# Finding the mass of NaNO3 in original mixture from preciptate

A 1.61-g sample of a mixture of $\ce{AgNO3}$ and $\ce{NaNO3}$ is treated with excess $\ce{Na2S}\:(\mathrm{aq})$. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.44 g. What is the percentage by mass of $\ce{NaNO3}$ in the original mixture?

I attempted this problem by using two equations: one with $\ce{AgNO3}$ and $\ce{Na2S}$, and the other with the other reactant and $\ce{Na2S}$. Then I wrote an equation in moles, using m and 1.61-m as the mass of the reactants. My answer was 2.34 g, which is greater than the original so I know I did something wrong.

The dried precipitate is $\ce{Ag2S}$. The number of moles of $\ce{Ag2S}$ is: $$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$ The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$
On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$
So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$