# Girolami method

The Girolami method,[1] named after Gregory Girolami, is a predictive method for estimating densities of pure liquid components at room temperature. The objective of this method is the simple prediction of the density and not high precision.

## Procedure

The method uses purely additive volume contributions for single atoms and additional correction factors for components with special functional groups which cause a volume contraction and therefore a higher density. The Girolami method can be described as a mixture of an atom and group contribution method.

### Atom contributions

The method uses the following contributions for the different atoms:

Element Relative volume
Vi
Hydrogen 1
Lithium to Fluorine 2
Sodium to Chlorine 4
Potassium to Bromine 5
Rubidium to Iodine 7.5
Cesium to Bismuth 9

A scaled molecular volume is calculated by

${\displaystyle V_{S}\,=\,\sum _{i}V_{i}}$

and the density is derived by

${\displaystyle d\,=\,{\frac {M}{5\cdot V_{S}}}}$

with the molecular weight M. The scaling factor 5 is used to obtain the density in g·cm−3.

### Group contribution

For some components Girolami found smaller volumes and higher densities than calculated solely by the atom contributions. For components with

it is sufficient to add 10% to the density obtained by the main equation. For sulfone groups it is necessary to use this factor twice (20%).

Another specific case are condensed ring systems like Naphthalene. The density has to increased by 7.5% for every ring; for Naphthalene the resulting factor would be 15%.

If multiple corrections are needed their factors have to be added but not over 130% in total.

## Example calculation

Component M
[g/mol]
Volume VS Corrections Calculated density
[g·cm−3]
Exp. density
[g·cm−3]
Cyclohexanol 100 (6×2)+(13×1)+(1×2)=26 One ring and a hydroxylic group = 120% d=1.2*100/5×26=0.92 0.962
Dimethylethylphosphine 90 (4×2)+(11×1)+(1×4)=23 No corrections d=90/5×23=0.78 0.76
Ethylenediamine 60 (2×2)+(8×1)+(2×2)=16 Two primary amine groups = 120% d=1.2×60/5×16=0.90 0.899
Sulfolane 120 (4×2)+(8×1)+(2×2)+(1×4)=24 One ring and two S=O bonds = 130% d=1.3×120/5×24=1.30 1.262
1-Bromonaphthalene 207 (10×2)+(7×1)+(1×5)=32 Two condensed rings = 115% d=1,15×207/5×32=1.49 1.483

## Quality

The author has given a mean quadratic error (RMS) of 0.049 g·cm−3 for 166 checked components. Only for two components (acetonitrile and dibromochloromethane) has an error greater than 0.1 g·cm −3 been found.

## References

1. ^ Gregory S. Girolami, A Simple "Back of the Envelope" Method for Estimating the Densities and Molecular Volumes of Liquids and Solids, Journal of Chemical Education, 71(11), 962-964 (1994)