A 0.4455 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 0.9496 g. What is the mass percentage of chlorine in the original compound? %
A 0.4455 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 0.9496 g. What is the mass percentage of chlorine in the original compound? %
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Sample Problem: Determining Mass Percentage of Chlorine in a Compound
#### Problem Statement:
A **0.4455 g** sample of a pure soluble **chloride** compound is dissolved in water, and all of the **chloride ion** is precipitated as **AgCl** by the addition of an excess of silver nitrate. The mass of the resulting **AgCl** is found to be **0.9496 g**.
What is the mass percentage of **chlorine** in the original compound? \[\ \%\]
#### Solution:
To find the mass percentage of chlorine in the original compound, follow these steps:
1. **Calculate moles of AgCl**:
\[ \text{Molecular weight of AgCl} = \text{atomic weight of Ag} + \text{atomic weight of Cl} \]
\[ = 107.87 \text{ g/mol} + 35.45 \text{ g/mol} = 143.32 \text{ g/mol} \]
2. **Determine moles of AgCl from mass given**:
\[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molecular weight of AgCl}} \]
\[ = \frac{0.9496 \text{ g}}{143.32 \text{ g/mol}} \]
\[ = 0.006625 \text{ mol} \]
3. **Determine moles of Cl**:
Since AgCl is composed of a 1:1 ratio of Ag to Cl, moles of Cl = moles of AgCl.
\[ \text{Moles of Cl} = 0.006625 \text{ mol} \]
4. **Calculate mass of Cl**:
\[ \text{Mass of Cl} = \text{moles of Cl} \times \text{atomic weight of Cl} \]
\[ = 0.006625 \text{ mol} \times 35.45 \text{ g/mol} \]
\[ = 0.2347 \text{ g} \]
5. **Calculate mass percentage of Cl** in the original sample:
\[ \text{Mass percentage of Cl} = \left( \frac{\text{mass of Cl}}{\text{mass of original compound}} \right)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2Ffdd72e07-f8e8-4594-b521-808c9af2ac83%2Fyk5cktf_processed.png&w=3840&q=75)
Transcribed Image Text:### Sample Problem: Determining Mass Percentage of Chlorine in a Compound
#### Problem Statement:
A **0.4455 g** sample of a pure soluble **chloride** compound is dissolved in water, and all of the **chloride ion** is precipitated as **AgCl** by the addition of an excess of silver nitrate. The mass of the resulting **AgCl** is found to be **0.9496 g**.
What is the mass percentage of **chlorine** in the original compound? \[\ \%\]
#### Solution:
To find the mass percentage of chlorine in the original compound, follow these steps:
1. **Calculate moles of AgCl**:
\[ \text{Molecular weight of AgCl} = \text{atomic weight of Ag} + \text{atomic weight of Cl} \]
\[ = 107.87 \text{ g/mol} + 35.45 \text{ g/mol} = 143.32 \text{ g/mol} \]
2. **Determine moles of AgCl from mass given**:
\[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molecular weight of AgCl}} \]
\[ = \frac{0.9496 \text{ g}}{143.32 \text{ g/mol}} \]
\[ = 0.006625 \text{ mol} \]
3. **Determine moles of Cl**:
Since AgCl is composed of a 1:1 ratio of Ag to Cl, moles of Cl = moles of AgCl.
\[ \text{Moles of Cl} = 0.006625 \text{ mol} \]
4. **Calculate mass of Cl**:
\[ \text{Mass of Cl} = \text{moles of Cl} \times \text{atomic weight of Cl} \]
\[ = 0.006625 \text{ mol} \times 35.45 \text{ g/mol} \]
\[ = 0.2347 \text{ g} \]
5. **Calculate mass percentage of Cl** in the original sample:
\[ \text{Mass percentage of Cl} = \left( \frac{\text{mass of Cl}}{\text{mass of original compound}} \right)
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