You have a sample of silver alloy and would like to know how much silver is in it. Dissolve the silver sample in nitric acid. Filter to remove any insoluble material. Add an excess of $\ce{KCl}$ solution, which causes the formation of silver chloride precipitate. Filter, dry, and weigh the precipitate. What do you estimate the percentage of silver in the original sample is if its mass was $5.2~\mathrm{g}$, and the mass of silver chloride was $6.4~\mathrm{g}$?

The atomic mass of silver is $M(\ce{Ag})= 107.87~\mathrm{\frac{g}{mol}}$. That of chlorine is $M(\ce{Cl}) = 35.45~\mathrm{\frac{g}{mol}}$.
My answer is $81\%$, because $5.2/6.4 \cdot 100 = 81$. However, I am quite unsure.


1 Answer 1


\begin{align}\ce{ 3Ag + 4HNO3 &-> 3AgNO3 + NO + 2H2O\\ AgNO3 + KCl &-> AgCl + KNO3\\ }\end{align} Moles of $\ce{AgCl} = 6.4/143.5 = 0.0446~\mathrm{mol}$

Mass of $\ce{Ag} = 0.0446 \cdot 108 = 4.816~\mathrm{g}$

Percentage of $\ce{Ag} = 4.816/5.2 \cdot 100=92.62\%$

  • $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$ Sep 19, 2014 at 4:44
  • $\begingroup$ @Martin that wa typo, division was right $\endgroup$
    – RE60K
    Sep 19, 2014 at 4:53
  • $\begingroup$ I noticed that :) $\endgroup$ Sep 19, 2014 at 4:54

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