3
$\begingroup$

I want to do a simulation of a mixture which consists a total of $512$ molecules of glycine, ammonia, and water molecules mixture in a cubic box of, say, length $\pu{2.6 nm}$; $\pu{2.6 nm}^3 = \pu{17.576 nm3}$ in terms of volume.

I want [1 glycine molecule + $x$ ammonia molecules + $y$ water molecules] which together comes to a total of $512$ molecules and in a box volume of $\pu{17.576 nm3}$ and molarity should be $\pu{15 M}$.

Actually here I am fixing the glycine molecule which is only one and varying only the number of ammonia and water molecules to get different molar concentrations, say, $\pu{5 M / 10 M / 15 M}$ and $\pu{18 M}$, where the total number of molecules is constant, i.e. $512$ molecules only (see the example below). Since a maximum of $\pu{18 M}$ solution can be prepared (since ammonia solubility is $\pu{31 g}$ in $\pu{100 mL}$), so, I am trying to study with varying the $\%$ composition of ammonia and water molecules.

In this case I want to get a concentration of $\pu{15 M}$. How to calculate the number of $x$ ammonia and $y$ water molecules that need to be taken in order to get $\pu{15 M}$, considering the conversion of number of molecules to mass, amount of substance, etc..

For example:

  • $1$ molecule of glycine + $52$ molecules of ammonia + $459$ molecules of water is $10 : 90$ mixture (total $512$ molecules);
  • $1$ molecule of glycine + $104$ molecules of ammonia + $408$ molecules of water is $20 : 80$ mixture (total $512$ molecules).

Then what would be the molarity of the above two mixtures ($1 : 90$, $20 : 80$) considering the same volume ($\pu{17.576 nm}$)?

$\endgroup$
  • 1
    $\begingroup$ Do you need molarity specifically? Often with molecular simulations, you will see molality used instead because it is easier to calculate when you dealing with a reasonably countable number of solvent molecules. $\endgroup$ – Tyberius Dec 16 '17 at 1:25
2
$\begingroup$

I would use a slightly different approach than andselisk, based on the volume on your box, and that I don't think you can neglect the volume of your ammonia molecules. Given $V(\text{sol}) = \pu{17.576E-27 m3}$ and $c(\ce{NH3}) = \pu{15.0E3 mol m-3}$ we can calculate the number of ammonia molecules we need: \begin{align} && c(\ce{NH3}) &= \frac{n(\ce{NH3})}{V(\text{sol})};\\ && n(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}};\\ &\to& c(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}\cdot V(\text{sol})}\\ &\Longleftrightarrow& N(\ce{NH3}) &= c(\ce{NH3})\cdot N_\mathrm{A}\cdot V(\text{sol}) \approx 159 \end{align}

Since you want to limit the number of molecules to $512$, we know that all other molecules have to be water: $$N(\ce{H2O}) = 512 - N(\ce{Gly}) - N(\ce{NH3}) = 352$$

Lastly, let's check whether the density is anywhere close to what we would expect, we take $M_\mathrm{m}(\ce{Gly}) = \pu{75 u}$, $M_\mathrm{m}(\ce{H2O}) = \pu{18 u}$, and $M_\mathrm{m}(\ce{NH3})= \pu{17 u}$. \begin{align} \rho(\text{sol}) &= \frac{m(\text{sol})}{V(\text{sol})}\\ &= \frac{\left( N(\ce{Gly})M_\mathrm{m}(\ce{Gly}) + N(\ce{H2O})M_\mathrm{m}(\ce{H2O}) + N(\ce{NH3})M_\mathrm{m}(\ce{NH3})\right)\frac{m_\mathrm{u}}{\pu{u}} }{V(\text{sol})}\\ \rho(\text{sol}) &\approx \pu{861 kg m-3} = \pu{0.861 g cm-3} \end{align}

Citing Wikipedia's entry of ammonia I would say you are quite close:

Solvent properties
Ammonia is miscible with water. [...] The maximum concentration of ammonia in water (a saturated solution) has a density of $\pu{0.880 g cm-3}$ and is often known as '.880 ammonia'. [...]

