Chapter 3, Problem 192IP

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has an LD₅₀ (the amount of substance that is lethal to 50.% of a population sample) of 10.μg 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 × 10⁻²¹ g, what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD₅₀ dosage for a person weighing 165 lb?

Interpretation Introduction

Interpretation: The empirical formula, molecular formula and the number of molecules of Tetrodotoxin are to be calculated.

Concept introduction: The empirical formula of an element is defined as the simplest formula that expresses the percentage composition of the element and the molecular formula of an element defines the total number of atoms present in a molecule.

To determine: The empirical formula, molecular formula and the number of molecules of Tetrodotoxin.

Answer to Problem 192IP

The empirical and molecular formula of Tetrodotoxin is ${\text{C}}_{\text{11}}{\text{H}}_{\text{17}}{\text{N}}_{\text{3}}{\text{O}}_{\text{8}}$ and the molecules of Tetrodotoxin in $L{D}_{50}$ dosage are $\underset{\_}{1.4\times 10^{18}molecules}$.

Explanation of Solution

Explanation

Given

The mass percent of carbon is $41.38\%$.

The mass percent of nitrogen is $13.16\%$.

The mass percent of hydrogen is $5.37\%$.

The remaining amount consists of oxygen. The mass percent of oxygen is calculated as,

$\text{Mass percent of oxygen}=\text{Total percent}-\left(\text{Mass percent of C}+\text{Mass percent of N}+\text{Mass percent of H}\right)$

Substitute the value of mass percent of carbon, nitrogen and hydrogen in the above expression, the mass percent of oxygen is calculated.

$\begin{array}{c}\text{Mass percent of oxygen}=\text{Total percent}-\left(\text{Mass percent of C}+\text{Mass percent of N}+\text{Mass percent of H}\right)\\ =100-\left(41.38+13.16+5.37\right)\\ =40.09\%\end{array}$

The mass percent of oxygen is $40.09\%$.

Therefore,

Mass of sample is $1\text{00 g}$

Mass of carbon is $\text{41}\text{.38 g}$

Mass of nitrogen is $\text{13}\text{.16 g}$

Mass of hydrogen is $\text{5}\text{.37 g}$

Mass of oxygen is $\text{40}\text{.09 g}$

The molar mass of carbon is $12\text{g/mol}$.

The molar mass of nitrogen is $14\text{g/mol}$.

The molar mass of hydrogen is $1\text{g/mol}$.

The molar mass of oxygen is $1\text{6 g/mol}$.

The number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the value of the given mass and the molar mass, to calculate the number of moles of carbon, nitrogen, hydrogen, and oxygen in the above equation.

For carbon,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{41.38\text{g}}{\text{12}\text{g/mol}}\\ =3.45\text{mol}\end{array}$

For nitrogen,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{13.16\text{g}}{\text{14}\text{.0}\text{g/mol}}\\ =0.939\text{mol}\end{array}$

For hydrogen,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{5.37\text{g}}{\text{1}\text{g/mol}}\\ =5.33\text{mol}\end{array}$

For oxygen,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{oxygen}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{40.09\text{g}}{\text{16}\text{g/mol}}\\ =2.51\text{mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent.

For carbon $\left(\text{C}\right)$,

$\text{C}=\frac{3.45\text{mol}}{0.939\text{mol}}=3.6\text{7}$

For nitrogen $\left(\text{N}\right)$,

$\text{N}=\frac{0.939\text{mol}}{0.939\text{mol}}=1$

For hydrogen $\left(\text{H}\right)$,

$\text{H}=\frac{5.33\text{mol}}{0.939\text{mol}}=5.67$

For oxygen $\left(\text{O}\right)$,

$\text{O}=\frac{2.51\text{mol}}{0.939\text{mol}}=2.67$

Multiply each of the numbers by $3$ in order to get whole number.

$\begin{array}{l}\text{Carbon}=\text{11}\\ \text{Nitrogen}=\text{3}\\ \text{Hydorgen}=\text{17}\\ \text{Oxygen}=\text{8}\end{array}$

So the empirical formula of Tetrodotoxin is ${\text{C}}_{\text{11}}{\text{H}}_{\text{17}}{\text{N}}_{\text{3}}{\text{O}}_{\text{8}}$.

