Question

a 0.4130 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 0.5221 g. what is the mass percentage of bromine in the original compound? %

Explanation:

Step1: Calculate the molar mass of AgBr

The molar mass of Ag (silver) is \(M_{Ag}= 107.87\space g/mol\), and the molar mass of Br (bromine) is \(M_{Br}=79.90\space g/mol\). The molar mass of \(AgBr\) is \(M = M_{Ag}+M_{Br}=107.87 + 79.90=187.77\space g/mol\)

Step2: Find the mass of Br in AgBr

In \(AgBr\), the mass fraction of \(Br\) is \(\omega=\frac{M_{Br}}{M_{AgBr}}\). Given \(m_{AgBr} = 0.5221\space g\), the mass of \(Br\) in \(AgBr\) (which is equal to the mass of \(Br\) in the original compound) is \(m_{Br}=m_{AgBr}\times\frac{M_{Br}}{M_{AgBr}}\)

Substitute the values: \(m_{Br}=0.5221\times\frac{79.90}{187.77}\)

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Step3: Calculate the mass percentage of Br

The mass percentage of \(Br\) in the original compound is \(\text{Mass}\%=\frac{m_{Br}}{m_{sample}}\times 100\%\)

Given \(m_{sample} = 0.4130\space g\) and \(m_{Br}=0.2221\space g\)

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Answer:

\(53.8\%\)