An aqueous ethylene glycol (HOCH_2CH_2OH, FW = 62.07 g/mol) solution with a mass

Question

An aqueous ethylene glycol (HOCH_2CH_2OH, FW = 62.07 g/mol) solution with a mass of 267.7 mg is titrated with 36.7 mL of 0.0831 M Ce^4+ in 4 M HCIO_4. The solution is held at 60 degree C for 15 minutes to oxidize the ethylene glycol to form acid (HCO_2H) and carbon dioxide. The excess Ce^4+ is titrated with 10.19 mL of 0.0459 M Fe^2+ to a ferrying end point. What is the mass percent of ethylene glycol in the unknown solution? Start by writing the balanced chemical equations for the reaction of Ce^4+ and ethylene glycol and for the reaction of Ce^4+ with Fe^2+ Calculate the total number of moles of Ce^4+ added to the solution. The excess number of moles of Ce^4+ can be determined from the amount of Fe^2+ added to the solution

Explanation / Answer

The balanced reaction between Ce^4+ and ethylene glycol is :

Ce^4+ + (CH2OH)2 ==> HCOOH + CO2 + Ce^2+

1mol Ce^4+ reacts with 1 mol ethylenen gycol

Ce^4+ + 2Fe^2+ <==> CE^2+ + 2Fe^3+

1 mol Ce^2+ reacts with 2 moles of Fe^2+

moles of Fe^2+ used to titrate excess Ce^4+ = 0.0459 M * 0.01019 L = 4.68*10^-4 moles

Moles of excess Ce^4+ = 4.68*10^-4 moles/2 = 2.34*10^-4 moles

Moles of Ce^4+ added intially = 0.0831 * 0.0367 L = 3.05*10^-3 moles

Moles of Ce^4+ that has reacted with ethylene glycol = 3.05*10^-3 moles- 2.34*10^-4 moles = 2.82*10^-3 moles

Moles of ethylene glycol present in the solution = 2.82*10^-3 moles

                                                                     =0.175gm

mass % = 0.175 *100/0.2677 = 65.4 %

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