms: \(165 \, \mathrm{lb} \times \frac{1 \, \mathrm{kg}}{2.205 \, \mathrm{lb}} = 74.83 \, \mathrm{kg}\) Now, find the LD50 dosage in micrograms (10 µg per kg): LD50 dosage: \(74.83 \, \mathrm{kg} \times 10 \, \mathrm{\mu g/kg} = 748.3 \, \mathrm{\mu g}\) Convert the dosage to grams and then to moles: Dosage in grams: \(748.3 \, \mathrm{\mu g} \times \frac{1 \, \mathrm{g}}{10^6 \, \mu \, \mathrm{g}} = 7.48 \times 10^{-4} \, \mathrm{g}\) Dosage in moles: \(\frac{7.48 \times 10^{-4} \, \mathrm{g}}{330.0 \, \mathrm{g/mol}} = 2.27 \times 10^{-6} \, \mathrm{mol}\) Finally, find the number of molecules in the lethal dosage: Number of molecules: \(2.27 \times 10^{-6} \, \mathrm{mol} \times 6.022 \times 10^{23} \, \mathrm{mol^{-1}} = 1.36 \times 10^{18}\) molecules.

Key concepts

Empirical Formula Calculation

Understanding empirical formulas is crucial for students studying chemistry. It represents the simplest ratio of elements in a compound. To find the empirical formula, we must first convert the mass percentages of each element to moles, as moles provide a way to quantify atoms based on Avogadro's number, which defines one mole as exactly 6.02214076×10²³ elementary entities (usually atoms or molecules).

In our exercise, we converted the mass percentages of tetrodotoxin to moles using the molar masses of carbon, nitrogen, hydrogen, and oxygen. Once we had the moles, we looked for the smallest value and used it to normalize the mole ratios to find the simplest whole number ratio.

For example, if we have 0.938 moles of nitrogen and this is the smallest quantity, we divide all other mole values by this number to achieve a ratio. If the resulting number is not a whole number, it's often a simple fraction away from a whole number, and rounding is usually appropriate or multiplying all ratios by the same factor to get whole numbers. This technique allowed us to identify the empirical formula for tetrodotoxin as C4H6N1O3.

Molecular Formula Calculation

The molecular formula is a step further from the empirical formula. While the empirical formula represents the simplest ratio, the molecular formula tells us the actual number of atoms of each element in a molecule of the compound.

To calculate the molecular formula, we need to determine the molar mass of the empirical formula and compare it to the experimental data provided, which in this case is the mass of three molecules of tetrodotoxin. By dividing the mass of three molecules by the mass of one molecule as defined by the empirical formula, we establish the relation between the molecular formula and the empirical formula.

Through the exercise's calculations, we determined the molecular formula to be double the empirical formula, ultimately arriving at C8H12N2O6, providing a complete depiction of the compound's molecular composition.

Molar Mass

Molar mass is defined as the mass of one mole of a substance, and it is a bridge between the macroscopic world we observe and the microscopic world of atoms and molecules. The unit commonly used is grams per mole (g/mol).

To calculate molar mass, we add together the atomic weights of each element in the compound's empirical formula, weighted by the number of atoms of each element. Each element's atomic weight is found on the periodic table, usually averaged to account for the distribution of its natural isotopes.

In our example, we calculated the molar mass of tetrodotoxin based on the empirical formula C4H6N1O3, yielding 140.0 g/mol. This particular calculation is essential when converting between the mass of a substance and the number of moles in that mass.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's the calculation of reactants and products in chemical reactions in chemistry.

In this case, we used stoichiometry to calculate the amount of tetrodotoxin that would be lethal, interpreting the molar mass in the context of a dose-response relationship. By converting the person's body mass into kilograms, then calculating the LD50 (lethal dose for 50% of the population) in micrograms, we converted this into the mass in grams. This mass was then converted into moles using the molar mass of tetrodotoxin as found in the molecular formula calculation. Finally, the moles were converted into the number of molecules using Avogadro's number.

This series of calculations is an application of stoichiometry, drawing on several critical concepts that we discussed earlier in the molar mass and molecular formula sections, demonstrating the interconnectivity of these concepts within chemistry.