The Handbook of Chemistry and Physics has a concentration table (91st Edition p. 8-53; currently 98th Edition, also online), and I'll reproduce a couple of values for comparison, at $\pu{20 ^\circ C}$.

\begin{array}{rrr} \text{Mass\%} & c / \pu{mol L-1} & \rho / \pu{g cm-3}\\\hline 5.0 & 2.868 & 0.9770\\ 9.0 & 5.080 & 0.9613\\ 12.0 & 6.695 & 0.9502\\ 18.0 & 9.823 & 0.9294\\ 24.0 & 12.826 & 0.9102\\ 28.0 & 14.764 & 0.8920\\\hline \end{array}

In conclusion, you are a few water molecules short, or your box is slightly too large, or your solution is not dense enough. I would expect some trade-offs when doing such a study and keeping the named constraints, so I guess you should have a close look at everything and settle for the optimum you can afford.

$\endgroup$
  • $\begingroup$ The last row (28 Mass%) would be well reproduced by 139 ammonia, 471 water in a cubic box of 2.5 nm. (18 Mass%) 93 NH3, 394 H2O, 2.5 pm. $\endgroup$ – Martin - マーチン Dec 18 '17 at 13:02
  • $\begingroup$ Can u kindly tell what is mu/u in the density equation..?? and one more thing... how to type the rho symbols, and subscripts/superscripts, equations here..?? $\endgroup$ – D.H.N Dec 18 '17 at 16:44
  • $\begingroup$ @D.H.N Have a look here and here for MathJax tutorials. The $m_\mathrm{u}/\pu{u}$ is only the conversion factor from the unified atomic mass unit, to the SI unit (kg). (And when you click the edit link, you can have a look at the source code, too.) $\endgroup$ – Martin - マーチン Dec 18 '17 at 16:52
  • $\begingroup$ @Tyberius I actually really dislike using \to instead of \implies, for me it simply has a completely different meaning. Why are you doing that anyway? Thanks for catching the spelling errors though. $\endgroup$ – Martin - マーチン Dec 19 '17 at 7:22
  • $\begingroup$ @Martin I tend to change \implies to \to because \implies doesn't show up in the app and I personally consider them as conveying the same idea. I can understand wanting to use them differently and that it might not help much in this case since \pu blocks a lot of the post anyway. I've just figured in cases where it was the one obstruction to reading the post on the app, I would change it and if the poster didn't think it was right it could be quickly reversed. $\endgroup$ – Tyberius Dec 19 '17 at 15:03
1
$\begingroup$

Using conventional macroscale quantities and SATP, one can express the desired concentration of aqueous ammonia using the amount $n$ and the volume $V$ of the solution:

$$C(\ce{NH3}) = \frac{n(\ce{NH3})}{V(\ce{H2O})} \tag{1}$$

Let's find the ratio between the numbers of molecules of ammonia ($N(\ce{NH3})$) and water ($N(\ce{H2O})$). By definition of a mole ($N_\mathrm{A}$ – Avogadro's number):

$$n(\ce{NH3}) = \frac{N(\ce{NH3})}{N_\mathrm{A}} \tag{2}$$

Volume of water required to achieve given concentration can be found using density $\rho$ and molecular weight $M$:

$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{n(\ce{H2O}) \cdot M(\ce{H2O})}{\rho (\ce{H2O})} = \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{\rho (\ce{H2O}) \cdot N_\mathrm{A}} \tag{3}$$

Assembling (2) and (3) in (1):

$$\frac{N(\ce{H2O})}{N(\ce{NH3})} = \frac{\rho (\ce{H2O})}{C(\ce{NH3}) \cdot M(\ce{H2O})} = \frac{\pu{1.0e3 g L-1}}{\pu{15 mol L-1} \cdot \pu{18.015 g mol-1}} = 3.70 \tag{4}$$

Now, knowing the ratio between ammonia and water, total number of molecules in system and assuming that single glycine molecule remains intact, one can count the exact numbers:

$$1 + N(\ce{NH3}) + 3.70 \cdot N(\ce{NH3}) = 512$$ $$\to N(\ce{NH3}) \approx 109 $$ $$\to N(\ce{H2O}) \approx 402 $$

By the way, 402 water molecules would only occupy $\pu{1.20e-26 m^3}$ (formula (3)), assuming this molecular assembly (cluster) behaves similarly to its macro counterpart (which is probably not exactly true). This means that you need a slightly smaller cube with a side of about $\sqrt[3]{\pu{1.20e-26 m^3}} = \pu{2.3 nm}$.

The same procedure can be repeated for other concentrations.

$\endgroup$
  • 1
    $\begingroup$ I've looked up the densities of ammonia solutions, and what I calculated is surprisingly close to what you would find in the theoretical setup. $\endgroup$ – Martin - マーチン Dec 18 '17 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.