Question

Complete the stoichiometry table using the starting values provided. reagent molecular weight mass (g) mmoles equivalents density (g/mL) - vol. (ML) glycerol tripleate 0.589 . sodium hydroxide 0.533 water 2.185 ethanol (95%) 0.816 2.117 theoretical product sodium oleate glycerol AM In your lab notebook, you should complete the written portion of the prelab which includes a title, date, reaction equation, reagent table and plan of procdure. I confirm that I have read the terms of the Ohio State University Department of Chemistry and Biochemistry Undergraduate Organic Chemistry Laboratory prelab/postlab submission system and agree to the policies and procedures therein. Submit

Answer 1

Glycerol trioleate: Mass = 0.589 g Molar mass = 885.43g/mol moles = mass/molar mass = 0.589/885.43 = 0.665*10^-3 moles mmoles = 0.665 mmol Density = 0.908 g/mL Volume = 0.589 / 0.908 = 0.649 mL Equivalents = 1 Sodium hydroxide: Mass = 0.533 g Molar mass = 40.0g/mol moles = 0.533/40.0 = 13.3*10^-3 moles mmoles = 13.3 mmol Equivalents = 13.3/0.665 = 20

Water: Mass Molar mass moles = density* volume = 1*2.185 = 2.19 g = 18.02 g/mol = 2.19/18.02 = 122*10^-3 moles = 122 mmol = 1.00 g/ml = 2.185 mL = 122/0.665 = 184 mmoles Density Volume Equivalents Ethanol: Mass Molar mass moles = density*volume = 0.816*2.117 = 1.73 g = 46.07 g/mol = 1.73/46.07 = 37.6*10^-3 moles = 37.6 mmol = 0.816 g/mL = 2.117 mL = 37.6/0.665 = 56.5 mmoles Density Volume Equivalents

Sodium oleate: Molar mass = 304.4 g/mol For 1 mole of Glycerol trioleate gives 3 moles of sodium oleate, Then for 0.665 mmol of Glycerol trioleate should give, mmoles = 0.665*3 mmol = 1.995 mmol moles = 1.995*10^-3 moles Mass = molar mass*moles = 304.4*1.995*10^-3 = 0.607 g Glycerol: Molar mass = 92.09 g/mol For 1 mole of Glycerol trioleate gives 1 moles of glycerol, Then for 0.665 mmol of Glycerol trioleate should give, mmoles = 0.665*1 mmol = 0.665 mmol moles = 0.665*10^-3 moles Mass = 92.09*0.665*10^-3 = 0.061 g

Please note that it is not indicated to use significant figure calculation rules hence I am not using them while above calculations.