Chapter 3, Problem 108AE

Interpretation Introduction

Interpretation: The empirical formula and molecular formula of tetrodotoxin is to be determined. The number of molecules of tetrodotoxin present in ${\text{LD}}_{\text{50}}$ dosage of a person weighing $\text{165}\text{lb}$ is to be stated.

Concept introduction: A simplified form of molecular formula is the empirical formula. Both molecular and empirical formula can be same for some compounds however they can also be different. The actual count of all atoms that a specific element has in the compound is only determined from molecular formula. This actual count is not given by empirical formula which gives the atom’s simplest ratio.

Answer to Problem 108AE

The empirical and molecular formula of tetrodotoxin is ${\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8$ and the number of molecules of tetrodotoxin present in ${\text{LD}}_{\text{50}}$ dosage of a person weighing $\text{165}\text{lb}$ is $1.412\times {10}^{18}$ .

Explanation of Solution

The given mass % of $\text{N}$ in the compound is 13.16%.

The given mass % of $\text{C}$ in the compound is 41.38%.

The given mass % of $\text{H}$ in the compound is 5.37%.

The given mass of three molecules of tetrodotoxin is $1.59\times {10}^{-21}\text{g}$ .

The given weight of person is $\text{165}\text{lb}$ .

The mass percentage of $\text{O}$ is calculated by the formula shown below.

  $\text{Mass\%}\text{of}\text{O}=100-\left(\text{Mass\%}\text{of}\text{N}+\text{Mass\%}\text{of}\text{C}+\text{Mass\%}\text{of}\text{H}\right)$(1)

Substitute the known values in equation (1).

  $\begin{array}{c}\text{Mass\%}\text{of}\text{O}=100-\left(13.16\%+41.38\%+5.37\%\right)\\ =40.09\%\end{array}$

Suppose the mass of the sample is $\text{100}\text{g}$ .

The percentage of oxygen in the given compound is $40.09\%$ . Therefore, in $\text{100}\text{g}$ of the given compound, the amount of oxygen present will be $\text{40}\text{.09}\text{g}$ .

The percentage of nitrogen in the given compound is $13.16\%$ . Therefore, in $\text{100}\text{g}$ of the given compound, the amount of nitrogen present will be $13.16\text{g}$

The percentage of carbon in the given compound is $41.38\%$ . Therefore, in $\text{100}\text{g}$ of the given compound, the amount of carbon present will be $41.38\text{g}$ .

The percentage of hydrogen in the given compound is $5.37\%$ . Therefore, in $\text{100}\text{g}$ of the given compound, the amount of hydrogen present will be $5.37\text{g}$

The number of moles is calculated by using the formula given below.

  $n=\frac{m}{M}$  (2)

Where,

• $n$is the number of moles.
• $m$is the given mass.
• $M$is the molecular mass.

The molar mass of $\text{C}$ is $\text{12}\text{.00 g/mol}$ .

The molar mass of $\text{N}$ is $\text{14}\text{.00 g/mol}$ .

The molar mass of $\text{O}$ is $\text{14}\text{.00 g/mol}$ .

The molar mass of $\text{H}$ is $\text{1}\text{.00 g/mol}$ .

Substitute the mass and molar mass of $\text{C}$ and $\text{N}$ in equation (2) to calculate their moles.

  $\begin{array}{c}n\left(C\right)=\frac{41.38\text{g}}{12.00\text{g/mol}}\\ =3.448\text{mol}\\ n\left(N\right)=\frac{13.16\text{g}}{14.00\text{g/mol}}\\ =0.94\text{mol}\end{array}$

Substitute the mass and molar mass of $\text{H}$ and $\text{O}$ in equation (2) to calculate their moles.

  $\begin{array}{c}n\left(\text{H}\right)=\frac{5.37\text{g}}{1.00\text{g/mol}}\\ =5.37\text{mol}\\ n\left(\text{O}\right)=\frac{40.09\text{g}}{16.00\text{g/mol}}\\ =2.50\text{mol}\end{array}$

The ratio of $\text{C:N:H:O}$ is calculated by dividing their moles with the lowest number of moles.

  $\begin{array}{c}\text{C:N:H:O}=\frac{3.448\text{mol}}{0.94\text{mol}}:\frac{0.94\text{mol}}{0.94\text{mol}}:\frac{5.37\text{mol}}{0.94\text{mol}}:\frac{2.50\text{mol}}{0.94\text{mol}}\\ \text{C:N:H:O}=3.66:1:5.71:2.66\end{array}$

Multiply the above ratio by $3$ to get whole numbers.

  $\begin{array}{c}\text{C:N:H:O}=10.98:3:17.13:7.98\\ \text{C:N:H:O}\approx 11:3:17:8\end{array}$

Therefore, the empirical formula of the tetrodotoxin is ${\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8$ .