Three molecules of Tetrodotoxin has a mass of $1.59\times {10}^{-21}\text{ g}$.

The molar mass is the mass of one mole or $6.022\times {10}^{23}$ molecules

The molar mass of Tetrodotoxin is calculated by using the expression,

$\text{Molar mass}=\frac{\text{Mass}}{\text{Number of Moles}}$

Substitute the value of mass and number of molecules of Tetrodotoxin the above expression, the molar mass of Tetrodotoxin is calculated.

$\begin{array}{c}\text{Molar mass of Tetrodotoxin}=1.59\times {10}^{-21}\text{ g}\times \left(\frac{6.022\times {10}^{\text{23}}\text{ molecules}}{\text{3 molecules}}\right)\\ =319.16\text{6 g/mol}\end{array}$

The empirical formula mass of Tetrodotoxin is calculated as,

$\begin{array}{c}{\text{C}}_{\text{11}}{\text{H}}_{\text{17}}{\text{N}}_{\text{3}}{\text{O}}_{\text{8}}=1\text{1}\left(\text{12 g/mol}\right)+\text{17}\left(\text{1}\text{.0 g/mol}\right)+\text{3}\left(\text{14 g/mol}\right)+\text{8}\left(\text{16 g/mol}\right)\\ =319.2\text{7 g}\end{array}$

Since both are almost equal the empirical formula is equal to the molecular formula.

So the molecular formula of Tetrodotoxin is ${\text{C}}_{\text{11}}{\text{H}}_{\text{17}}{\text{N}}_{\text{3}}{\text{O}}_{\text{8}}$.

$L{D}_{50}$ for Tetrodotoxin is $1\text{0 μg}$ per kg of body mass.

The conversion of $\text{ μg}$ to g is done as.

$1\text{ μg}={10}^{-6}\text{ g}$

Therefore, the conversion of $1\text{0 μg}$ to g is done as,

$10\text{ μg}=10\times {10}^{-6}\text{ g}$

Weight of person is $16\text{5 lb}$.

The conversion of $\text{lb}$ to kg is done as.

$1\text{ lb}=0.454\text{ kg}$

Therefore, the conversion of $16\text{5 lb}$ to kg is done as,

$165\text{ lb}=165\times 0.454\text{ kg}$

The weight of person is calculated as,

$\begin{array}{c}\text{Weight of person}=\text{165 lb}\times \left(\frac{\text{0}\text{.454 kg}}{\text{1 lb}}\right)\\ =74.9\text{1 kg}\end{array}$.

The mass of $L{D}_{50}$ dosage is calculated as,

$\begin{array}{c}\text{Mass}\text{of}L{D}_{50}=10\times {10}^{-6}\text{ g}\times \left(\frac{\text{74}\text{.91 kg}}{\text{1 kg}}\right)\\ =74.91\times {10}^{-5}\text{ g}\end{array}$.

One mole of Tetrodotoxin has a mass of $319.2\text{7 g}$ and $6.022\times {10}^{23}$ molecules.

So molecules of Tetrodotoxin are calculated by using the expression.

$\text{Number of molecules}=\frac{\text{Mass}}{\text{Molar Mass}}\times 6.022\times {10}^{\text{23}}\text{ molecules}$

Substitute the value of Mass and Molar mass of Tetrodotoxin in the above expression.

$\begin{array}{c}\text{Number of molecules}=\frac{\text{Mass}}{\text{Molar Mass}}\times 6.022\times {10}^{\text{23}}\text{ molecules}\\ =\frac{74.91\times {10}^{-5}\text{ g}}{319.2\text{7 g}}\times 6.022\times {10}^{\text{23}}\text{ molecule}\\ =\underset{\_}{1.4\times 10^{18}molecules}\end{array}$

Hence, the molecules of Tetrodotoxin in $L{D}_{50}$ dosage are $\underset{\_}{1.4\times 10^{18}molecules}$.

Conclusion

Conclusion

The empirical and molecular formula of Tetrodotoxin is ${\text{C}}_{\text{11}}{\text{H}}_{\text{17}}{\text{N}}_{\text{3}}{\text{O}}_{\text{8}}$ and the molecules of Tetrodotoxin in $L{D}_{50}$ dosage are $\underset{\_}{1.4\times 10^{18}molecules}$.