Q155CP - Chemistry: Molecular Nature Of Matter And Change

Question

The anticancer drug Platinol (Cisplatin), Pt(NH₃)₂Cl₂, reacts with the cancer cell’s DNA and interferes with its growth. (a) What is the mass % of platinum (Pt) in Platinol? (b) If Pt costs \(19/g, how many grams of Platinol can be made for \)1.00 million (assume that the cost of Pt determines the cost of the drug)?

Step-by-Step Solution

Answer

a) The mass % of platinum is $\mathbf{65}\mathbf{.}\mathbf{02}\mathbf{\%}$

b) There are approx. 53000 g platinol can be made for $1 million

Step 1: Calculating the mass percentage of platinum in platinol

On listing the atomic mass of each element,

$\mathbf{Pt}\mathbf{=}\mathbf{195}\mathbf{.}\mathbf{08}\mathbf{u}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{14}\mathbf{u}\mathbf{,}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{CI}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{45}\mathbf{u}$ 

On adding the atomic mass,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{Pt}\mathbf{+}\mathbf{2}\left(N+3H\right)\mathbf{+}\mathbf{2}\mathbf{CI} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{195}\mathbf{.}\mathbf{08}\mathbf{+}\mathbf{2}\left(14+3\left(1.01\right)+2\left(35.45\right)\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{300}\mathbf{.}\mathbf{04}\mathbf{amu} \end{gathered}$$ 

On solving the mass percentage of platinum,

$$\begin{gathered} \mathbf{\%}\mathbf{mass}\mathbf{ }\mathbf{Pt}\mathbf{=}\frac{\mathbf{195}\mathbf{.}\mathbf{08}}{\mathbf{300}\mathbf{.}\mathbf{04}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{65}\mathbf{.}\mathbf{02}\mathbf{\%} \end{gathered}$$ 

Step 2: Calculating the amount of platinol made of $1 M

For 1 g the cost is $19 then in 1M, i.e., $1000000 platinol made will be 

 $\frac{\mathbf{1}\mathbf{g}}{\mathbf{\$}\mathbf{19}}\mathbf{\times }\mathbf{1000000}\mathbf{=}\mathbf{52631}\mathbf{.}\mathbf{58}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{pl}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{tinol}$