Calculate the mass of chromium found in the 2.50 g sample of stainless steel

The question is incomplete and missing important information which is provided below:

"Stainless steel is a corrosion resistant alloy composed of nickel, chromium and iron. A lab technician was assigned to determine the mass of chromium in a 2.50 g sample of stainless steel. The sample was first dissolved in acid to convert the chromium metal in the sample to chromium(II) ions. The acidified chromium(II) ions were then titrated with 17.69 mL of 0.125 mol/L perchlorate solution according to the unbalanced equation: ClO4- + Cr2+ --> Cl- + Cr3+. Calculate the mass of chromium found in the 2.50 g sample of stainless steel."

Now with the full question, we can first balance the equation and determine how many moles of Cr were in the sample to begin with. To balance the equation, break the equation down into its two half reactions, balance electrons and charge, then combine the reactions:

ClO₄⁻ + 4H₂O + 8e⁻ → Cl⁻ + 8OH⁻

(Cr²⁺ → Cr³⁺ + 1e⁻) x 8

ClO₄⁻ + 8Cr²⁺ + 4H₂O → Cl⁻ + 8Cr³⁺ + 8OH⁻

We know that 17.69 mL of 0.125 M ClO₄⁻ is required to titrate the Cr²⁺:

0.125 mol/L x 0.01769 L = 0.00221 mol ClO₄⁻

0.00221 mol ClO₄⁻ x 8 mol Cr²⁺/mol ClO₄⁻ = 0.01769 mol Cr²⁺

With moles of chromium, we can now use the molecular weight of chromium to determine the mass of chromium present:

0.01769 mol x 52 g/mol = 0.92 g chromium

In a 2.50 g sample of stainless steel, there is 0.92 g of chromium.

Answer is to avoid experimental error