HI GENIUS! C) The percent purity of a malonic acid reagent must be calculated. A

Question

HI GENIUS!

C) The percent purity of a malonic acid reagent must be calculated. A student weighs out 0.1511 grams of malonic acid, CH2(CO2H)2, and quantitatively transfers the solid to a 250 mL Erlenmeyer flask. To the flask 50.00 mL of distilled water is added to dissolve the malonic acid. After dissolution, the student uses a 50 mL volumetric pipet to add 50 mL of 0.200 M Ce4+, which is allowed to react with the malonic acid. After complete reaction, the excess Ce4+ is back titrated with 14.58 mL of 0.100 M Fe2+ solution. Calculate the percent purity malonic acid.

CH2(CO2H)2 + 2 H2O + 6 Ce4+ 2 CO2 + HCO2H + 6 Ce3+ + 6 H+

Ce4+ + Fe2+ Ce3+ + Fe3+

THANK YOU IN ADVANCE!

Explanation / Answer

Moles of Fe+2= 0.1*14.58/1000= 0.001458 used for reacting with Ce+4 which is excess

Moles of 0.2M Ce4+ added initially = 50*0.2/1000=0.01 moles

Moles of ce+4 added for determination of malonic acid = 0.01-0.001458= 0.008542 moles

As per the reaction (1), 6 moles of Ce+4 reacts with 1 mole of malonic acid

Malonic acid used for reaction (1)= 0.008542/6= 0.001424

Molecular weight of malonic acid CH2(CO2H)2= 12+2+2*(12+32+1)= 84

Mass of malonic acid =0.001424*84=0.1195gm

Mass of actual sample =0.1511

Purity of the sample =100*0.1195/0.1511 =79%

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