We did a lab where we used spectrophotometry to find the % iron inan unknown sam

Question

We did a lab where we used spectrophotometry to find the % iron inan unknown sample. We used a standard iron solution that Icalculated to be 0.005013 M because I weighed 0.0280 g iron anddiluted it to 100 mL.

We used 4 different volumes of our standard iron solution; 25 mL,10 mL, 5 mL and 2 mL. All solutions were diluted to 100 mL.

I got absorbances that increased with increasing volume (the 25 mLsample was 1.700, the 2 mL sample was 0.186. When I calculate theconcentrations for the graph, however, I get an opposite trend andmy absorbance vs. concentration graph is an inverse relationshipwhen it should be direct (beers law).

For example, I calculted the concentration of the 25 mL sample tobe 0.02004 M by using M1V1 = M2V2...M1 being 0.005013 V1 being .1Land V2 being .025L.

I don't know what I'm doing wrong. Any advice is appreciated.

Explanation / Answer

I agreed that based on your info the standard iron solution is0.005013 M Table of values for calibration curve V1(L)           C1(M)                V2(L)        C2(M)                    absborbance 0.025        0.005013                  0.100      0.001253 0.010        0.005013                  0.100      0.0005013 0.005         0.005013                 0.100      0.00025 0.002         0.005013                   0.100      0.00010 Your mistake is using wrong V1 and V2 You took 25 mL of standard solution , that is V1 @ aconc of 0.005013 M and dilute to 100 mL(V2) @ conc you have tofind

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---