Question

A stainless steel alloy is to be analyzed for its chromium content. A 3.00 g sample of the steel is used to produce 250.0 mL of a solution containing Cr2O72−. A 10.0-mL portion of this solution is added to BaCl2(aq). When the pH of the solution is properly adjusted, 0.145 g BaCrO4(s)precipitates.

What is the percent Cr, by mass, in the steel sample?​

Answer 1

Molar mass of Cr = 52.00 g / mol

Molar mass of BaCrO ₄ = 137.34 + 52.00 + ( 4 16.00 ) = 253.34 g / mol

Let's calculate mass of Cr present in 10.0 ml diluted solution from the mass of BaCrO ₄

One molecule of BaCrO ₄ contain one Cr atom.

1 mol BaCrO ₄ 1 mol Cr

i e 253.34 g BaCrO ₄ 52.00 g Cr

0.145 g BaCrO ₄ ( 52.00 0.145 / 253.34 ) g Cr

0.145 g BaCrO ₄ 0.02976 g Cr

We have calculated, 10 ml diluted solution contain 0.02976 g Cr.

10 ml diluted solution 0.02976 g Cr

250 ml diluted solution ( 0.02976 250 / 10 ) g Cr

250 ml diluted solution 0.744 g Cr

Now, we can calculate % of Cr by mass in the steel sample.

% Cr by mass in the steel sample = ( Mass of Cr / Mass of sample )   100

= ( 0.744 g / 3.00 g )   100

= 24.8 %

ANSWER : % Cr by mass in the steel sample = 24.8 %