# Given the mass of precipitate, how can I calculate the relative amount of the original dissolved compounds?

A simple mixture of $\ce{NaCl}$ and $\ce{NaBr}$ weighing $0.180~ \mathrm{g}$ is treated with $\ce{AgNO3}$ solution to give $0.3715~ \mathrm{g}$ of precipitate. Calculate the content of $\ce{NaCl}$ and $\ce{NaBr}$ in the mixture.

The answer given in book are $\ce{NaCl}$ = $0.0682~ \mathrm{g}$ and $\ce{NaBr}$ = $0.1118~ \mathrm{g}$.

The answers I got are $0.1519~ \mathrm{g}$ $\ce{NaCl}$ and $0.0281~ \mathrm{g}$ $\ce{NaBr}$. Please tell me where I made mistake.

Working:

$$\ce{NaCl + AgNO3 -> NaNO3 + AgCl}$$ $58.5~ \mathrm{g}$ $\ce{NaCl}$ give $143.5~ \mathrm{g}$ $\ce{AgCl}$

$0.3715~ \mathrm{g}$ $\ce{AgCl}$ give $58.5/143.5 \cdot 0.3715~ \mathrm{g}$ $\ce{NaCl}$ = $0.1519~ \mathrm{g}$ $\ce{NaCl}$

$\ce{NaBr}$ = $0.180 ~ \mathrm{g}$ - $0.1519~ \mathrm{g}$ = $0.0281~ \mathrm{g}$ $\ce{NaBr}$

• Silver bromide is insoluble as well. – Abel Friedman Oct 23 '14 at 13:29
• I think that NaCl only react with AgNO3 and NaBr does not react. If you think I am wrong please correct me – pcforgeek Oct 23 '14 at 13:32
• No, that's not correct. Ag+ and Br- give an insoluble precipitate as well; all silver halides (with the exception of AgF) are poorly soluble: NaCl + AgNO3 -> NaNO3 + AgCl and NaBr + AgNO3 -> NaNO3 + AgBr. – Abel Friedman Oct 23 '14 at 13:36
• Then please tell me how to solve this problem. – pcforgeek Oct 23 '14 at 13:39
• @pcforgeek you to construct a system of equations and solve it. – permeakra Oct 23 '14 at 14:23

Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:
$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$
Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.