# Acac

General Chemistry I Lab Work - Wiki Pages

## Sample Calculation for Fe(acac)3 Experiment

Here is a sample calculation for the determination of iron in Fe(acac)3. Each sample studied required three solutions:

1. Red Fe(acac)3 sample dissolved in HNO3, diluted to 100 mL
2. Removed 10 mL, diluted to 100 mL
3. Removed 10 mL, diluted to 100 mL again

Use your Beer's law graph to get the concentration of Fe in solution "3", and then work backwards to get the concentration in "2" and then in "1", which represents all of the Fe in the original sample.

Part 1: Suppose that your Beer's law graph gives 3.28 ppm Fe for solution "3" above. Work backwards to get the concentration of Fe in solution "2" by the usual dilution formula:

(10 mL) (CBD) = (100 mL) (3.28 ppm)
Solve to get CBD = 32.8 ppm, for soln "2"

Part 2: Now get the concentration of Fe in solution "1" by the same dilution formula:

(10 mL) (CBD) = (100 mL) (32.8 ppm)
Solve to get CBD = 328 ppm, for soln "1"
Note that 328 ppm means 328 grams iron for every 1,000,000 grams of solution

Part 3: Finally, all of the Fe in solution "1" came from the original Fe(acac)3. Thus the mass of iron in the entire 100 mL is

(100 mL soln) (1 g soln / mL soln) (328 g Fe / 1,000,000 g soln) = 0.0328 g Fe
The final result is that the Fe(acac)3 contained 0.0328 g of iron.
NOTE: The iron in Fe(acac)3 is not iron metal! It is in the form of Fe3+, so we're not actually looking at Fe but rather an iron ion.

Part 4: Calculate the percent of iron in the original Fe(acac)3 sample weighed out.

% iron in Fe(acac)3 = 100 x 0.0328 g / (mass of Fe(acac)3 used)
For example, if the mass of red Fe(acac)3 weighed out was 0.2050 grams then
% iron in Fe(acac)3 = 100 x 0.0328 g / 0.2050 g = 16.0 % iron

## Explosion movie

General Chemistry I lecture demonstration of November 22, 2006, the explosion of 0.2 grams of hydrogen.

A 6-megabyte movie of the explosion is here, in the AVI format.

Be sure to crank up the volume!