Problem 19 - Analytical Chemistry

Question

Ammonium ions can be analyzed by precipitating with \(\mathrm{H}_{2} \mathrm{PtCl}_{6}\) as \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{6}\) and then igniting the precipitate to platinum metal, which is weighed \(\left[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{6} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Pt}+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{g})+\right.\) \(\left.2 \mathrm{Cl}_{2}(\mathrm{~g})\right] .\) Calculate the percent ammonia in a \(1.00-\mathrm{g}\) sample that yields \(0.100 \mathrm{~g} \mathrm{Pt}\) by this method.

Step-by-Step Solution

Answer

The percent ammonia in the sample is approximately 1.75%.

Step 1: Write the Balanced Chemical Equation

The reaction provided is: \[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{6} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Pt} + 2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{g}) + 2 \mathrm{Cl}_{2}(\mathrm{~g})\]This indicates that one mole of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_6\) yields one mole of Pt.

Step 2: Find Molar Masses

Calculate the molar masses required for conversion.- Molar mass of Pt = 195.08 g/mol - Molar mass of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_6\) = 18.04 (each NH4) + 195.08 (Pt) + 6(35.45) (each Cl) = 470.37 g/mol.

Step 3: Use Given Yield to Find Moles of Pt

Given that 0.100 g of Pt was produced, use the molar mass of Pt to determine moles:\[\text{Moles of Pt} = \frac{0.100 \text{ g}}{195.08 \text{ g/mol}} \approx 0.000513 \text{ moles of Pt}\]

Step 4: Relate Moles of Pt to Moles of Ammonium Salt

Since the stoichiometry of the reaction shows a 1:1 ratio:\[ 0.000513 \text{ moles of Pt} = 0.000513 \text{ moles of } (\mathrm{NH}_4)_2 \mathrm{PtCl}_6\]

Step 5: Convert Moles of Ammonium Salt to Mass

Convert the moles of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_6\) back to grams:\[\text{Mass of } (\mathrm{NH}_4)_2 \mathrm{PtCl}_6 = 0.000513 \times 470.37 \text{ g/mol} \approx 0.241 \text{ g}\]

Step 6: Calculate Mass of Ammonia in Sample

There are 2 moles of \(\mathrm{NH}_4^+\) per mole of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_6\), hence 2 moles of NH3 per mole.\[ 0.000513 \text{ moles of } (\mathrm{NH}_4)_2 \mathrm{PtCl}_6 = 0.001026 \text{ moles of NH}_3\]Molar mass of NH3 is 17.03 g/mol, so:\[\text{Mass of NH}_3 = 0.001026 \times 17.03 \approx 0.0175 \text{ g}\]

Step 7: Calculate Percent Ammonia in the Sample

Use the mass of NH3 to find the percent in the original 1.00 g sample:\[\text{Percent NH}_3 = \left(\frac{0.0175 \text{ g}}{1.00 \text{ g}}\right) \times 100 \approx 1.75\%\]

Key Concepts

Gravimetric Analysis

Gravimetric analysis is a method in quantitative chemical analysis where the amount of an analyte is determined by measuring a change in mass. In this particular exercise, gravimetric analysis involves converting the ammonium ions into a solid precipitate by using a chemical reaction with hexachloroplatinic acid Chemical reactions can be carried out so that the analyte forms a compound of known composition, which then can be ignited or weighed directly. In our exercise, the ammonium ion is precipitated as

• For this exercise, the mass change helps us determine the nitrogen component by igniting the precipitate and converting it into pure platinum (Pt).

The chemical stoichiometry provides information on the reactants and the products, and gravimetric methods can offer a high degree of accuracy. Remember: - The point of gravimetric analysis is to transform a studied substance into a pure and accurately measurable form. - The substance formed must be simple and pure compounds that are easily isolated.

Molar Mass Calculation

Molar mass calculations are essential in the quantitative analysis to convert between grams of a substance and moles of a substance. Molar mass is the mass of one mole of a chemical compound.To find the molar mass of a compound, you simply add the molar masses of the individual elements contained in the compound. Here’s how you do it for \In this exercise:

• Molar mass of Pt (platinum) = 195.08 g/mol.
• Molar mass of (NH₄)₂PtCl₆ =\(2(14.01 + 1.01 \times 4)+ 195.08 + 6 \times 35.45 = 470.37\) g/mol.

With these values, we determined moles from a given mass. This step sets the groundwork for finding out more about the quantity of a substance produced or consumed in a reaction.Understanding how to calculate molar mass aids you in systematically analyzing chemical reactions, which involves both the substance's quantity and composition.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions using balanced chemical equations. It acts as the foundation for predicting the outcomes of chemical processes.For the given exercise:- The reaction is\((\mathrm{NH}_4)_2 \mathrm{PtCl}_6 \rightarrow \mathrm{Pt} + 2 \mathrm{NH}_4 \mathrm{Cl} + 2 \mathrm{Cl}_2\). Each reactant and product is directly connected in the reaction by moles, serving as conversion factors.

• Here, understanding the 1:1 mole ratio between platinum and the ammonium salt is crucial. This shows that one mole of platinum produced implies that one mole of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_6\) reacted.

Another thing to note:- In stoichiometry, you need the balanced chemical equation as the roadmap – it tells you how moles of reactants are transformed to moles of products.- It allows the calculation of definite ratios which aid in the prediction and verification of the mass of products formed compared to reactants consumed.Understanding stoichiometry is about recognizing these relationships and applying them to explore the quantities involved in chemical reactions.

Percent Composition

Percent composition is determining the percentage by mass of each element in a compound. This helps you know how much of each element is present in a mixture or compound.In our exercise, we calculated the percent ammonia in a 1.00-g sample by first finding the mass of the actual ammonia formed:- By reacting, we found the mass of \(\mathrm{NH}_3\) to be approximately 0.0175 g.Then calculated \\[\text{Percent NH}_3 = \left( \frac{0.0175}{1.00} \right) \times 100 = 1.75\%\]

• The concept of percent composition allows us to compare the mass contribution of each element in a compound or a particular component within a mixture.

Knowing percent composition is useful in fields such as formulation chemistry, where understanding the precise makeup of components in a formula is key.