Problem 14 - Analytical Chemistry

Question

A 523.1 -mg sample of impure \(\mathrm{KBr}\) is treated with excess \(\mathrm{AgNO}_{3}\) and \(814.5 \mathrm{mg} \mathrm{AgBr}\) is obtained. What is the purity of the KBr?

Step-by-Step Solution

Answer

The purity of the KBr sample is 98.74%.

Step 1: Understand the Reaction

The reaction between potassium bromide (KBr) and silver nitrate (AgNO3) produces silver bromide (AgBr) as a precipitate. The balanced chemical equation for the reaction is:\[ \text{KBr} + \text{AgNO}_3 \rightarrow \text{AgBr} + \text{KNO}_3 \]

Step 2: Convert Mass to Moles for AgBr

First, find the molar mass of AgBr. Silver (Ag) has a molar mass of 107.87 g/mol and bromine (Br) has a molar mass of 79.90 g/mol, so:\[ \text{Molar mass of AgBr} = 107.87 + 79.90 = 187.77 \text{ g/mol} \]The mass of AgBr obtained is 814.5 mg, which is 0.8145 g. Convert this mass to moles:\[ \text{Moles of AgBr} = \frac{0.8145}{187.77} = 0.00434 \text{ moles} \]

Step 3: Relate Moles of KBr to Moles of AgBr

Since the reaction between KBr and AgNO3 is in a 1:1 ratio, the moles of KBr that reacted are equal to the moles of AgBr produced. Thus, the moles of KBr that reacted is 0.00434 moles.

Step 4: Calculate Mass of Pure KBr

The molar mass of KBr is the sum of the molar masses of K (39.10 g/mol) and Br (79.90 g/mol):\[ \text{Molar mass of KBr} = 39.10 + 79.90 = 119.00 \text{ g/mol} \]Convert moles of pure KBr back to mass:\[ \text{Mass of pure KBr} = 0.00434 \times 119.00 = 0.5165 \text{ g} \] or 516.5 mg.

Step 5: Calculate Purity

Purity is calculated by dividing the mass of pure KBr by the mass of the impure sample and multiplying by 100%.\[ \text{Purity} = \left(\frac{516.5}{523.1}\right) \times 100 = 98.74\% \]

Key Concepts

Precipitation Reactions

Precipitation reactions are a fascinating type of chemical process where two soluble salts in a solution react to form an insoluble solid, known as the precipitate. In this process, the ions from the different solutions exchange partners. As a result, a new combination forms, which is no longer soluble, and so it precipitates out of the solution.
In the given exercise, potassium bromide ( KBr ) reacts with silver nitrate ( AgNO_3 ). This swap between partners yields silver bromide ( AgBr ), an insoluble precipitate, and potassium nitrate ( KNO_3 ), which remains dissolved in the solution. The precipitation of AgBr is important in our exercise because it's used to determine how much pure KBr was originally present in the impure sample.

• Common uses include water softening, in which water-hardening ions precipitate out.
• A helpful analogy is sugar water: when enough sugar is added, it no longer stays dissolved, similar to a precipitate.

This type of reaction is crucial in various fields, from analytical chemistry to environmental science, for isolating and identifying substances.

Molar Mass Calculation

Molar mass plays a crucial role in quantitative analysis. It is the mass of one mole of a substance, typically expressed in grams/mole (g/mol). Understanding how to calculate it allows us to relate grams of a compound to moles, which enables us to interpret chemical reactions quantitatively.
For our example involving AgBr, first we determine the molar mass by adding together the atomic masses of silver (Ag) and bromine (Br). Silver has a molar mass of 107.87 g/mol, and bromine is 79.90 g/mol. Therefore, the molar mass of AgBr is calculated as:
\[ 107.87 + 79.90 = 187.77 \text{ g/mol} \]

• Atomic masses can be found on the periodic table or in chemical handbooks.
• Always ensure precision in these calculations for accurate results.

This calculation helps us convert the mass of AgBr obtained in reaction into moles, by dividing the mass obtained (in grams) by the molar mass. This step is foundational for finding how much of the original reactant was present, paving the way to calculate purity.

Purity Determination

Purity determination is significant in fields that require high-quality substances, such as pharmaceuticals and materials science. It tells us how much of the desired compound is in a sample, expressed as a percentage.
To find purity, we take the mass of the pure compound calculated from the reaction data and divide it by the initial mass of the impure sample. Then, multiply by 100 to convert it into a percentage. For KBr, we calculated the mass of pure KBr as 516.5 mg. The starting impure sample weighed 523.1 mg, leading to a purity calculation of:
\[ \left( \frac{516.5}{523.1} \right) \times 100 = 98.74\% \]

• This calculation assumes all impurities do not react or alter the main reaction.
• High purity is often required for precise scientific measurements.

The result indicates the percentage of the sample that is pure, guiding decisions in quality control and further refining processes.