VII. 2.00 g of an unknown mono-ester is hydrolyzed by refluxing with 50.0 mL of

Question

VII. 2.00 g of an unknown mono-ester is hydrolyzed by refluxing with 50.0 mL of 0.500 M NaOH ntil the reaction is complete. The reaction mixture is cooled and an indicator is added to the still basic olution. Titration to end point with 0.200 M HCI required 48.1 mL.. a. How many moles of hydroxide remained after hydrolysis of the ester? Show all work including omplete balanced equation for the reaction. b. How many moles of hydroxide were consumed in the hydrolysis? Show all work. c. Calculate the saponification equivalent of the unknown ester. Show all work. d. Which ester below is the most likely candidate for the unknown? Why? i. ethyl ethanoate i. Phenyl benzoate ii. Isopentyl acetate

Explanation / Answer

a.

mol of HCl = MV = 0.2*48.1*10^-3 = 0.00962

mol of NaOH left = mol of HCl = 0.00962 mol left

b)

mol of NAOH total MV = 0.5*50*10^-3 = 0.025

mol of NaOH used in raction = 0.025 - 0.00962 = 0.01538 mol of NaOH

c)

saponification equiv. of unkown ester --> mass of NaOH required to saponify 1 g of fat

mass

2 g of fat --> 0.01538 mol

equiv -> 2/0.01538 = 130.039 g of fat / 1 mol of NaOH  

130.039 *40 = 5201.56 g of NaOH per g of fat

1 mol of NaOH = 130.039

d.

the most likely:

130.039 g/mol approx:

MW of isopentyl acetate = 130.1849

nearest value is this one

choose isopentyl acetate

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