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 $\ce{Ag^+ + SCN- -> AgSCN}$.
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.