Question

A 0.9157 -g mixture of \(\mathrm{CaBr}_{2}\) and NaBr is dissolved in water, and \(\mathrm{AgNO}_{3}\) is added to the solution to form AgBr precipitate. If the mass of the precipitate is \(1.6930 \mathrm{~g}\), what is the percent by mass of \(\mathrm{NaBr}\) in the original mixture?

Short answer

The percent by mass of NaBr in the original mixture is approximately 54.38%.

Step-by-step solution

Calculate molar mass of AgBr

To solve the problem, we first need the molar mass of the precipitate, silver bromide (AgBr). The molar mass of Ag = 107.87 g/mol, and that of Br = 79.90 g/mol. Thus, the molar mass of AgBr is: \[ 107.87 + 79.90 = 187.77 \text{ g/mol} \]

Find moles of AgBr precipitate

Using the mass of the AgBr precipitate (1.6930 g) and its molar mass (187.77 g/mol), we can find the moles of AgBr formed:\[ \text{Moles of AgBr} = \frac{1.6930 \text{ g}}{187.77 \text{ g/mol}} \approx 0.009018 \text{ mol} \]

Relate moles of AgBr to moles of original bromides

Since each mole of AgBr formed corresponds to one mole of Br⁻ ion, and the precipitate results from both \( \text{CaBr}_2 \) and NaBr: - Total moles of Br⁻ available is equal to total moles of AgBr formed, which is approximately \( 0.009018 \) mol.

Establish equations for the original mixture

Let \( x \) be the mass of NaBr and \( 0.9157 - x \) be the mass of \( \text{CaBr}_2 \). We have two unknowns and must establish equations:1. \( x + (0.9157 - x) = 0.9157 \text{ g} \) (total mass equation)2. \( \frac{x}{102.89} + 2 \times \frac{0.9157 - x}{199.89} = 0.009018 \text{ mol} \) (total moles of \( \text{Br}^- \) from both salts).

Calculate x, the mass of NaBr

Solve the equation from step 4 for \( x \):\[ \frac{x}{102.89} + \frac{2(0.9157-x)}{199.89} = 0.009018 \]Multiply through by the least common multiple to clear fractions:\[ 199.89x + 2 \times 102.89 \times (0.9157 - x) = 199.89 \times 102.89 \times 0.009018 \]

Final Calculation

Simplify and solve the equation:\[ 199.89x + 2 \times 102.89(0.9157) - 2 \times 102.89x = 185.28 \]This simplifies to:\[ 199.89x - 205.78x = 185.28 - 0.188348 \]\[ -5.89x = -0.002932 \]\[ x \approx 0.498 \text{ g (mass of NaBr)} \]

Calculate percent by mass of NaBr

Using the mass of NaBr we calculated, find the percent by mass:\[ \text{Percent by mass} = \left( \frac{0.498}{0.9157} \right) \times 100 \approx 54.38 \% \]

Key concepts

Molar Mass Calculation

Understanding molar mass is essential in various chemical calculations. Molar mass, often expressed in grams per mole ( ext{g/mol} ), is the mass of one mole of a substance. For instance, to find the molar mass of silver bromide ( ext{AgBr} ), we simply add the atomic masses of silver ( 107.87 ext{ g/mol} ) and bromine ( 79.90 ext{ g/mol} ). This gives us the molar mass of AgBr as 187.77 ext{ g/mol} . Calculating molar mass is like building blocks; each element contributes its atomic mass to form the overall molar mass of the compound. Mastering this concept allows you to calculate other critical parameters, such as the number of moles in a sample.

Precipitation Reaction

A precipitation reaction occurs when two solutions are mixed, and an insoluble solid forms. This solid is known as a precipitate. In the context of our problem, when AgNO_3 is added to a solution containing CaBr_2 and NaBr , the Ag^+ ions from AgNO_3 react with Br^- ions to form silver bromide ( AgBr ) precipitate. Key features of precipitation reactions include:

• The formation of a solid from two aqueous solutions.
• Color change indicating the presence of a precipitate.
• Balancing chemical equations to predict product formation.

These reactions are vital in analytical chemistry for isolating a compound from a solution, as seen with the AgBr precipitate formation.

Mass Percent Calculation

Mass percent is a way to express the concentration of a component in a mixture. It represents the mass of a specific substance relative to the total mass of the mixture, expressed as a percentage.To calculate the percent by mass of NaBr in our problem, you first need the mass of NaBr and the total mass of the sample. You then use the formula:\[ \text{Percent by mass} = \left( \frac{\text{mass of } \text{NaBr}}{\text{total mass of mixture}} \right) \times 100 \]In this example, using the mass of 0.498 ext{ g} for NaBr and 0.9157 ext{ g} for the entire mixture results in a mass percent of roughly 54.38 ext{%}. Mass percent calculations simplify understanding the relative composition of different substances in a mixture, crucial for chemical analysis.

Moles and Molarity

Moles provide a bridge between atomic and macroscopic chemical quantities. A mole corresponds to 6.022 imes 10^{23} particles (Avogadro's number) and assists in relating mass to the number of entities of a substance. Molarity, on the other hand, is a measure of the concentration of a solute in a solution. It is expressed in terms of moles per liter ( ext{mol/L} ). To find the moles from the mass of AgBr in our problem, we divide its mass by its molar mass, yielding approximately 0.009018 ext{ mol} . Understanding these concepts involves:

• Using molar mass to convert between mass and moles.
• Calculating molarity for various solutions in chemical reactions.
• Applying stoichiometry to predict yields and concentration changes.

These principles are foundational to solving quantitative problems in chemistry effectively.