# Calculating the apparent molecule length from density of the bulk [closed]

I'm interested to estimate the apparent length of the water molecule assuming it could be approximated as a cube. This is an intentional simplification.

## Calculating molecule length from volume in density equation

I used this equation to find the volume for a mole of water molecules with $$\varrho$$ (water density, assumed $$\pu{0.997 g cm^{-3}}$$), $$m$$ (mass), and $$V$$ (volume).

For the determination of the molar volume, I substituted mass $$m$$ by molar mass $$M$$ and rearranged

$$\varrho = m / V$$

into the equivalent form

$$V_{\mbox{molar}} = \frac{M} {\varrho} = \frac {\pu{18.01 g mol^{-1}}} {\pu{0.997 g cm^{-3}}} = \pu{18.06 cm^3 mol^{-1}}$$

Because $$\pu{1 m^3} = \pu{10^6 cm^3}$$, the molar volume equates to $$\pu{18.06 \times 10^{-6} m^3 mol^{-1}}$$.

Given Avogadro's number, I computed the volume of the individual volume with

$$V_{\mbox{molecule}} = \frac{ \pu{18.06 \times 10^{-6} m^3 mol^{-1}}} { \pu{6.022 \times 10^{23} mol^{-1}} } = \pu{2.999 \times 10^{-29} m^3}$$

Then I took the cube root to find the length, as I want $$m^3 \rightarrow m$$:

$$\ell(\text{molecule}) = \sqrt[3]{\pu{2.999 \times 10^{-29} m^3}} = \pu{3.10 \times 10^{-10} m} = \pu{0.31 nm}.$$

#### What I need help with

This isn't taking intermolecular forces into account, but I want to know if it's an ok method for getting a rough idea of the molecule length, and if it's done correctly.

• Density often is denoted by $\rho$ ($\rho$) or $\varrho$ ($\varrho$ in LaTeX / mhchem syntax available to you here on chemistry.se), which is different to $p$ then typically used about pressure. But this is smallish compared to the initial equation, because density is not mass times volume, but mass divided by volume (e.g., water, 1 g per cubic centimetre, or about a [metric] ton per cubic metre. The length you obtain is about twice as wide as the (intramolecular) H-H distance (about 0.15 nm). – Buttonwood Jun 20 at 20:29
• Assuming molecules as small cubes is a simplification and it depends a lot on the circumstances if this is acceptable, or not. It may be a start of a model later to be refined (e.g. limiting the number of interactions modelling crystal growth) for water's bent, or benzene's flat shape. For future reference: if you use an equality sign, then both numbers and dimensions must equate, too. This shows e.g., when they cancel out (dimension analysis) even if this demands a bit more to type / write. – Buttonwood Jun 20 at 21:58
• Over time (and repeated exposure) you will pick up skills new to you. Once you have enough reputation, for example, you may e.g. access to the edits made by other users. This equally is a source of inspiration (and eventually, training) by reference for how to format questions and answers in the community of chemistry.se (maybe you know the expression of «when in Rome, do as the Romans do»). – Buttonwood Jun 21 at 19:47
• Strange this question was closed for the reason given but from things you have said @buttonwood my question is answered (about if it can be used roughly to estimate length), would gladly have given you the answer if your stuff wasn’t in a comment. – Nickotine Jun 22 at 11:40
• It actually is fascinating how close you get with this absolutely simplistic approach. I#m not sure why this was closed, I wouldn't have. But I'm also not confident enough to overrule this decision. Also (but not entirely sure how helpful this is): researchgate.net/post/… – Martin - マーチン Jun 24 at 21:51