A 0.5008 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8881 g. What is the mass percentage of bromine in the original compound?
A 0.5008 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8881 g. What is the mass percentage of bromine in the original compound?
Chemistry
10th Edition
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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![**Problem: Calculating Mass Percentage of Bromine**
In this exercise, a sample weighing **0.500 g** of a pure soluble bromide compound is dissolved in water. An excess of silver nitrate is added, allowing the bromide ion to precipitate as silver bromide (AgBr). The mass of the resulting AgBr is measured to be **0.8881 g**.
**Task:**
Determine the mass percentage of bromine in the original compound.
**Solution Steps:**
1. Calculate the moles of AgBr produced using its molar mass.
2. Determine the moles of bromine based on the stoichiometry of the reaction (1:1 ratio of Br⁻ to AgBr).
3. Find the mass of bromine using its atomic mass.
4. Calculate the mass percentage of bromine in the original compound using the formula:
\[
\text{Mass percentage of bromine} = \left(\frac{\text{Mass of bromine}}{\text{Total mass of the compound}}\right) \times 100
\]
**Instructions:**
Enter the calculated value in the provided input box and submit your answer.
*Button Options:*
- **Submit Answer**
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You have 9 more attempts remaining to solve this question.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F144ac4d4-58bf-4a39-b1bc-5bb4c77f8379%2Fa10888c5-7dc6-4a4e-952f-7ff279755e23%2Fe7gfaqv.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Calculating Mass Percentage of Bromine**
In this exercise, a sample weighing **0.500 g** of a pure soluble bromide compound is dissolved in water. An excess of silver nitrate is added, allowing the bromide ion to precipitate as silver bromide (AgBr). The mass of the resulting AgBr is measured to be **0.8881 g**.
**Task:**
Determine the mass percentage of bromine in the original compound.
**Solution Steps:**
1. Calculate the moles of AgBr produced using its molar mass.
2. Determine the moles of bromine based on the stoichiometry of the reaction (1:1 ratio of Br⁻ to AgBr).
3. Find the mass of bromine using its atomic mass.
4. Calculate the mass percentage of bromine in the original compound using the formula:
\[
\text{Mass percentage of bromine} = \left(\frac{\text{Mass of bromine}}{\text{Total mass of the compound}}\right) \times 100
\]
**Instructions:**
Enter the calculated value in the provided input box and submit your answer.
*Button Options:*
- **Submit Answer**
- **Retry Entire Group**
You have 9 more attempts remaining to solve this question.
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