Q135CP - Chemistry: Molecular Nature Of Matter And Change

Question

Dimercaprol ( ) was developed during World War I as an antidote to arsenic-based poison gas and is used today to treat heavy-metal poisoning. It binds the toxic element and carries it out of the body. 

(a) If each molecule binds one arsenic (As) atom, how many atoms of As could be removed by 175 mg of dimercaprol? 

(b) If one molecule binds one metal atom, calculate the mass %of each of the following metals in a metal-dimercaprol combination: mercury, thallium, chromium.

Step-by-Step Solution

Answer

1. The number of atoms of As could be removed by 175 mg of dimercaprol is $\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{20} \mathbf{arsenic}\mathbf{ }\mathbf{atoms}$.
2. i)The mass % of mercury in a metal-dimercaprol combination is   $\mathbf{61}\mathbf{.}\mathbf{75}\mathbf{\%}$.

ii) The mass % of thallium in a metal-dimercaprol combination is $\mathbf{=}\mathbf{62}\mathbf{.}\mathbf{20}\mathbf{\%}$.

iii) The mass % of chromium in a metal-dimercaprol combination is $\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{51}\mathbf{\%}$.

Step 1: Introduction to the Concept

The mole is the amount of substance in a system that includes the same number of elementary entities as there are atoms in 0.012 kilograms of carbon 12; it is represented by the sign "mol."

Step 2: Explanation of number of As atoms in dimercaprol

Initially let us compute for the moles of 175 mg dimercaprol.

Here, Dimercaprol’s molecular mass$\mathbf{=}\mathbf{124}\mathbf{.}\mathbf{23}\frac{\mathbf{g}}{\mathbf{mol}}$ 

Moles of dimercaprol

  $$\begin{gathered} \mathbf{=}\mathbf{175}\mathbf{ }\mathbf{mg}\mathbf{\times }\frac{\mathbf{1}\mathbf{g}}{\mathbf{1000}\mathbf{ }\mathbf{mg}}\mathbf{\times }\frac{\mathbf{1}\mathbf{mol}}{\mathbf{124}\mathbf{.}\mathbf{23}\mathbf{ }\mathbf{g}} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{mol} \end{gathered}$$

Let’s determine the number of atoms of arsenic that will attach to ,

Here, the Number of Arsenic atoms

 $$\begin{gathered} \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{-3} \mathbf{mol}\mathbf{\times }\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}}{\mathbf{1}\mathbf{ }\mathbf{mol}}\times \frac{\mathbf{1}\mathbf{ }\mathbf{arsenic}\mathbf{ }\mathbf{atom}}{\mathbf{1}\mathbf{ }\mathbf{dimercaprol}\mathbf{ }\mathbf{molecule}} \\ \mathbf{=}\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{20 }\mathbf{arsenic}\mathbf{ }\mathbf{atoms} \end{gathered}$$

Step 3: Explanation

i) Combination of molecular mass of dimercaprol and mercury

 $\mathbf{124}\mathbf{.}\mathbf{23}\mathbf{+}\mathbf{200}\mathbf{.}\mathbf{59}\mathbf{=}\mathbf{324}\mathbf{.}\mathbf{82}\frac{\mathbf{g}}{\mathbf{mol}}$

Let’s compute the percent mass of mercury,

 $$\begin{gathered} \mathbf{\%}\mathbf{mass}\mathbf{=}\frac{\mathbf{200}\mathbf{.}\mathbf{59}}{\mathbf{324}\mathbf{.}\mathbf{82}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{61}\mathbf{.}\mathbf{25}\mathbf{\%} \end{gathered}$$

ii) Combination of molecular mass of dimercaprol and Thallium

  $\mathbf{124}\mathbf{.}\mathbf{23}\mathbf{+}\mathbf{204}\mathbf{.}\mathbf{38}\mathbf{=}\mathbf{328}\mathbf{.}\mathbf{61}\frac{\mathbf{g}}{\mathbf{mol}}$  

Let’s compute the percent mass of Thallium,

$$\begin{gathered} \mathbf{\%}\mathbf{mass}\mathbf{=}\frac{\mathbf{204}\mathbf{.}\mathbf{38}}{\mathbf{328}\mathbf{.}\mathbf{61}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{62}\mathbf{.}\mathbf{20}\mathbf{\%} \end{gathered}$$ 

iii) Combination of molecular mass of dimercaprol and Chromium

  $\mathbf{124}\mathbf{.}\mathbf{23}\mathbf{+}\mathbf{52}\mathbf{.}\mathbf{00}\mathbf{=}\mathbf{176}\mathbf{.}\mathbf{23}\frac{\mathbf{g}}{\mathbf{mol}}$

Let’s compute the percent mass of Chromium,

 $$\begin{gathered} \mathbf{\%}\mathbf{mass}\mathbf{=}\frac{\mathbf{52}\mathbf{.}\mathbf{00}}{\mathbf{176}\mathbf{.}\mathbf{23}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{51}\mathbf{\%} \end{gathered}$$