## Appendix A  
Sample calculations for each metric:  
Assuming: 6:1 ethanol to oil ratio, 1% NaOH, 80 mL of oil, 20 mL of glycerol (1.26 g/mL density), 71 mL of biodiesel (0.9 g/mL density)  
**Find: each green metric for the sample**  
Solution:  
Weight = volume \(\times\) density  
Oil weight = 80 mL \(\times\) 0.92 g/mL = 73.6 g of oil  
Ethanol weight = 33.39 mL \(\times\) 0.79 g/mL = 26.4 g of ethanol  
Catalyst weight = 73.6 g of oil \(\times\) 1% NaOH= 0.736 g of NaOH (\(wt/wt\))  
Glycerol weight = 20 mL \(\times\) 1.26 g/mL = 25.2 g of glycerol  
Biodiesel weight = 71 mL \(\times\) 0.9 g/mL = 63.9 g of biodiesel  
\(\underline{E\textit{-Factor}}\):  
Mass in the process = Oil + ethanol + catalyst = 73.6 g + 26.4 g + 0.736 g = 100.736 g  
Products = Glycerol + biodiesel = 25.2 g + 63.9 g = 89.1 g  
Waste= 100.736 g \(-\) 89.1 g = 11.636 g  
\(\underline{E\textit{-factor}}\) = \(\underline{T\textit{Total}\textit{neste}\textit{(kg)}}\)  
\(\underline{E\textit{-factor}}\) = 0.13  
\(\underline{Atom\textit{ Economy}}\):  
\[\begin{array}{l}AE=\frac{\textit{MW}\,\textit{of}\,\textit{product}\, \,C}{\textit{MW}\,\textit{of}\,\,A+\textit{MW}\,\textit{of}\,\,B}\times 100 \end{array}\]  
Molecular weight: biodiesel (856 g/mol), ethanol (46.07 g/mol), catalyst (39.997 g/mol), oil (922 g/mol)  
\[\begin{array}{l}AE=\frac{856\,\text{g/mol}}{(922\,\frac{\text{g}}{\text{mol }}+39.997\,\frac{\text{g}}{\text{mol}}+46.07\,\frac{\text{g}}{\text{mol}})}= 84.9\%\end{array}\]  
\(\underline{Atom\textit{ Economy}}\) = 85%  
\(\underline{Atom\textit{ Efficiency}}\):  
\(\underline{Atom\textit{ efficiency}}\) = \(\underline{AE\times\%}\) _yield_  
Assuming: _yield_ = 71 mL biodiesel/80 mL oil = 88.75 %  
\[\begin{array}{l}Atom\,\textit{efficiency}=84.9\%\,\times 88.75\%=75.3\% \end{array}\]  
\(\underline{Atom\textit{ efficiency}}\) = 75%  
**Mass Intensity (_MI_) and Mass Productivity (_MP_):**  
\[MI=\frac{\text{Total\ mass\ used\ in\ a\ process\ (kg)}}{\text{Mass\ of\ final\ product\ (kg)}}\]  
Mass used in the process (kg) = 0.100736 (from E-factor calculation)  
Mass of final product (kg) = 0.0891  
\[MI=\frac{0.100736\ \text{kg}}{0.0891\ \text{kg}}=1.13\ \text{kg}/\text{kg}\]  
\[MP=\frac{1}{MI}\times 100\]  
\[MP=\frac{1}{1.13}\times 100=88.5\%\]  
**MI = 1.13 kg/kg and _MP = 88.5 %_**  
**Reaction Mass Efficiency:**  
\[RME=\frac{\text{Mass\ of\ product\ (C+D)\ (kg)}}{\text{Mass\ of\ A\ (kg)+ mass\ of\ B\ (kg)}}\]  
or  
\[RME=\frac{1}{1+E_{f}}\]  
\[RME=\frac{0.0891\ \text{kg}}{0.100736\ \text{kg}}=88\%\]  
or  
\[RME=\frac{1}{1+E_{f}}=\frac{1}{1+0.13}=88\%\]  
**RME = 88%**  
**Atom Utilization:**  
\[AU\%=\frac{\text{Mass\ of\ the\ final\ product}}{\text{Total\ mass\ of\ all\ substances\ produced}}\]  
\[AU\%=\frac{63.9\ g}{89.1\ g}=72\%\]  
**\(AU\) = 72%**  
**Solvent and Catalyst Environmental Impact Parameter:**  
\[f=\frac{\sum\text{mass\ of\ reactants\ and\ postreaction\ solvents\ and\ materials}+\text{mass\ of\ catalyst\ used}}{\text{Mass\ of\ final\ product}}\]  
\[f=\frac{73.6+26.4+0.736+25.2}{63.9}=1.97\]  
**\(f\) = 1.97 kg/kg**  
**Stoichiometric Factor:**  
\[SF=1+\frac{(AE)\sum\text{Mass\ of\ excess\ \frac{\ reagents}{ chemicals}}}{\text{Expected\ product\ mass\ at\ 100\%\ yield}}\]  
The ideal stoichiometry is a 3:1 alcohol to oil molar ratio. In these sample calculations, the used molar ratio is 6:1. Therefore, the ideal ethanol amount should be 16.7 mL (13.2 g), but 33.39 mL (26.4 g) was used instead.  
Expected product mass at 100% yield= 13.2 + 73.6 + 0.736 = 87.536 g  
\[SF=1+\frac{(0.849)(13.2)}{87.536}=1.12\]  
**SF = 1.12 kg/kg**