March 29, 2017

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A student wanted to determine the amount of copper in a sample of copper ore. The student
dissolved a 2.500 ± 0.001 g piece of copper ore in about 75 mL of nitric acid; then, a
complexing agent was added. The student transferred this solution to a volumetric flask and
used de-ionized water to fill to a final volume of 100.00 ± 0.08 mL. From that volumetric
flask, the student transferred 15.0 mL of solution into a 250.00 ± 0.08 mL volumetric flask
and diluted to the mark with de-ionized water. This final solution (from the 250.00 mL
volumetric flask) showed an absorbance of 0.545 ± 0.004 at 610 nm in the spectrometer.
The student used a series of aqueous copper solutions of known concentration to generate a
calibration curve at 610 nm and obtained a best-fit line of y = (3,256)x + 0.007.
What is the % copper by mass in the ore sample from above with the correct significant figures?

  • College Gen.Chemistry - ,

    The straight line of y = 3,256x + 0.007 is the calibration curve of Absorbance on the y axis vs concn on the x axis. So insert 0.545 for y (that's the absorbance) and solve for x (that's the concn). I assume that is concn in grams/250 mL although nowhere is that information available. Let's call that number of grams = z.
    So z/250 = grams/mL and that x 15 mL = w grams in that 15 mL aliquot. That of course is the amount in the original 100 mL. So (g Cu/mass sample)*100 = (w grams/2.50 gram sample)*100 = %Cu. You're responsible for the correct number of s.f.

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