DATA TABLE: Trial 1 / Trial 2 Volume of Fe 2+ solution (mL): 25 / 25 Volume of C

Question

DATA TABLE: Trial 1 / Trial 2

Volume of Fe 2+ solution (mL): 25 / 25

Volume of Ce 4+ solution used to reach equivalence point (mL): 26 / 25.5

addition information:

MATERIALS:

0.100 M (NH4)2Ce(NO3)6 in 1.0 M H2SO4

(NH4)2Fe(SO4)2•6H2O solution, unknown.

we measured out precisely 25.00 mL of a ferrous ammonium sulfate solution of unknown molar concentration and transfered it to a 250 mL beaker

0.100 M (NH4)2Ce(NO3)6 in 1.0 M H2SO4

(NH4)2Fe(SO4)2•6H2O solution, unknown.

DATA ANALYSIS 1. Calculate the molar amount of Ce4 used to reach the equivalence point of the reaction. 2. Calculate the molar amount of iron in the sample of ferrous ammonium sulfate solution. 3. Calculate the molar concentration of the ferrous ammonium sulfate solution. The ferrous ammonium sulfate solution that you tested was prepared by dissolving 40.0 g of solid (NH4)2Fe(SO4)2.6H20 in 1.00 liter of solution. This substance is often impure. a. Calculate the theoretical percent Fe in a pure sample of(NH4)2Fe(SO4)2"6H20. b. Calculate the percent Fe in the sample that you tested. c. Compare your experimental percent Fe to the theoretical percent Fe. How pure was your 4. sample? Use a calculation to support your assertion.

Explanation / Answer

Trial 1

1. Volume of Fe2+ solution = 25 mL

Volume of Ce4+ solution used to reach equivalence point = 26 mL

The molarity of (NH4)2Ce(NO3)6 in 1.0 M H2SO4 = 0.1 M  

1. The molar amount of Ce4+ used to reach the equivalent point of the reaction = 26 mL * 0.1 mmol/mL = 2.6 mmol = 0.0026 mol

2. The molar amount of iron in the sample of ferrous ammonium sulfate solution = 0.0026 mol

(Note: At equivalence point, the no. of moles of both Fe2+ and Ce4+ are the same.)

3. Therefore, the molar concentraiton of ferrous ammonium sulfate solution = 0.0026 mol/25*10-3 L = 0.104 M

4. a) The molar mass of (NH4)2Fe(SO4)2•6H2O = 392 g/mol

The molar mass of Fe = 56 g/mol

Therefore, the theoretical percent Fe in the pure sample of (NH4)2Fe(SO4)2•6H2O = (56/392)*100 = 14.3%

b) The molar mass of (NH4)2Fe(SO4)2•6H2O = 392 g/mol

The no. of moles of (NH4)2Fe(SO4)2•6H2O = 40/392 = 0.102 mol

The molar mass of Fe = 56 g/mol

i.e. The mass of Fe = 0.102 mol*56 g/mol = 5.714 g

Therefore, the theoretical percent Fe in the sample tested = (5.714/40)*100 = 14.3%

c) The theoretical percent of Fe is the same in both the pure sample as well as the tested one.

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