Question

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

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

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