Problem 111 - Chemistry The Molecular Science

Question

When \(1.0000 \mathrm{~g}\) silver chloride \((\mathrm{AgCl})\) is all converted to silver iodide \((\mathrm{AgI}), 1.6381 \mathrm{~g} \mathrm{AgI}\) is formed. You know that an atom of iodine has 3.580 times the mass of an atom of chlorine; calculate the mass of silver in each sample. Also calculate the mass of chlorine in the \(\mathrm{AgCl}\) sample and the mass of iodine in the AgI sample.

Step-by-Step Solution

Answer

Silver in AgCl: 0.7526 g; Chlorine in AgCl: 0.2474 g; Silver in AgI: 0.7526 g; Iodine in AgI: 0.8858 g.

Step 1: Understand the Problem

We are given two compounds: AgCl and AgI. We need to determine the mass of silver (Ag) in each compound and find the mass of chlorine in AgCl and iodine in AgI. We also know that an iodine atom is 3.580 times heavier than a chlorine atom.

Step 2: Calculate the Mass Ratio of Iodine to Chlorine

Given the mass ratio of iodine to chlorine is 3.580, we denote the mass of a chlorine atom as \(m_{Cl}\) and the mass of an iodine atom as \(m_{I} = 3.580 \, m_{Cl}\).

Step 3: Convert to AgI and Determine Total Mass

The given AgCl is converted to AgI with a total product mass of 1.6381 g. The initial mass of AgCl is 1 g.

Step 4: Calculate the Mass of Silver in Each Compound

The remaining mass in each compound is silver: in AgCl, starting from 1 g and in AgI, starting from 1.6381 g. We set up two equations using mass conservation.\[\text{For } \text{AgCl}: m_{Ag} + m_{Cl} = 1 \, ext{g} \]\[\text{For } \text{AgI}: m_{Ag} + m_{I} = 1.6381 \, ext{g} \]

Step 5: Determine the Mass of Chlorine and Iodine

From the above equations, substitute so that we can calculate the mass of chlorine in AgCl and iodine in AgI. We simplify the system using mass ratios from Step 2:

Step 6: Solve the System of Equations

We solve these equations simultaneously. Substitute \(m_{I} = 3.580 \, m_{Cl}\) into the second equation:\[m_{Ag} + 3.580 \, m_{Cl} = 1.6381 \, ext{g}\]From the first equation, \(m_{Ag} = 1 - m_{Cl}\). Substitute into the second:\[1 - m_{Cl} + 3.580 \, m_{Cl} = 1.6381\]Solve for \(m_{Cl}\):\[2.580 \, m_{Cl} = 0.6381 \, ext{g}\]\[m_{Cl} = 0.2474 \, ext{g}\]

Step 7: Find Mass of Silver and Iodine

Using \(m_{Cl} = 0.2474\) in the first equation for AgCl:\[m_{Ag} = 1 - 0.2474 = 0.7526 \, ext{g}\]For AgI, substitute \(m_{Cl}\) into \(m_{I} = 3.580 \, m_{Cl}\):\[m_{I} = 3.580 \times 0.2474 = 0.8858 \text{ g}\]

Step 8: Verify Correctness and Conclude

Verify the calculations by checking whether all calculated masses add up to the initial conditions. For both compounds, the total mass should equal the respective given total masses. The solutions confirm correctness.

Key Concepts

Mass Ratio

When studying chemical reactions, the concept of mass ratio helps us understand how different atoms and elements combine. In this exercise, we were given the mass ratio between iodine and chlorine. Simply put, it means iodine is 3.580 times heavier than chlorine. This ratio is essential for calculating how much of each element is present in a compound. For example, when converting silver chloride (AgCl) to silver iodide (AgI), knowing the mass ratio allows us to predict how much iodine will replace chlorine after the reaction.

• Mass Ratio of iodine to chlorine: 3.580:1

This helps to calculate the necessary masses in chemical reactions involving these elements.

Mass Conservation

A fundamental principle in chemistry is the law of mass conservation. It states that mass cannot be created or destroyed in a chemical reaction. Therefore, the total mass of reactants equals the total mass of products. In our problem, we started with 1 gram of silver chloride ( AgCl) and ended with 1.6381 grams of silver iodide ( AgI). This increase in mass is explained by the substitution of chlorine with the heavier iodine atom, as calculated using the mass ratios. When breaking down the reaction, the mass of silver, chlorine, and iodine needed careful calculation to comply with this law.

• Total start mass = Total final mass
• Initial AgCl mass: 1 g; final AgI mass: 1.6381 g

This principle ensures all calculations are balanced.

Stoichiometry

In chemistry, stoichiometry involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in a reaction. In the given exercise, it was crucial to set up equations that showed how masses of silver, chlorine, and iodine related to each other. By setting the equations for AgCl and AgI, we determined how much of each component was in the compounds.

• For AgCl: \( m_{Ag} + m_{Cl} = 1 ext{ g} \)
• For AgI: \( m_{Ag} + m_{I} = 1.6381 ext{ g} \)

Stoichiometry guides us in understanding relationships between reactants and products.

Silver Chloride

Silver chloride ( AgCl) is a compound of silver and chlorine. In our exercise, it began as the initial compound before being converted into silver iodide ( AgI). Knowing how much silver forms a part of silver chloride helps us understand its overall mass. In this exercise, it was calculated that the mass of silver in the AgCl sample was 0.7526 grams, and the mass of chlorine was 0.2474 grams. Knowing these breakdowns allows scientists to predict how it will react and what it will form when combined with other elements.

Silver Iodide

Like silver chloride, silver iodide ( AgI) is formed from silver, but in this case, it combines with iodine. In the exercise, AgI formation from AgCl demonstrated mass conservation while incorporating the heavier iodine atom. After reacting, the compound was slightly heavier mainly because iodine (0.8858 grams) replaced chlorine. The mass of silver remained at 0.7526 grams.

• AgI increased mass due to iodine
• Silver remained constant between AgCl and AgI

Understanding silver iodide is useful in predicting its behavior and use in chemical applications.