Problem 112 - Chemistry and Chemical Reactivity

Question

A The cancer chemotherapy drug cisplatin, \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) can be made by reacting (NH_) \(_{2} \mathrm{PtCl}_{4}\) with ammonia in aqueous solution. Besides cisplatin, the other product is \(\mathrm{NH}_{1} \mathrm{Cl}\) (a) Write a balanced equation for this reaction. (b) To obtain \(12.50 \mathrm{g}\) of cisplatin, what mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{4}\) is required? What volume of \(0.125 \mathrm{M}\) \(\mathrm{NH}_{3}\) is required? (c) Cisplatin can react with the organic compound pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N},\) to form a new compound. $$\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{aq})+x \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq}) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x}(\mathrm{s})$$ Suppose you treat \(0.150 \mathrm{g}\) of cisplatin with what you believe is an excess of liquid pyridine \((1.50 \mathrm{mL}\) \(d=0.979 \mathrm{g} / \mathrm{mL}) .\) When the reaction is complete, you can find out how much pyridine was not used by titrating the solution with standardized HCl. If 37.0 mL. of \(0.475 \mathrm{M} \mathrm{HCl}\) is required to titrate the excess pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) what is the formula of the unknown compound \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x} ?\)

Step-by-Step Solution

Answer

(a) \((\mathrm{NH}_2)_2 \mathrm{PtCl}_4 + 2\mathrm{NH}_3 \rightarrow \operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 + 2\mathrm{NH}_4 \mathrm{Cl}\). (b) 15.54 g of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\); 666.6 mL NH₃. (c) \(\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}_2(\mathrm{C}_5\mathrm{H}_5\mathrm{N})_2\)."

Step 1: Write the Balanced Reaction Equation

For part (a), we need to balance the chemical reaction of forming cisplatin. The reaction involves \((\mathrm{NH}_2)_2 \mathrm{PtCl}_4\) reacting with \(\mathrm{NH}_3\) to form cisplatin \(\operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\) and \(\mathrm{NH}_4 \mathrm{Cl}\). The balanced equation is: \[(\mathrm{NH}_2)_2 \mathrm{PtCl}_4 + 2\mathrm{NH}_3 \rightarrow \operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 + 2\mathrm{NH}_4 \mathrm{Cl}\]

Step 2: Calculate Molar Masses

For part (b), calculate the molar masses of \( \operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 \) and \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\). Molar mass of cisplatin is approximately \(300.05 \text{ g/mol}\). Molar mass of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\) is approximately \(372.98 \text{ g/mol}\).

Step 3: Mass of (NH₄)₂PtCl₄ Required

With the target mass of cisplatin being \(12.50\, \text{g}\), calculate the moles of cisplatin required: \( \frac{12.50}{300.05} = 0.04166\, \text{mol}\). Since 1 mole of cisplatin is made from 1 mole of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\), we need the same amount in moles of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\). Therefore, the mass required is \(0.04166 \times 372.98 = 15.54\, \text{g}\).

Step 4: Volume of NH₃ Required

Calculate the moles of \(\mathrm{NH}_3\) required. Since the balanced equation shows 2 moles of \(\mathrm{NH}_3\) are needed per mole of cisplatin, we require \(2 \times 0.04166 = 0.08332\, \text{mol}\) of \(\mathrm{NH}_3\). Given the molarity of \(\mathrm{NH}_3\) is 0.125 M, we use \(\text{Volume} = \frac{\text{Moles}}{\text{Molarity}}\) to find the volume: \(\text{Volume} = \frac{0.08332}{0.125} = 0.6666\, \text{L} = 666.6\, \text{mL}\).

Step 5: Calculate Excess Pyridine

For part (c), determine how much pyridine remains unreacted. Given 37.0 mL of 0.475 M \(\mathrm{HCl}\), the moles of \(\mathrm{HCl}\) used in titration are \(37.0 \times 0.475 / 1000 = 0.01758 \text{ mol}\). This equals the moles of excess pyridine.

