Question

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 $\mathbf{10}\mathit{\mu }\mathit{g}$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 $\mathbf{1}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathit{g}$, what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD₅₀ dosage for a person weighing 165 lb?

Short answer

The empirical formula and the molecular formula of tetrodotoxinis C₁₁H₁₇N₃O₈ and$14\times {10}^{18}$of tetrodotoxin would be the LD₅₀ dosage for a person weighing 165 lb.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it isequal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Determination of carbon, nitrogen, hydrogen, and oxygen

Tetrodotoxin has a carboncontent of 41.38%, a nitrogen content of 13.16%, and a hydrogencontent of 5.37%.

The remaining portion is made up of oxygen.

So, percent mass of oxygen

$$\begin{gathered} =100-\left(41.38+13.16+5.37\right) \\ =40.09\% \end{gathered}$$

Tetrodotoxin comprises 41.38 g carbon, 13.16 g nitrogen, 5.37 g hydrogen, and 40.09 g oxygen in a 100 g sample.

Molar of carbon =$\frac{Mass}{Atomicmasss}$

$$\begin{gathered} =\frac{41.38g}{12.01{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}} \\ =3.45moles \end{gathered}$$

Molar of nitrogen = $\frac{13.16g}{14.01{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}$

= 0.939moles

Molar of hydrogen =$\frac{5.37g}{1.008{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}$

= 5.33moles

Molar of oxygen = $\frac{40.09g}{16.00{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}}$

= 2.51moles

Determination of empirical formula of tetrodotoxin

Taking the moles of carbon, nitrogen, hydrogen, and oxygen into account:

Carbon Nitrogen Hydrogen Oxygen

3.45 0.93 5.33 2.51

Dividing by the smallest number 0.939 dividing by the smallest number

3.67 1 15.68 2.67

11 3 17 8

is the result of multiplying each of the integers by three.

Tetrodotoxin's empirical formula is C₁₁H₁₇N₃O₈

Determination of molecular formula of tetrodotoxin

From the given, Mass of 3 molecules of tetrodotoxin =$1.59\times {10}^{-21}g$

Therefore 1 mole =$6.022\times {10}^{23}molecules$

Here, the molar mass oftetrodotoxin =$1.59\times {10}^{-21}g\times \left(\frac{6.022\times {10}^{23}molecules}{3molecules}\right)$

$=319.166\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Empirical formula for mass =$11\left(12.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\right)+17\left(1.008\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\right)+3\left(14.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\right)+8\left(16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\right)$

= 319.27g

So, Empirical formula is almost equal to molecular formula.

Therefore the molecular formula isC₁₁H₁₇N₃O₈

Determination of molecules of tetrodotoxin

From the given, LD₅₀ fortetrodotoxin is$10\mu g$per kg of body mass.

Here, the weight of person =$165lb\times \left(\frac{0.454kg}{1lb}\right)$

= 74.91kg

TheLD₅₀ dosage is =$10\times {10}^{-6}g\times \left(\frac{74.91kg}{1kg}\right)$

$=74.91\times {10}^{-5}g$

Molecules oftetrodotoxin =$74.91\times {10}^{-5}g\times \left(\frac{6.022\times {10}^{23}MOLECULES}{319.27g}\right)$

$14\times {10}^{18}molecules$

Therefore number of molecules of tetrodotoxin is$14\times {10}^{18}molecules$