How to determine the mass of ammonium nitrate needed and enthalpy of solution? I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.

I have the formula:

$$Q=ms\Delta T$$

I looked up the enthalpy of solution for ammonium nitrate and found it to be $\pu{25.41 kJ/mol}$.

I also have data that if I add $\pu{15 g}$ of $\ce{NH4NO3}$ the temperature will decrease by $\pu{25 ^\circ C}$, if I add $\pu{10 g}$ the temperature will decrease by $\pu{20 ^\circ C}$, and if I add $\pu{5 g}$ the temperature will decrease $\pu{13 ^\circ C}$.

Clearly then, to get my solution to be $\pu{3 ^\circ C}$, I would need to add a mass somewhere in between $10$ and $\pu{15 g}$.  In the lab I could just add random amounts between until the solution settled on $\pu{3 ^\circ C}$. Shouldn't there be a mathematical way to solve this rather than by random guessing?

I tried to use $Q=ms\Delta T$ by substituting the enthalpy of solution (in J) for $Q$, the specific heat of water for $s$, and $\pu{20 ^\circ C}$ for $\Delta T$ (since I am going from $23$ to $\pu{3 ^\circ C}$). However, solving $$\pu{25410 J} = m (\pu{4.186 J// ^\circ C})(\pu{20 ^\circ C})$$ gives me $\pu{303.66 g}$ which is clearly nowhere near the $\pu{10 - 1 g}$ that is sufficient to change the temperature.

How can I accurately determine the mass using a formula?
I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.


The result will depend on the amount of water where you're dissolving the salt. For instance, the data you present refers to different solvent mass. And by the way, for your temperature difference ($\pu{20^\circ C}$) the problem is already solved: a $\pu{20^\circ C}$ temperature decrease, will require $\pu{10 g}$ of the salt.

It is important to distinguish the mass of the salt from the mass of the solvent.
\begin{align}
Q &amp;= m_\mathrm{salt} \times \Delta_\mathrm{solution} H^\circ_\mathrm{salt}\\
Q &amp;= m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T\\
\therefore m_\mathrm{salt} &amp;= \frac{m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T}{\Delta_\mathrm{solution} H^\circ_\mathrm{salt}}
\end{align}

Hope this helps.