I couldn't find a topic quite like mine when I used the search function, but if there is a post about this already, you can just redirect me there.
A Pre-Lab question for my upcoming Iron Titration Lab is:
Using data generated by following the procedure in the lab, you calculate that there is 3.315 x 10-5 mol of Fe2+ in an average 10.00 mL sample of your stock solution. Calculate how many mg of Fe2+ were present in a single tablet. Note that you initially dissolved two tablets to make the 100.0 mL of stock solution.
We went over this type of calculation in class, the example we used was:
Molecular Weight: 278 FeSO47H2O/mol
1 mole Fe2+ = 1 mole FeSO47H2O
If there are 2.40x10-4 mol Fe2+ we calculated that there are 334mg FeSO47H2O/tablet:
(2.40x10-4 mol Fe2+ / 10mL) * (100.0mL / 2 Tablets) * (1m FeSO47H2O / 1m Fe2+) * (278.0g FeSO47H2O / 1m FeSO47H2O) * (1000mg / 1g) = 334mg FeSO47H2O / Tablet
I used this example, given by my professor, to find the Pre-Lab calculation:
(3.315 x 10-5 mol Fe2+ / 10mL) * (100.0mL / 2 Tablets) * (1m FeSO47H2O / 1m Fe2+) * (278.0g FeSO47H2O / 1m FeSO47H2O) * (1000mg / 1g) = 46.1mg FeSO47H2O / Tablet
The available answers are:
18.51 mg
7.448 mg
10.92 mg
0.9256 mg
8.422 mg
9.256 mg
8.711 mg
I don't really know what I am doing wrong, so thank you for the help.