Problem 66 - Chemistry An Atoms First Approach

Question

You are given a \(1.50-g\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{mL}\) of water and then add an excess of 0.500 \(M\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.64 \(1 \mathrm{g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

Step-by-Step Solution

Answer

a. In the solution, you would see the following species at the atomic-molecular level: Sodium ions (Na⁺), Nitrate ions (NO₃⁻), Silver ions (Ag⁺), and Chloride ions (Cl⁻). b. The balanced net ionic equation for the reaction that produces the solid is: Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s) c. To calculate the percent sodium chloride in the original unknown mixture, first, find the moles of AgCl formed (≈ 0.00447 mol) and the moles of NaCl in the mixture (also ≈ 0.00447 mol). The mass of NaCl in the original mixture is ≈ 0.261 g. The percent of sodium chloride in the original mixture is ≈ 17.4%.

Step 1: Identify the species present in the solution at the atomic-molecular level

In the solution, we have sodium nitrate (NaNO₃), sodium chloride (NaCl), and silver nitrate (AgNO₃). Sodium nitrate and sodium chloride are dissolved in water, and silver nitrate is added. Sodium nitrate and sodium chloride are strong electrolytes, meaning they completely dissociate into ions in a solution. Thus, the solution will have the following ions: - Sodium ions (Na⁺) - Nitrate ions (NO₃⁻) - Silver ions (Ag⁺) - Chloride ions (Cl⁻)

Step 2: Write the balanced net ionic equation for the reaction that produces the solid

A precipitate will form when a silver ion (Ag⁺) reacts with a chloride ion (Cl⁻). The molecular equation for this reaction is: AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq) Now, we write the full ionic equation: Ag⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) → AgCl (s) + Na⁺ (aq) + NO₃⁻ (aq) Cancel the spectator ions (ions that are the same on both sides of the equation): Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s) This is the net ionic equation for the reaction that produces the solid.

Step 3: Calculate the mass of sodium chloride in the original mixture

We are given that 0.641 g of white solid is formed. This white solid is silver chloride (AgCl). We can use the stoichiometry of the net ionic equation to determine the mass of sodium chloride (NaCl) in the original mixture. First, let's find the moles of AgCl formed: moles of AgCl = mass of AgCl / molar mass of AgCl molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol moles of AgCl = 0.641 g / 143.32 g/mol ≈ 0.00447 mol From the net ionic equation, the mole ratio of NaCl to AgCl is 1:1. Therefore, the moles of NaCl in the original mixture are equal to the moles of AgCl formed: moles of NaCl = 0.00447 mol Now, we can calculate the mass of NaCl in the original mixture: mass of NaCl = moles of NaCl × molar mass of NaCl molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol mass of NaCl = 0.00447 mol × 58.44 g/mol ≈ 0.261 g

Step 4: Calculate the percent sodium chloride in the original unknown mixture

We are given that the original mixture has a mass of 1.50 g. The mass of sodium chloride in the original mixture is 0.261 g. We can now calculate the percent of sodium chloride in the original mixture: percent sodium chloride = (mass of NaCl / mass of mixture) × 100% percent sodium chloride = (0.261 g / 1.50 g) × 100% ≈ 17.4% Thus, the percentage of sodium chloride in the original unknown mixture is approximately 17.4%.

Key Concepts

Net Ionic Equation

Understanding the net ionic equation is a crucial aspect of chemical reactions in solution. When certain chemicals are mixed, not all the species present participate directly in the reaction. Some remain unchanged and are called 'spectator ions'. The net ionic equation represents only the chemical species that undergo a change.

For the reaction involving sodium chloride and silver nitrate, the net ionic equation excludes the sodium and nitrate ions, focusing on the formation of silver chloride precipitate. The equation, \( Ag^+ (aq) + Cl^- (aq) \rightarrow AgCl (s) \), conveys essential information on the actual chemistry taking place—the reaction between silver and chloride ions forming a solid. Students should focus on understanding how to identify and remove spectator ions to write clear and concise net ionic equations.

Mole Ratio

The mole ratio is the heart of stoichiometry and is derived from the coefficients of a balanced chemical equation. It tells us the relative amount of reactants and products in a chemical reaction. For instance, in the reaction between sodium chloride and silver nitrate, the mole ratio of NaCl to AgCl, according to the balanced net ionic equation, is 1:1. This tells us that one mole of sodium chloride reacts with one mole of silver ions to produce one mole of silver chloride.

Understanding mole ratios enables students to perform stoichiometric calculations, which are vital in determining quantities of reactants or products in a chemical reaction. By grasping this concept, one can easily convert between moles of different substances involved in a reaction.

Percent Composition

Percent composition refers to the percentage by mass of each element in a compound or a mixture. It is an essential concept in chemistry for the analysis of substances and can reveal the purity of a substance or the composition of a mixture, as in the exercise provided.

In the given solution, calculating the percent composition involved determining the mass of sodium chloride initially present in the mixture. Using the mole ratio and the mass of the precipitate formed, we can calculate the mass of NaCl, and subsequently, its proportion in the mixture. The formula \( \text{percent composition} = (\frac{\text{mass of component}}{\text{total mass of mixture}}) \times 100\% \) was used to determine that the mixture contains approximately 17.4% sodium chloride.

Students should note that understanding percent composition not only helps in solving textbook problems but is also vital in practical chemistry where the composition of mixtures and compounds needs to be quantified.