Question

An aqueous ethylene glycol (HOCH2CH2OH, FW = 62.07 g/mol) solution with a mass of 253.1 mg is titrated with 49.5 mL of 0.0895 M Ce4+ in 4 M HCIO4. The solution is held at 60°C for 15 minutes to oxidize the ethylene glycol to formic acid (HCO2H) and carbon dioxide. The excess Ce4+ is titrated with 10.81 mL of 0.0439 M Fe2+ to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution? Number

Answer 1

HOCHCHGH + Ethylene Glycol ce 4+ - EXCESS H oH & calg). formic acid Cett + Fe 2+ ( Remaining) - Fest ferroin - - No of equivalents of Ethylene glycol = No of equivalent of cett - No of equivalent of Fet. No of equivalente = Hoo of moles x n-factor Given Concentration of cett = 0.0895M =0 Volume of cett = v = 49.5mL = 0.0495L No. of moles of cent = 0.0895M X 0.0495L = 0.00443 mola No of equivalents of cett = 0.00443x1 = 0.00 443 Volume of Fe²+ = = 10.81mL = 0.010812 Concentration of Feet = q = 0.0439 No of mole of Fe2+ = 8xV2 - 0.0439M X 0.01081L - 0.00047 45 mol y No of equivalent of Fe2+ = 0.0004745*1 = 0.0004745 - Noof equivalents of ethylene glycol = 0.00443 - 0.0004 745 = 0.003955 Noº of moles of ethylene glycol = 0.003955 mole - Molar mass of ethylene glycol = 62.07 g/mole Amount of ethylene glycol = 0.003955 mole x 62.078/mol = 0-3459 19 = looomg I Amount of ethylene glycol = 245ing Total mass of solution = 253.img

Mass% of ethylene - glycol Amount of ethylene glycol x 100 Tot af mass of solution 245 mg xloo 2531ng 96.79% Answer