I\'m having trouble figuring out how to do the calculations for a \"determinatio

Question

I'm having trouble figuring out how to do the calculations for a "determination of manganese in a steel sample" lab. In LoggerPro, we were supposed to have a line of best fit and to use that equation to determine the permanganate ion concentration in the solution containing the dissolved sample. We also need to compute the mass of manganese present in the 50 mL flask and % manganese in the steel sample. It seems that my line of best fit is y=802x+.0252 and the absorbance of the unknown manganate sample came out to be 0.286. Values are as follows:

wavelength nm ( ) Absorbance 405 0.07 425 0.06 445 0.06 465 0.11 485 0.21 505 0.35 525 0.47 545 0.45 565 0.26 585 0.09 605 0.05 625 0.04 525 0.47 527 0.46 529 0.45 531 0.43 533 0.41

Explanation / Answer

The equation of the best fit line is

y = 802x + 0.252

where y denotes the absorbance of permanganate (MnO4-) and x is the molar concentration of MnO4-. The absorbance of the unknown sample of MnO4- is 0.286. Put y = 0.286 in the equation and obtain

0.286 = 802x + 0.252

====> 802x = 0.286 – 0.252 = 0.034

====> x = 0.034/802 = 4.2394*10-5

The concentration of MnO4- in the unknown sample is 4.2394*10-5 M.

We have 50 mL of the sample solution; hence, mole(s) of Mn in the 50 mL sample solution is (50 mL)*(1 L/1000 mL)*(4.2394*10-5 M) = 2.1197*10-6 mole.

The atomic mass of Mn is 54.938 g/mol; therefore, mass of Mn present in the sample = (2.1197*10-6 mole)*(54.938 g/mol) = 1.16452*10-4 g = (1.16452*10-4 g)*(1000 mg/1 g) = 0.116452 mg ? 0.116 mg (ans).

We need to know the mass of the sample to calculate the percent Mn in the sample.

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