Question

Cisplatin \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\), a compound used in cancer treatment, is prepared by reaction of ammonia with potassium tetrachloroplatinate: $$ \mathrm{K}_{2} \mathrm{PtCl}_{4}+2 \mathrm{NH}_{3} \longrightarrow 2 \mathrm{KCl}+\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} $$ How many grams of cisplatin are formed from \(55.8 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{PtCl}_{4}\) and \(35.6 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) if the reaction takes place in \(95 \%\) yield based on the limiting reactant?

Short answer

38.33 grams of cisplatin are formed.

Step-by-step solution

Calculate Molar Masses

The first step is to determine the molar masses of the compounds involved. The molar mass of \(\mathrm{K}_2\mathrm{PtCl}_4\) is calculated as follows:- K: \(2 \times 39.1 = 78.2\) g/mol- Pt: \(1 \times 195.08 = 195.08\) g/mol- Cl: \(4 \times 35.45 = 141.8\) g/molThus, the molar mass of \(\mathrm{K}_2\mathrm{PtCl}_4\) is \(78.2 + 195.08 + 141.8 = 415.08\) g/mol.The molar mass of \(\mathrm{NH}_3\) is:- N: \(1 \times 14.01 = 14.01\) g/mol- H: \(3 \times 1.01 = 3.03\) g/molTherefore, the molar mass of \(\mathrm{NH}_3\) is \(14.01 + 3.03 = 17.04\) g/mol.

Convert Mass to Moles

Next, convert the given masses of the reactants to moles:For \(55.8\, \mathrm{g}\) of \(\mathrm{K}_2\mathrm{PtCl}_4\):\[\text{Moles of } \mathrm{K}_2\mathrm{PtCl}_4 = \frac{55.8}{415.08} = 0.1345 \text{ mol}\]For \(35.6\, \mathrm{g}\) of \(\mathrm{NH}_3\):\[\text{Moles of } \mathrm{NH}_3 = \frac{35.6}{17.04} = 2.0891 \text{ mol}\]

Identify Limiting Reactant

The balanced chemical equation:\[\mathrm{K}_2 \mathrm{PtCl}_4 + 2 \mathrm{NH}_3 \rightarrow 2 \mathrm{KCl} + \mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\]shows that 1 mole of \(\mathrm{K}_2\mathrm{PtCl}_4\) reacts with 2 moles of \(\mathrm{NH}_3\).Calculate the amount of \(\mathrm{K}_2\mathrm{PtCl}_4\) needed using \(2.0891\) moles of \(\mathrm{NH}_3\):\[\text{Moles of } \mathrm{K}_2\mathrm{PtCl}_4 = \frac{2.0891}{2} = 1.0446 \text{ moles}\]Since only \(0.1345\) moles of \(\mathrm{K}_2\mathrm{PtCl}_4\) are available, \(\mathrm{K}_2\mathrm{PtCl}_4\) is the limiting reactant.

Theoretical Yield of Cisplatin

Using the limiting reactant \(\mathrm{K}_2\mathrm{PtCl}_4\), the theoretical yield is calculated as 1 mole of \(\mathrm{K}_2\mathrm{PtCl}_4\) produces 1 mole of \(\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\):\[\text{Moles of } \mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2 = 0.1345 \text{ mol}\]Calculate the molar mass of \(\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\):- Pt: \(195.08\) g/mol- N: \(2 \times 14.01 = 28.02\) g/mol- H: \(6 \times 1.01 = 6.06\) g/mol- Cl: \(2 \times 35.45 = 70.9\) g/molThe molar mass is \(195.08 + 28.02 + 6.06 + 70.9 = 300.06\) g/mol.Theoretical yield in grams:\[0.1345 \times 300.06 = 40.35 \text{ grams}\]

Calculate Actual Yield

Considering a 95% yield,\[\text{Actual yield} = 40.35 \times 0.95 = 38.333\text{ grams}\]Thus, the actual yield of cisplatin formed is approximately \(38.33\) grams.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product that can be formed. It's like having a recipe with limited ingredients: you run out of an ingredient, and you can't make any more of the dish.

When determining the limiting reactant, you compare the amount of each reactant that is available to the amount required by the balanced chemical equation. For cisplatin synthesis, the equation is: \[ \mathrm{K}_2 \mathrm{PtCl}_4 + 2 \mathrm{NH}_3 \rightarrow 2 \mathrm{KCl} + \mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2.\] In this reaction, 1 mole of \(\mathrm{K}_2 \mathrm{PtCl}_4\) reacts with 2 moles of \(\mathrm{NH}_3\). By evaluating the moles of each reactant, you compare which one runs out first. When \(\mathrm{K}_2 \mathrm{PtCl}_4\) is depleted, the reaction stops. Thus, it is the limiting reactant.

Theoretical Yield

The theoretical yield is a calculation that predicts the maximum amount of product obtainable from a given amount of limiting reactant. It's based on perfect reaction conditions where all of the limiting reactant is converted into the product.

For the synthesis of cisplatin, once you have identified the limiting reactant (\(\mathrm{K}_2 \mathrm{PtCl}_4\) in this case), you can calculate how much \(\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\) should be produced. You start by using the stoichiometry from the balanced equation: as \(0.1345\) moles of \(\mathrm{K}_2 \mathrm{PtCl}_4\) yields the same moles of cisplatin, the theoretical yield is directly tied to this ratio.

Theoretical yield is an ideal number, as in practice, reactions may never reach this yield due to various factors such as incomplete reactions or side reactions.

Percent Yield

Percent yield provides insight into the efficiency of a reaction by comparing the actual yield to the theoretical yield. It's calculated with the following formula: \[ \text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%.\] In the context of the cisplatin synthesis, with an actual yield of \(38.33\) grams and a theoretical yield of \(40.35\) grams, the percent yield is about \(95\%\).

This means the reaction was quite efficient, though some losses were inevitable, possibly due to material sticking to labware or incomplete reactions. Percent yield helps chemists refine their processes by indicating where improvements can be made.

Molar Mass Calculation

Understanding molar mass is crucial in converting between the mass of a substance and the moles of that substance. The molar mass is the mass of one mole of a compound and includes the combined atomic masses of all atoms in the molecule.

For \(\mathrm{K}_2 \mathrm{PtCl}_4\), its molar mass calculation is: - Potassium (K): \(2 \times 39.1 = 78.2\) g/mol - Platinum (Pt): \(195.08\) g/mol - Chlorine (Cl): \(4 \times 35.45 = 141.8\) g/mol Adding these, the molar mass is \(415.08\) g/mol.

Similarly, for \(\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2\): - Platinum (Pt): \(195.08\) g/mol - Nitrogen (N): \(2 \times 14.01 = 28.02\) g/mol - Hydrogen (H): \(6 \times 1.01 = 6.06\) g/mol - Chlorine (Cl): \(2 \times 35.45 = 70.9\) g/mol The total is \(300.06\) g/mol.

These calculations are the foundation for mole-to-mass conversions that are essential in predicting yields.