To determine the yield of your ASA synthesis, you can quantify the amount of unreacted SA

Question

Weigh out 0.15 g of your crude ASA into a clean tared 7 mL vial. Record exact mass.
2) Add 5 mL ethanol.
3) Cap vial and mix well.
4) Transfer the above solution into a 50 mL volumetric flask using a transfer pipette.
5) Use wash bottle to thoroughly rinse vial multiple times, transferring each rinse into the volumetric flask
6) Bring flask volume to the 50.00 mL mark with D.I. water.
7) Place 1.00 mL of this solution into a clean, dry 7 mL vial. Add 3.00 mL 0.025 M Fe(NO3)3. Cap and mix well.
8) Fill cuvette 2/3 to 3/4 full with the solution from step 7).
9) Read the absorbance at 517 nm.
10) Determine the mass of SA in the crude analyzed and then the percentage of ASA (ASA yield) in the crude. ϵ = 2147 M-1cm-1 of the iron-salicylate complex, SA-Fe(H2O)4+.

MM of ASA= 180.16 g/mol
MM SA= 138.12 g/mol

My absorbance= .141
So far all I have is (using beers law)
.141= 2141 M-1 cm-1(1)(c)
c= .141/2141= .000065673 M
.000065673 * 50 mL= .0032836516 moles SA?
I am very lost, any and all help would be great! Please and thank you!

Answer 1

I think you meant to type in 2147. Your M is ok. However, mols = M x L (not mL) so 6.567E-5 x 0.05 = 3.284E-6 and that times molar mass SA gives grams SA in the 0.15g sample. If you subtract 0.15 - grams SA that gives you grams ASA and percent yield is (g ASA/0.15)*100 = ?

Answer 2

Thank you!