Question

A \(0.8612-\mathrm{g}\) sample of a mixture of \(\mathrm{NaBr}\), \(\mathrm{NaI}\), and \(\mathrm{NaNO}_{3}\) is analyzed by adding \(\mathrm{AgNO}_{3}\) and precipitating a \(1.0186-\mathrm{g}\) mixture of \(\mathrm{AgBr}\) and AgI. The precipitate is then heated in a stream of \(\mathrm{Cl}_{2}\), which converts it to \(0.7125 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{NaNO}_{3}\) in the sample.

Short answer

% w/w NaNO3 in the sample is calculated by finding the mass of NaNO3 remaining after determining the masses of NaBr and NaI from the conversions.

Step-by-step solution

Understanding the Problem

The problem involves a mixture containing three sodium salts: NaBr, NaI, and NaNO3. We know the total mass of this mixture is 0.8612 g. By reacting this mixture with AgNO3, only NaBr and NaI will form precipitates, AgBr and AgI, while NaNO3 will remain in solution.

Identify and Apply Stoichiometry Relationships

First, when the mixture is treated with AgNO3, the reaction forms AgBr from NaBr and AgI from NaI:\[ \text{NaBr (aq) + AgNO}_{3} \rightarrow \text{AgBr (s) + NaNO}_{3} \]\[ \text{NaI (aq) + AgNO}_{3} \rightarrow \text{AgI (s) + NaNO}_{3} \]All non-iodide and bromide parts of the Na salts do not form precipitates.

Transition from AgBr and AgI to AgCl

The method then involves heating the precipitate mixture of AgBr and AgI in Cl2 to form AgCl:\[ \text{AgBr} + \text{Cl}_{2} \rightarrow \text{AgCl} + \text{Br}_2 \]\[ \text{AgI} + \text{Cl}_{2} \rightarrow \text{AgCl} + \text{I}_2 \]From the mass of the AgCl formed (0.7125 g), we can deduce the original masses of AgBr and AgI since Ag only forms AgCl with a molar ratio of 1:1 during conversion.

Use Molar Mass to Separate AgBr and AgI

The molar masses are approximately: AgBr = 187.77 g/mol, AgI = 234.77 g/mol, and AgCl = 143.32 g/mol. Using conservation of Ag:\[ \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} = \frac{\text{mass of AgBr}}{\text{molar mass of Ag}} + \frac{\text{mass of AgI}}{\text{molar mass of Ag}} \]Translate this relationship using the initial total mass of 1.0186 g of AgBr and AgI.

Calculate Mole Fractions and Mass of NaNO3

Using the mole fraction and mass relationships derived in Step 4, determine the individual masses of NaBr and NaI. Then, the percentage mass of NaNO3 is: \[ \text{Mass of NaNO3} = 0.8612 \text{ g} - (\text{mass of NaBr} + \text{mass of NaI}) \].

Calculate the %w/w of NaNO3

Finally, compute the % w/w for NaNO3:\[ \% \text{w/w NaNO}_{3} = \left( \frac{\text{mass of NaNO}_{3}}{0.8612\text{ g}} \right) \times 100 \% \].

Key concepts

Molar Mass

Molar mass is a central concept in chemistry, crucial for converting between the mass of a substance and the amount in moles. It essentially tells us how much one mole of a chemical substance weighs. Molar mass is measured in grams per mole (g/mol). For example, the molar mass of AgCl is approximately 143.32 g/mol. This means that one mole of AgCl weighs 143.32 grams.
Understanding molar mass allows us to perform stoichiometric calculations, where we can determine the mass of reactants and products in a chemical reaction. In the given exercise, we are transitioning between different silver halides and eventually forming AgCl. Calculating the molar masses of AgBr, AgI, and AgCl allows us to use their respective molecular weights to find out how much of each compound participates in or results from a reaction.
It works through a simple formula:

• Molar mass of compound = Sum of atomic masses of all atoms in the formula
• Example: Molar mass of \( ext{AgBr} = 107.87 + 79.90 = 187.77 \text{ g/mol}\)

Using these molar masses, you can calculate the mass of any one compound if you have the mass of another related compound, as they all are connected through their stoichiometric coefficients.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two soluble salts (ionic compounds) react in aqueous solution to form an insoluble solid, called a precipitate. In the exercise, when \( ext{AgNO}_3\) is added to the solution containing \( ext{NaBr}\) and \( ext{NaI}\), a precipitation reaction occurs. The silver ions \( ext{Ag}^+\) react with bromide (Br\(^-\)) and iodide (I\(^-\)) ions to form AgBr and AgI.
These reactions are a common way to isolate or identify substances in a mixture by removing the unwanted soluble ions:

• \( ext{NaBr (aq) + AgNO}_{3} (aq) \rightarrow \text{AgBr (s) + NaNO}_{3} (aq)\)
• \( ext{NaI (aq) + AgNO}_{3} (aq) \rightarrow \text{AgI (s) + NaNO}_{3} (aq)\)

In both reactions, a solid silver halide is precipitated out of the aqueous mixture, simplifying the process of further chemical analysis or separation. Precipitation helps us understand which components are part of the original mixture and simplifies their quantitative analysis.

Chemical Calculations

Chemical calculations can seem daunting, but they follow a logical sequence that eases the understanding process. These calculations allow us to quantify components in a chemical reaction, such as determining how much of a reactant is needed or how much of a product is formed. In the given exercise, the chemical calculations primarily involved using the law of conservation of mass and stoichiometric relationships.
To perform such calculations, it’s necessary to:

• Identify what you are solving for (e.g., mass of NaNO\(_3\)).
• Write down the known quantities (e.g., total mass, molar masses of compounds).
• Apply stoichiometry to find unknowns by setting equations based on molar relationship and mass conservation, ensuring units are consistent.

Through stoichiometric calculations, one can find out the mass of components not directly precipitated, like NaNO\(_3\). For example, using the known total mass, you calculate the mass of each compound after the reaction, subtracting known values from totals provided. These calculated values then help derive quantities and concentrations, given in percentages like the w/w (")percentage by weight") calculation concerning NaNO\(_3\) in this problem.