Ask Question
12 September, 16:15

A 0.4657 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 0.8878 g. What is the mass percentage of bromine in the original compound?

+4
Answers (1)
  1. 12 September, 16:30
    0
    The mass percentage of bromine in the original compound is 81,12%

    Explanation:

    Step 1: Calculate moles AgBr

    moles AgBr = mass AgBr / molar mass AgBr

    = 0.8878 g / 187.77 g/mol

    = 0.00472812 moles AgBr



    Since 1 mol AgBr contains 1 mol Br-

    Then the amount of moles Br - in the original sample must also have been 0.00472812 moles

    Step 2: Calculating mass Br-

    mass Br - = molar mass Br x moles Br-

    = 79.904 g/mol x 0.00472812 mol

    = 0.377796 g Br-



    There were 0.377796 g Br - in the original sample

    Step 3: Calculating mass percentage Br-

    ⇒mass percentage = actual mass Br - / total mass x 100%

    % mass Br = 0.377796 g / 0.4657 g x 100 %

    = 81.12%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers