Percent of silver by mass in an unknown sample Question
  If all of the silver in a $7.563~\mathrm{g}$ sample of a mineral, which was dissolved by using nitric acid, is precipitated as silver thiocyanate ($\ce{AgSCN}$) by using $82.46~\mathrm{mL}$ of a $0.2500~\mathrm{M}$ potassium thiocyanate solution, what is the percent silver (by mass) in the sample?


My Attempt
I'm pretty sure the relevant chemical equation here is $\ce{Ag^+ + SCN- -&gt; AgSCN}$.

Molarity is mols solute over L solution, so the number of mols of $\ce{KSCN}$ used is $0.25~\mathrm{M} \times 0.08246~\mathrm{L} = 0.020615~\mathrm{mol\ KSCN}$. According to the chemical equation above, the molar ratio is 1 to 1 so there is $0.020615~\mathrm{mol\ Ag}$  involved. That accounts for all of the silver in the original mineral sample, I believe. Therefore, there are $2.22370~\mathrm{g\ Ag}$, so the percent silver by mass is $2.22370~\mathrm{g\ Ag} \div 7.563 ~\mathrm{g} = 0.294$. Put into percent, that's $29.4\%~\mathrm{Ag}$ by mass.

I'm unsure about this because  I totally ignored the nitric acid.
Noah--

You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:

82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)

0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)

2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass

I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.

And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.