Question

suppose a 1.51 g sample of unknown x containing k 2 c 2 o 4 kx2cx2ox4 was dissolved in 25 ml of di water and 25 ml 3 m h 2 s o 4 hx2sox4 and then titrated with 0.0655 m k m n o 4 kmnox4 . the sample required 26.75 ml of titrant to reach the endpoint. determine the mass percent k 2 c 2 o 4 kx2cx2ox4 in sample unknown x.

Answer 1

The  mass percent K2C2O4  in sample unknown is 48.2 % as determined by the formulas.

if we write the reaction:

• 2MnO4– + 16H+ + 5C2H42----> 2Mn2+ + 10CO2 + 8H2O
• Based on this, we will see that 2 moles of KMnO4 can react with five mole of oxalic acid moles of KMnO4 = extent of KMnO4 x awareness of KMnO4 / 1000 = 0.0655 x 26.75/1000 = 0.0017521 moles
• so 0.0017521  moles of KMnO4 can neutralize the = 5 x 0.0017521 /2 = 0.004380 moles of potassium oxalate so moles of potassium oxalate is 0.004380
• mass of potassium oxalate = moles of potassium oxolate x molecular weigh of potassium oxolate
• As given compound need to be K2C2O4 (now no longer KC2O4)
• so molecular weight of K2C2O4 = 166.22 g/mole
• mass of potassium oxalate = 166.22 x 0.004380 =0.7280 g
• so solution is 0.7280g
• mass percentage of potassium oxalate is = mass of potassium oxalate x 100 / pattern mass of unknown
• = 0.7280 x 100 /1.51 =48.2 %, so mass percentage is 48.2%.

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