Mass Percent Calculations

Question

A 5.211g solid sample containing only KNO2 (FW = 85.104 g/mol) and KNO3 (FW = 101.103 g/mol) is dissolved in 500.0mL of water. A 50.00 mL aliquot of the solution is treated with 45.00 mL of 0.1199 M Ce4+in strong acid to oxidize the nitrite (NO−2) ion and form the nitrate (NO3−) ion. After 5 min, the excess Ce4+ is back-titrated with 29.35 mL of 0.03635 M Fe2+.

Calculate the mass percent of KNO2 in the solid.

Answer 1

In the first reaction, the NO₂^- (from KNO₂) is oxidized to NO₃^- by Ce⁴+:

NO₂^- + 2Ce⁴⁺ + H₂O ==> NO₃^- + 2Ce³⁺ + 2H⁺

In the next reaction, the excess Ce⁴⁺ is back titrated with Fe²⁺:

Ce⁴⁺ + Fe²⁺ ==> Ce³⁺ + Fe³⁺

How much excess Ce⁴⁺ is present?

mols Fe²⁺ used = 29.35 mls x 1 L / 1000 mls x 0.03635 mols / L = 0.001067 mols Fe²⁺

mols Ce⁴⁺ = 0.001067 mols Fe²⁺ x 1 mol Ce⁴⁺ / mol Fe²⁺ = 0.001067 mols Ce⁴⁺ in excess

How much Ce⁴⁺ was present initially?

mols Ce⁴⁺ added = 45 mls x 1 L / 1000 mls x 0.1199 mols / L = 0.005396 mols Ce⁴⁺

How much Ce⁴⁺ reacted with NO₂^-?

0.005396 mols - 0.001067 mols = 0.004329 mols Ce⁴⁺ to oxidize all of the NO₂^-

How many mols NO₂^- were present?

0.004329 mols Ce⁴⁺ x 1 mol NO₂^- / 2 mols Ce⁴⁺ = 0.002165 mols NO₂^- originally present

This is present as KNO₂ and it is present in a 50.00 ml aliquot from a 500.00 ml sample. Correcting for this...

0.002165 mols KNO₂ / 50.00 mls x 500.00 mls = 0.02165 mols KNO₂ in original sample

Mass KNO₂ = 0.02165 mols x 85.104 g / mol = 0.1842 g KNO₂

Mass % KNO₂ = 0.1842 g / 5.211 g (x100%) = 3.542%

(be sure to check the math and redox equations)