Question

An aqueous ethylene glycol (HOCH2CH2OH, FW = 62.07 g/mol) solution with a mass of 279.1 mg is titrated with 42.9 mL of 0.0939 M Ce4 in 4 M HClO4. 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 11.89 mL of 0.0417 M Fe2 to a ferroin end point. What is the mass percent of ethylene glycol in the unknown solution?

Answer 1

mol of Ce4 used = MV = 42.9*0.0939 = 4.02831 mmol of Ce4

mol of Ce4 excess = use formula:

Fe2+ + Ce4+ Fe3+ + Ce3+

so ratio is 1:1

mmol of Fe2 = MV = 11.89*0.0417 = 0.495813 mmol of Fe+2

then, 0.495813 mmol of Ce+4 where initially left

so

actual Ce+4 used in reaction = initial - leftover = 4.02831 -0.495813 = 3.532497 mmol of Ce+4 used in oxidation

then, mmol of EG = 1:1 ratio with respec tto ce4+

3.532497 mmol of EG

mass of EG = mmol*MW = (3.532497)*62.07 = 219.26 grams of EG

%composition = mass of EG / tota mass of sample * 100%

%EG composition = 219.26 / 279.1 *100 = 78.559 %