Question

I am trying to calculate the mass of my unknown based on this data This was...

I am trying to calculate the mass of my unknown based on this data

This was measured at 505nm

unknown conc=3.12E-5M

unknown absorbance=.290

Other measurements of known solutions with .2132g of FeSO4(NH4)2SO4⋅6H2O, MW = 392.14)

Absorbance Concentration
.04 4.50E-06
.077 8.70E-06
.16 1.74E-06
.315 3.48E-05
.472 5.44E-05

y=mx+b

y=7430.4x+.058

Procedure:

Unknown solution of Fe2+. Each student analyzes their own unknown. Your unknown will be provided in a 250-mL volumetric flask. The unknown already contains sufficient sulfuric acid. Dilute to volume with deionized water and mix thoroughly. Next, pipet a 25-mL aliquot of your diluted unknown solution into a clean 100-mL volumetric flask, add the color-forming reagents as described in part b for the standard solutions, dilute to volume and mix thoroughly.

(part b.:

To each flask add 1 mL of the hydroxylamine hydrochloride solution, 10 mL of the sodium acetate solution and 20 mL of the 1,10-phenanthroline solution.)

update: Unknown is Fe2+ so the molar mass is 55.845 g/mol

Homework Answers

Answer #1

As per the equation of the data

The absorbance and concentration are related as

y = absrobance = 7430.4x+.058

Where x = concentration of the diluted solution for which we have measured the absrobance

Now the x concentration of solution gives = 0.290 absrobance

Putting values

0.290 = 7430.4x+.058

x = 3.122 X 10^-5 M

Let the concentration of starting stock = M1

25mL of this is diluted to 100mL

M1 X 25 = 100 X 3.122 X 10^-5 M

M1 = 1.2489 X 10^-4 M

Molarity = Mass of solute / Molecualar weight of solute X volume of solution in litres

1.2489 X 10^-4 = Mass of Fe+2 / 55.845 X 1 L

Mass of Fe+2 = 0.00697 grams

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT