September 29, 2020, 07:15:23 AM
Forum Rules: Read This Before Posting

### Topic: Help with lab - determining the amount of iron in cereal.  (Read 3061 times)

0 Members and 1 Guest are viewing this topic.

#### Smiles89

• Very New Member
• • Posts: 2
• Mole Snacks: +0/-1 ##### Help with lab - determining the amount of iron in cereal.
« on: October 14, 2014, 10:13:01 AM »
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? - This is the background information of the experiment. - This the calculation page we are given. - This the question exactly. It refers to using the A value I calculated above.
« Last Edit: October 14, 2014, 10:50:25 AM by Smiles89 »

#### Smiles89

• Very New Member
• • Posts: 2
• Mole Snacks: +0/-1 ##### Re: Help with lab - determining the amount of iron in cereal.
« Reply #1 on: October 14, 2014, 10:40:07 AM »
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?
« Last Edit: October 14, 2014, 01:21:04 PM by Smiles89 »