Question

A solid chloride sample weighing 0.3147 g required \(43.75 \mathrm{mL}\) of \(0.05273 \mathrm{M}\) AgNO, to reach the \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) end point. a. How many moles \(\mathrm{Cl}^{-}\) ion were present in the sample? (Use Eqs. 2 and \(3 .\) ) b. How many grams Cl- ion were present? (Use Eq. 4.) c. What was the mass percent \(\mathrm{Cl}^{-}\) ion in the sample? (Use Eq. 5.)

Short answer

a. The number of moles of Cl- ions present in the sample can be found using the formula n(Cl-) = n(AgNO₃), where n(AgNO₃) = 0.05273 M x 43.75 mL x (1L/1000mL). After calculation, n(Cl-) = \(2.303 \times 10^{-3}\) moles. b. The mass of Cl- ions present in the sample can be found using the formula m(Cl-) = n(Cl-) x M(Cl-), where M(Cl-) = 35.45 g/mol. After calculation, m(Cl-) = 0.08164 g. c. The mass percentage of Cl- ions in the sample can be found using the formula Mass percentage (Cl-) = (m(Cl-) / 0.3147 g) x 100. After calculation, Mass percentage (Cl-) = 25.93%.

Step-by-step solution

Calculate the number of moles of Cl- ions

First, we need to determine how many moles of Cl- ions were present in the sample. We'll use the given volume and molarity of AgNO₃ solution to find the number of moles of AgNO₃ (since it reacts in a 1:1 ratio with Cl- ions). We'll use the formula n = c x V, where n is the number of moles, c is the concentration, and V is the volume. n(AgNO₃) = c(AgNO₃) x V(AgNO₃) n(AgNO₃) = 0.05273 M x 43.75 mL x (1L/1000mL) Now, calculate the moles of AgNO₃. Since it reacts in a 1:1 ratio with Cl- ions, the number of moles of Cl- ions will be equal to the number of moles of AgNO₃. n(Cl-) = n(AgNO₃)

Calculate the mass of Cl- ions

To find the mass of Cl- ions, we'll use the formula m = n x M, where m is the mass, n is the number of moles, and M is the molar mass of Cl- ions. The molar mass of Cl- ions is 35.45g/mol. m(Cl-) = n(Cl-) x M(Cl-) Now, calculate the mass of Cl- ions.

Calculate the mass percentage of Cl- ions in the sample

To find the mass percentage of Cl- ions in the sample, we'll use the following formula: Mass percentage = (mass of Cl- ions / total mass of the sample) x 100 The total mass of the sample is given as 0.3147 g. Mass percentage (Cl-) = (m(Cl-) / 0.3147 g) x 100 Now, calculate the mass percentage of Cl- ions in the sample.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles, which allows chemists to predict the quantities of substances consumed and produced in a reaction.

For instance, stoichiometry uses balanced chemical equations to calculate the amount of reactants needed to form a certain amount of product, or to find out what amount of product can be expected from a given quantity of reactants. In our exercise, we're using stoichiometry to understand the reaction between a chloride sample and AgNO₃ to form AgCl and how it relates to the Cl^- content in the sample.

Key Steps in Stoichiometry:

• Write and balance the chemical equation for the reaction.
• Convert all given information into moles (if not already provided in moles).
• Use the mole ratio from the balanced equation to find the number of moles of the desired compound.
• Convert moles back to grams if necessary, using the molar mass of the compound.
• Calculate the percent composition, yield, or purity if required.

Chemical Analysis

Chemical analysis involves determining the composition and quantities of chemical substances in a sample. There are many techniques used for this purpose, including titration, which is highlighted in the given exercise.

In titration, a solution with a known concentration (the titrant) is used to determine the concentration of an analyte present in a sample. The point at which the reaction between the titrant and the analyte is complete is known as the end point, indicated by a change in color or the formation of a precipitate. In our exercise, the titration of a chloride sample with AgNO₃ reached the end point when Ag₂CrO₄ formed, indicating the chloride has been completely reacted.

Importance of Precise Measurement:

• Accurately measuring the volume of titrant dispensed is crucial for correct calculations.
• The concentration of the titrant must be known with precision.
• The end point determination should be precise to ensure the correct stoichiometric ratio.

Molarity and Concentration

Molarity and concentration are essential concepts in stoichiometry and chemical analysis. Molarity (M) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute divided by the volume of solution in liters. In the context of our exercise, we're using the molarity of the AgNO₃ solution to determine the amount of Cl^- ions in the sample.

The molarity allows us to directly relate the volume of the AgNO₃ solution used in the titration to the number of moles of Cl^- ions through the 1:1 reaction ratio. Once we have the number of moles of Cl^- ions, we can use the concept of molar mass to find the mass in grams and eventually the mass percent of Cl^- in the original sample.

Key Points on Concentration:

• Molarity is particularly useful in titrations and preparing solutions with desired properties.
• Changes in temperature and volume can affect the concentration of a solution.
• Understanding molarity is crucial for accurate stoichiometric calculations in chemical reactions.