Problem 68 - Chemistry The Molecular Science

Question

Cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), a drug used in the treatment of cancer, can be made by the reaction of \(\mathrm{K}_{2} \mathrm{PtCl}_{4}\) with ammonia, \(\mathrm{NH}_{3}\). Besides cisplatin, the other product is \(\mathrm{KCl}\) (a) Write a balanced equation for this reaction. (b) To obtain \(2.50 \mathrm{~g}\) cisplatin, calculate what masses (in grams) of \(\mathrm{K}_{2} \mathrm{PtCl}_{4}\) and ammonia you need.

Step-by-Step Solution

Answer

Balanced Equation: \( \mathrm{K}_{2} \mathrm{PtCl}_{4} + 2 \mathrm{NH}_{3} \rightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} + 2 \mathrm{KCl} \). Masses needed: \(3.45\) g of \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) and \(0.285\) g of \( \mathrm{NH}_{3} \).

Step 1: Write the Formula for Reactants and Products

Identify the chemical formula for each reactant and product in the reaction. We have reactants as \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) and \( \mathrm{NH}_{3} \), and products as \( \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} \) and \( \mathrm{KCl} \).

Step 2: Balance the Equation

To balance the equation, ensure that each element has the same number on both sides. The initial unbalanced equation is: \[ \mathrm{K}_{2} \mathrm{PtCl}_{4} + \mathrm{NH}_{3} \rightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} + \mathrm{KCl} \]. To balance:\[ \mathrm{K}_{2} \mathrm{PtCl}_{4} + 2 \mathrm{NH}_{3} \rightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} + 2 \mathrm{KCl} \].

Step 3: Calculate Molar Masses

For calculations in part (b), determine the molar masses of the compounds involved: \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) is 414.10 g/mol, \( \mathrm{NH}_{3} \) is 17.03 g/mol, and \( \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} \) is 300.05 g/mol.

Step 4: Use Stoichiometry to Find Required Masses

To calculate the masses of \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) and \( \mathrm{NH}_{3} \):- We need 1 mole of \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) to get 1 mole of \( \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} \).- From the balanced equation: 1 mole \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) (414.10 g) produces 1 mole of cisplatin (300.05 g).Using proportion for 2.50 g cisplatin: \( \frac{2.50 \, \mathrm{g} \; \mathrm{cisplatin}}{300.05 \, \mathrm{g}/\text{mol} } = 8.33 \times 10^{-3} \, \text{mol cisplatin}\)Therefore, mass of \( \mathrm{K}_{2} \mathrm{PtCl}_{4} \) needed:\( 8.33 \times 10^{-3} \, \text{mol} \times 414.10 \, \mathrm{g}/\text{mol} = 3.45 \, \mathrm{g}\)For \( \mathrm{NH}_{3} \), the mole ratio is 2:1 with cisplatin:\( 8.33 \times 10^{-3} \, \text{mol cisplatin} \times 2 = 1.67 \times 10^{-2} \, \text{mol NH}_{3}\)Therefore, mass of \( \mathrm{NH}_{3} \) needed:\( 1.67 \times 10^{-2} \, \text{mol} \times 17.03 \, \mathrm{g}/\text{mol} = 0.285 \, \mathrm{g}\)

Key Concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They display the reactants on the left and the products on the right. For the formation of cisplatin from potassium tetrachloroplatinate (K\(_2\)PtCl\(_4\)) and ammonia (NH\(_3\)), the reactants and products are represented by their chemical formulas. Through chemical equations, we can see which substances transform into other substances under chemical reactions.
For our reaction, we start with K\(_2\)PtCl\(_4\) and NH\(_3\) as reactants. These undergo a transformation to produce cisplatin, Pt(NH\(_3\))\(_2\)Cl\(_2\), and potassium chloride (KCl) as products. This translation of chemicals into equations allows us to understand the quantitative relationships between the substances involved.
Recognizing these symbols and formulating equations are fundamental steps in solving stoichiometry problems effectively, as they set the stage for balancing and further calculations.

Molar Mass Calculation

Molar mass is the weight of one mole of a given substance and is expressed in grams per mole (g/mol). Understanding how to calculate molar masses is essential for determining the quantities of reactants needed in a chemical reaction. The molar mass helps to convert between grams and moles, enabling precise calculations.
To find the molar mass of a compound:

• Add together the atomic masses of all the atoms in its formula.
• Use the periodic table to find each element's atomic mass.

For example, in our reaction:

• K\(_2\)PtCl\(_4\) has a molar mass of 414.10 g/mol.
• NH\(_3\) has a molar mass of 17.03 g/mol.
• Cisplatin, Pt(NH\(_3\))\(_2\)Cl\(_2\), has a molar mass of 300.05 g/mol.

These molar mass calculations are critical steps that allow us to assess the correct amounts of chemicals to use and help predict product formation accurately.

Balanced Chemical Reaction

Balancing chemical reactions ensures that the same number of each type of atom is present on both sides of the equation. This is a reflection of the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
Initially, a chemical reaction may not be balanced naturally. For example, in the unbalanced form of our reaction:

• K\(_2\)PtCl\(_4\) + NH\(_3\) → Pt(NH\(_3\))\(_2\)Cl\(_2\) + KCl

This equation does not have equal numbers of chlorine atoms on both sides. Balancing involves adjusting the coefficients (the numbers before molecules) to ensure atoms are conserved.

• For our balanced equation: K\(_2\)PtCl\(_4\) + 2 NH\(_3\) → Pt(NH\(_3\))\(_2\)Cl\(_2\) + 2 KCl.

Balancing equations is a crucial skill as it allows chemists to calculate precise stoichiometry. By knowing the exact proportions of reactants and products, we can proceed with gram-to-gram calculations directly informed by the balanced chemical equation.