Question

Lab: Spectroscopy and Qualitative Analysis in the Determination of an Unknown Iron Salt

Procedure

Part A:

Obtain one of the provided kits and inspect for completeness (see available list in the laboratory). Ask your instructor for replacement items if necessary. Select one of the available unknown iron salts. All of the iron salts (which may or may not be hydrated) are in the +2 oxidation state and will have one of the following anions: Cl^-, Br^-, SO4^2- and CH3COO^- (acetate). Finally, obtain ~ 30 mL of the standard iron(II) solution (record the concentration).

2. Using the 10 mL beaker, weigh out ~ 0.0400 to 0.0500 g (±0.0001 g) of your salt. Record the mass dispensed. Transfer the salt to a 100.00 mL volumetric flask using a small amount of water. Rinse the beaker 3 to 4 times; transferring the washings to the flask each time. Allow the iron salt to completely dissolve and then dilute to the mark.

3. Using the pipette from your kit, transfer 1.00 mL of the solution from step 2 into a 100.00 mL volumetric flask. To the flask add 1 mL each of: 1% hydroxylamine hydrochloride solution, 1 M sodium acetate and 1% 1,10-phenanthroline solution from the provided burettes. Dilute to the mark and mix thoroughly. This is the unknown solution you will measure the absorbance of.

4. Rinse and fill a burette with the standard iron solution obtained in step 1. Prepare the standard solutions as follows: accurately dispense – recording initial and final burette readings (±0.02 mL) – approximately 1.00, 2.00, 3.00 and 4.00 ml of the standard Fe^2+ solution into separate volumetric flasks. As in step 3, add 1 mL each of: 1% hydroxylamine hydrochloride solution, 1 M sodium acetate and 1% 1,10-phenanthroline solution from the provided burettes. Dilute to the mark and mix thorougly.

5. Using the final 100.00 mL volumetric flask, prepare the blank solution by adding 1 mL each of: 1% hydroxylamine hydrochloride solution, 1 M sodium acetate and 1% 1,10-phenanthroline solution from the provided burettes. Dilute to the mark and mixthoroughly.

6. To analyze your samples ask the lab demonstrators for instruction on the proper use of the spectrophotometer. Obtain the absorbance of each of the standards and the unknown at 508 nm.

Part B:

Qualitative Analysis

Transfer some of your salt (tip of a scoopula) into a beaker and dissolve it in ~ 30 mL distilled water. If you are finding it difficult to dissolve your salt, add two to three drops of 6M HNO3 and mix.Transfer equal portions (about 2-3 cm in height) of this solution into two test tubes:

1. Chloride and Bromide. Add 3 drops of 6M HNO3 to the solution in the first test tube and shake well. Add 5 drops of 0.1M AgNO3 and mix well. If a precipitate has formed upon the addition of silver nitrate a positive test for chloride or bromide has occurred. If no precipitate formed you do not have either the chloride or bromide salt and can move on to the sulfate test. To differentiate between chloride and bromide, add 10 drops of 6M NH4 OH to the mixture (do not mix). If two distinct layers form immediately, then the chloride anion is present.

2. Sulfate. Add 3 drops of 6M HNO3 to the solution in the second test tube and shake well. Add 5 drops of 0.1M BaCl2. The formation of a white precipitate indicates the presence of SO4^2-.

3. Acetate. Place a small amount of the solid salt (~ the size of a pea) into a test tube. Add 20 drops of concentrated H2SO4 (CARE!!!) and 20 drops of 1-pentanol from the dropper bottles in the fume hood to the test tube and mix with a glass stirring rod. Place the test tube in the water bath provided (making a note of which one is yours) for about 5 minutes. Remove the test tube from the bath and check the odour of the solution by bringing your nose slowly to the test tube and waving your hand over the top of the test tube towards your nose. You can also pour it onto a watch glass to make it easier to pick up the odour. A fruity odour (it often reminds people of fake banana smell) indicates the presence of the acetate ion.

_______My Data Below

Concentration of standard iron(II) solution = 50.0 ppm

Weight of unknown salt: 0.0413 g (+- 0.0001 g)

Fe^2+ Solution (mL)	Volume (final)-Volume (initial) = Change in Volume (+- 0.04 mL)
1.00	1.01
2.00	1.89
3.00	3.02
4.00	4.01

Fe^2+ Solution (mL)	Calculated Concentration (ppm)	Absorbance (a.u.
1.00	0.505	0.079
2.00	0.945	0.198
3.00	1.51	0.304
4.00	2.005	0.410
Unknown solution	1.35	0.267

The positive ion test determined that the unknown salt was sulphate (SO4^2-).

Question: Using the mass of the sample analyzed, the dilutions you performed and the positive anion test determine: the identity of your unknown salt, its molecular weight and the water of hydration.

Answer 1

The concentration of Fe²⁺ in the unknown salt = 1.35 ppm.

We know that 1 ppm = 1 mg/L; therefore,

1.35 ppm = 1.35 mg/L.

Note that the Fe²⁺ salt was prepared from the solid sample by a one step dilution by taking 1.00 mL of the prepared salt and diluting to 100.0 mL.

Therefore, the concentration of Fe²⁺ in the second solution = (1.35 mg/L)*(100.0 mL)*(1 mL) = 135.0 mg/L

The total volume of the sample prepared = 100.0 mL; therefore, the amount of Fe²⁺ present in the sample = (135.0 mg/L)*(100.0 mL)*(1 L)/(1000 mL) = 13.5 mg.

The qualitative test confirmed the presence of SO₄^2-.

Fe²⁺ and SO₄^2- combine in a 1:1 molar ratio as

Fe²⁺ + SO₄^2- --------> FeSO₄ ……(1)

The atomic/molar masses are

Fe: 55.845 g/mol

SO₄^2-: 96.06 g/mol

FeSO₄: 151.908 g/mol

H₂O: 18.015 g/mol.

Millimoles of Fe²⁺ corresponding to 13.5 mg = (13.5 mg)/(55.845 g/mol) = 0.2417 mmole.

As per the stoichiometric equation,

0.2417 mmole Fe²⁺ = 0.2417 mmole SO₄^2- = 0.2417 mmole FeSO₄.

Mass of FeSO₄ corresponding to 0.2417 mmole = (0.2417 mmole)*(151.908 g/mol) = 36.7162 mg.

Mass of the salt taken = 0.0413 g = (0.0413 g)*(1000 mg)/(1 g) = 41.3 mg.

Mass of water in the salt = (41.3 – 36.7162) mg = 4.5838 mg.

Millimoles of water corresponding to 4.5838 mg water = (4.5838 mg)/(18.015 g/mol)

= 0.2544 mmole.

Simple ratio of the number of millimoles of FeSO₄ and H₂O = 0.2417:0.2544 = 1.00:1.05 ≈ 1:1.

Therefore, the molecular formula of the iron compound is FeSO₄.H₂O and the molecular mass is (1*55.845 + 1*96.06 + 1*18.015) g/mol = 166.92 g/mol (ans).