Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Smiles89 on October 14, 2014, 10:13:01 AM
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Hi guys,
Hi was looking to get some help with an experiment. Currently I trying to determine the amount iron in cereal and was hoping to get some help with one of the questions I have been given.
First off I am given the recorded absorbance across the region 470 – 490 nm for the 3.00 mg L–1 iron(III) standard solution constructed in Part 1 above yielded a λmax of 478 nm with an absorbance of 0.674 AU on an instrument with a 1 cm path length.
I have calculated the molar absorptivity to be e= 0.225 L mg^-1 cm^-1 from e = A/bc (a variant of Beer's Law).
I then have been given a set of values for %transmittance of 80, 60, 50, 40, 30, and 20. I have calculated their A values using A = -log (%T/100).
However they want me to find:
(1) the amount of [Fe3+] required (mg L–1);
(2) the quantity Fe3+ required in 50 mL (mg); and
(3) Volume Fe3+ standard solution required (mL)
For (1) I am guessing that I need to use A = ebc and solve for c? (2) I am lost on and same with (3). Namely because of what they say at the bottom 3rd of the calculation page. If I already have A=ebc where I know A, e, b, and then they give us c what exactly I am using to find (2) and (3) for each solution?
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FhYecfSN.png&hash=8d84dea352f0e546deacfa0661a9992a4f468317) - This is the background information of the experiment.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FnttdFUA.png&hash=7c852575275bf0a0aa8f922d1b9c0d7a44c77e73) - This the calculation page we are given.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2Fc8iHNeZ.png&hash=17d2a38913610742d637e93864a2403e47a988c0) - This the question exactly. It refers to using the A value I calculated above.
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For example
Say for 20% T:
A = - log (%T/100) = - log (20/100)
=0.69897 = 0.699
Then using c = A/bc:
c = 0.699/(0.225 x 1)
= 3.11
Does that mean the amount for (2) is 3.11 mg L^-1? And then how do I get (3)?
Is it a matter of going c = 3.11; where 0.15 mg in 50mL = 3ppm. So 3.11mg/50000L = 0.0000622 = 62.2 ppm?
Or would I be using 30 mL instead? So 3.11/30000L = 0.104 = 103.6 ppm?