Step 6: Initial Moles and Remaining Moles of Pyridine

Calculate the initial moles of pyridine. Given 1.50 mL pyridine and density = 0.979 g/mL, mass = \(1.50 \times 0.979 = 1.4685\, \text{g}\). Molar mass of \(\mathrm{C}_5 \mathrm{H}_5 \mathrm{N}\) is \(79.10\, \text{g/mol}\). Thus, initial moles = \(\frac{1.4685}{79.10} = 0.01857\, \text{mol}\). Remaining moles = unreacted = \(0.01758\, \text{mol}\).

Step 7: Calculate Reacted Moles of Pyridine

Determine the moles of pyridine that reacted: Initial moles (0.01857) minus unreacted moles (0.01758) = 0.00099 mol pyridine reacted with 0.150 g cisplatin.

Step 8: Determine Formula of the Unknown Compound

We know \(0.150\, \text{g}\) of cisplatin reacted. This is \(0.150 / 300.05 = 0.0005\, \text{mol}\) of \(\text{cisplatin}\). We have \(0.00099\, \text{mol}\) pyridine reacted. Therefore, \(x = \frac{0.00099}{0.0005} = 2\). Hence, the compound's formula is \(\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}_2(\mathrm{C}_5\mathrm{H}_5\mathrm{N})_2 \).

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. This is a fundamental concept in chemistry. Understanding how materials interact with each other helps us predict the outcomes of mixing different chemicals.

In a typical reaction, bonds between atoms in the reactants are broken, and new bonds are formed to create the products. For instance, in the reaction forming cisplatin, \[(\mathrm{NH}_4)_2 \mathrm{PtCl}_4 + 2\mathrm{NH}_3 \rightarrow \operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 + 2\mathrm{NH}_4 \mathrm{Cl}\], we see the transformation of reactants into different substances.

Reactions can be classified in several ways:

• Synthesis: Simple substances combine to form a more complex compound.
• Decomposition: A complex molecule breaks down into simpler ones.
• Single replacement: An element in a compound is replaced by another element.
• Double replacement: The exchange of elements between two compounds.

Understanding reactions help in predicting product quantities and the materials needed for chemical processes.

Molar Mass

Molar mass is a crucial concept that links the microscopic world of atoms to measurable quantities. It represents the mass of one mole of a substance, providing a bridge between grams and molecules.

To calculate the molar mass, add the masses of all atoms within a molecule. For cisplatin, \(\operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\), the molar mass is approximately 300.05 g/mol, calculated from the sum of its atomic components.

Here's how you can calculate it:

• Identify the atoms: Sum the atomic weights of all atoms in the molecule.
• Use periodic table: Atomic weights are available from the periodic table. E.g., \mathrm{Pt} = 195.08, \mathrm{N} \approx 14.01.

This calculation is essential in reaction stoichiometry, allowing chemists to convert between mass and moles when looking at chemical equations.

Chemical Equations

Chemical equations represent reactions using symbols and formulas. They provide a concise way to convey what's occurring on an atomic level.

Balancing chemical equations ensures the same number of each type of atom appears on both sides of the equation. This obeys the law of conservation of mass, which states that mass cannot be created or destroyed in chemical reactions.

Take our reaction:\[(\mathrm{NH}_4)_2\mathrm{PtCl}_4 + 2\mathrm{NH}_3 \rightarrow \operatorname{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 + 2\mathrm{NH}_4 \mathrm{Cl}\] Here's how to balance an equation:

• Write the unbalanced equation: Start with the correct formulas for reactants and products.
• Count atoms for each element: Ensure the same number of atoms on both sides.
• Adjust coefficients: Use coefficients to adjust the number of molecules.
• Repeat: Recheck and reorganize as necessary until balanced.

Balanced equations are crucial for stoichiometric calculations as they show the proportional relationships between different substances in a reaction.

Chemistry Education

Chemistry education focuses on imparting knowledge about chemical principles and their applications in real-world scenarios. Learning chemistry involves both theoretical understanding and practical skills development.

The essentials of chemistry education include:

• Understanding fundamental concepts: Grasp the basics like atoms, molecules, reactions, and equations.
• Practical lab skills: Learn to handle chemicals safely, measure accurately, and conduct experiments.
• Problem-solving abilities: Develop the ability to apply concepts to solve chemical equations and predict reaction outcomes.
• Creative thinking: Innovate and design experiments or solutions for chemical problems.

Effective chemistry education empowers students to explore and analyze scientific questions, preparing them for various scientific fields and everyday decision-making.