Question

An aqueous ethylene glycol (HOCH2CH2OH, FW=62.07 g/mol) solution with a mass of 220.7 mg is titrated with 45.3 mL of 0.0671 M Ce4 in 4 M HClO4. The solution is held at 60 °C for 15 min to oxidize the ethylene glycol to formic acid (HCO2H) and carbon dioxide. The excess Ce4+ is titrated with 11.73mL of 0.0449 M Fe2+ to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution?

mass percent =__________%

Answer 1

Given FW= 62-04 glmal Mass = 220.7 mg is Hitrated with 4513 m/ of 0.0671m cen in um Helona - The solution is held at 60% for min to oxidize the Ethylene glycol to formie acid (Hogw) and Carbon dioxide. The Excess Com is fitrated with 1:43 ml of 0.0449 MFe 2+ calculator of the no. of modes of excess cent 1000 mL of anooma mpenta oon99 mol sezt 11.73 m2 of 0.0uua impent = 0.0449 mol 1000 ot x 11.73ml fet i. = 5.26x105 mol of felt treat > cet se 34 Reaction between cettaud Fe 2+ ceht = fe2f imol = imod 5.26x10kmed. = 5.26x10+4 met

- The number of males of Excss 1041 6.26410kmal calculation of no. of males of rett Haid ovidises Ethylene glayal 1000 ml of 0.0611M:2044 = 0.0671 mol ce 44 : 45.3 m2 of 0.0671M ceht = 0.0671 mol x 451392cett 1000 m - 3.039 x 10 molceut No.of moles of creat used for cridation of Ethylene glycol. . . no. of moles & Faceps (e4+ 3.034x 10 mol 2.of males of cent used foo oridation of Ethylene glycol = 3-039 4107279/ -pood moles of everest |= 3.039 x 10**med - 5.26 xrolmad = 2.51x1013 Ethylene glycol = 2.51x103

calculation of noiol moles of_thy lene glycol; -the products of oxidation of glycol in acid . medium (4M H dow) are HCOOH 4 (02 CH₂OH 1 CH2OH + 2H20 - > HCOOH + CO2+ 8H*78e ( 264 104 45 040 changes (10ct 10 10 497 [084 185) CH2OH +2120 -> HCOOH + CO2+8H78e loxidation CH₂OH of celle-cest] CH₂OH ton + 2H₂O + 8ce4ty HOO H+ (O+ 8H7+ 8ce3+ CH2OH - CH₂-OH Emil = imod 2.51*10*3 mol = 12.51*10*3mol) = 3.13 x 10-4 md. . CH2-OH CH₂-OH

Mass of CH₂-OH į mole x molar mass CH₂OH = 3.13810-4 mol x 62507 gimol -0.01947 =19.47 mg Mass of solution = 22014 mg mass % 9140 19:47 mg 220.7 mg X 100% - 008821 X100% =8.82% :fross von d ethylene_alget =8-82 %?