The molecular formula of a compound is related to the empirical formula as shown below.

  $\text{Molecular}\text{formula}={\left(\text{Empirical}\text{formula}\right)}_n$(3)

Where,

• $n$ is the number of empirical formula units.

Empirical formula units can be calculated as shown below.

  $n=\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}$  (4)

Empirical formula mass can be calculated by the formula shown below.

  $M^{\prime }=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{O}}\times M_{\text{O}}+n_{\text{N}}\times M_{\text{N}}$(5)

Where,

• $n_{\text{C}}$is the number of atoms of carbon.
• $M_{\text{C}}$is the molar mass of carbon.
• $n_{\text{H}}$is the number of atoms of hydrogen.
• $M_{\text{C}}$is the molar mass of hydrogen.
• $n_{\text{O}}$is the number of atoms of oxygen.
• $M_{\text{O}}$is the molar mass of oxygen.
• $n_{\text{N}}$is the number of atoms of nitrogen.
• $M_{\text{N}}$ is the molar mass of nitrogen.
• $M^{\prime }$is the empirical formula mass of the tetrodotoxin.

Substitute the known values in equation (5).

  $\begin{array}{c}{M}^{\prime }=11\times 12.0\text{g/mol}+17\times 1.00\text{g/mol}+8\times 16.00\text{g/mol}+3\times \text{14}\text{.00 g/mol}\\ =319\text{g/mol}\end{array}$

Molar mass is the mass of compound containing $6.022\times {10}^{23}$ molecules.

Therefore, the mass of $6.022\times {10}^{23}$ molecules of tetrodotoxin are calculated as shown below.

Mass of $3$ molecules is $1.59\times {10}^{-21}\text{g}$ .

Therefore, the mass of $1$ molecule will be $\frac{1.59\times {10}^{-21}}{3}\text{g}$ .

That is, the mass of $6.022\times {10}^{23}$ molecule will be $\frac{1.59\times {10}^{-21}}{3}\times 6.022\times {10}^{23}\text{g}$ .

Hence, the molar mass is $\text{319}\text{g}$

Substitute the molecular mass and empirical formula mass in equation (4).

  $\begin{array}{c}n=\frac{319\text{g/mol}}{319\text{g/mol}}\\ =1\end{array}$

Substitute the value of empirical formula units in equation (3) to find the molecular formula.

  $\begin{array}{c}\text{Molecular}\text{formula}={\left({\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8\right)}_1\\ ={\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8\end{array}$

Therefore, the molecular formula of the tetrodotoxin is ${\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8$ .

The conversion of mass of the person in kg is shown below.

  $\begin{array}{c}\text{2}\text{.2046}\text{lb}=\text{1}\text{kg}\\ \text{lb}=\frac{\text{1}}{\text{2}\text{.2046}}\text{kg}\\ \text{165}\text{lb}=\frac{\text{1}}{\text{2}\text{.2046}}\times 165\text{kg}\\ =\text{74}\text{.83 kg}\end{array}$

The conversion of $\text{μg}$ into g is shown below.

  $\text{1}\text{μg}={\text{10}}^{-6}g$

Mass of ${\text{LD}}_{\text{50}}$ present in $\text{1}\text{kg}$ of the body is $\text{10}\text{.0}\text{μg}$ .

That is,mass of ${\text{LD}}_{\text{50}}$ present in $\text{1}\text{kg}$ of the body is ${10}^{-5}\text{g}$ .

Therefore, mass of ${\text{LD}}_{\text{50}}$ present in $\text{74}\text{.83 kg}$ of the body is $\text{74}\text{.83 kg}\times {10}^{-5}\text{g}$ .

Hence, the mass of ${\text{LD}}_{\text{50}}$ present in $\text{74}\text{.83 kg}$ of the body is $\text{7}\text{.483 kg}\times {10}^{-4}\text{g}$ .

The moles of tetrodotoxin is calculates by substituting the calculated mass and its molar mass in equation (2).

  $\begin{array}{c}n=\frac{\text{7}\text{.483 kg}\times {10}^{-4}\text{g}}{319\text{g/mol}}\\ =2.345\times {10}^{-6}\text{mol}\end{array}$

The number of molecules present in $1$ mole is $6.022\times {10}^{23}$ .

Therefore, number of molecules present in $2.345\times {10}^{-6}$ mole is $2.345\times {10}^{-6}\times 6.022\times {10}^{23}$ .

Hence, the number of molecules of tetrodotoxin is $1.412\times {10}^{18}$ .

Conclusion

The empirical and molecular formula of tetrodotoxin is ${\text{C}}_{11}{\text{N}}_3{\text{H}}_{17}{\text{O}}_8$ and the number of molecules of tetrodotoxin present in ${\text{LD}}_{\text{50}}$ dosage of a person weighing $\text{165}\text{lb}$ is $1.412\times {10}^{18}$ .