Question

Chem 2582L Calculation Worksheet OH + 2 но он "H₂C a 0125 01331 0 - 6 C-5x, 7. I'llg/ml = m .7 m = 7.779 Suppose: 1. A chemist starts with 7 ml of ethylene glycol as the limiting reagent 2. 2 equivalents of acetyl is needed for every equivalent of cthyine glycol 3. The chemist actually synthesized 15.5 g of crude diacetyl ethylene glycol 4. After purification 14.4 g of diacetyl ethylene glycol were isolated Mali ODD Mmales Ethylene Acetyl Chloride Hydrochloric Diacetyl Glycol Acid ethylene glycol 5 mobs Mass 183 mole Volume 7 ml Imore Molecular ? ba 9/mol . As aime Weight Mmol ? Das ? Density ? . 19 ml? Boiling Point ? ? Melting Point ? ? 0.125 mol 0.331 mol ? 7.279 26.23 g Imoles What is the theoretical yield of diacetyl ethylene glycol? What is the percent yield of diacetyl ethylene glycol? What is the percent recovery of diacetyl ethylene glycol? What is the theoretical yield of Hydrochloric acid? How many mmoles of diacetyl ethylene glycol are produced in the given reaction? How many reagents are involved in the given reaction?

Answer 1

The reaction is, woro" z --> 2001 toplommoyen Molecular weight of ethylene glycol = 62.07 g lemot Densitys of ethylene glycol = long lem? Volume of ethylene glycol reacted 7 ml So, mass of ethylene glycol reacted = 1.11 x 7 g 27.77g 3o, moles of ethylene glycol reacted = molecular weight mans 7.77 mol 62.07 - 0.1252 mol From the given date, mass of acetyl chloride reacted -20 Molecular weight of acetyl chloride = 78.49 g/mol So, moles of acetyl chloride reacted - 2612 mot - 0:333 mol since ethylene glycol is the limitings reagent, 0.1252 mot ethylene glycol should produce 0.1252 mot of diacetyl ethylene glycol.

Now molecular weight of diacetyl ethylene glysol = 146:14 og mot . So, theoretical gield of ethylene glycol, Theonelical yield = 0. 1252 x 146 14 g - 18.296 g F 18. 3g X100 And percentage yield of diacetylene glycol is given by Reported gield x 100 Percentage yield - Calentated yield Guide geld of diacetyle illeglene glycol = 15.5g So, percentage yield a 15 x 100 - 84.7%. Prisfied yield of diacetyl ethylene glycol =14:49 So, percent recovery of deacetyl ethylene glycol - 14.1 100 = 78.7 % Now, .333 notes of acetyl chloride should give 0.333 moles of hydrochloric acid.

Molecular weight of hydrochloric acid = 36.5 g/mol .30, theoretical yield of hydrochloric acid = 0.333 * 36.57 = 12.15 g - 12 Now amount of diacetyl ethylene glycol, peroduced is 15:59 And molecular weight of diecityl ethylene glysed = 146 14 g/mol. So, no. of moles of diacetyl ethylene glycol produced an 15.5 mol 14 6.14 - 0.106 mol = 106 mmol involved in the given reaction No. of magents are two.