At 25°C in a 50 mL Erlenmeyer flask were combined 1.050 g of solid iron (III) ch

Question

At 25°C in a 50 mL Erlenmeyer flask were combined 1.050 g of solid iron (III) chloride hexahydrate, 1.4 mL of liquid acetylacetone (C5H8O2), 10.0 mL of 0.500 M aqueous solution of sodium acetate, 10.0 mL of water, and a magnetic stirring bar. After stirring for 15 minutes at 25°C, 0.350 g of solid tris(acetylacetonato)iron (III), Fe(C5H7O2)3, was isolated by vacuum filtration. What was the percent yield of tris(acetylacetonato)iron (III)? At 25°C, the densities of acetylacetone and water are 0.975 g/mL and 0.997 g/mL, respectively. The unbalanced equation for the reaction is shown below.

Fe3+(aq) + C5H8O2(aq) + C2H3O2-(aq) Fe(C5H7O2)3(s) + HC2H3O2(aq)

Explanation / Answer

Balance the given chemical equation.

Fe3+ (aq) + 3 C5H8O2 (aq) + 3 C2H3O2- (aq) --------> Fe(C5H7O2)3 (s) + 3 HC2H3O2 (aq)

As per the stoichiometric equation,

1 mole Fe3+ = 3 moles C5H8O2 = 3 moles C2H3O2- = 1 mole Fe(C5H7O2)3

We have 1.4 mL of C5H8O2 and the density of C5H8O2 is 0.975 g/mL; therefore, mass of C5H8O2 taken = (1.4 mL)*(0.975 g/mL) = 1.365 g.

Atomic mass of Fe3+ = 55.845 g/mol.

Molar mass of C5H8O2 = (5*12.01 + 8*1.008 + 2*15.9994) g/mol = 100.1128 g/mol.

Molar mass of Fe(C5H7O2) = [1*55.845 + 3*(5*12.01 + 7*1.008 + 2*15.9994)] g/mol = 353.1594 g/mol.

Mole(s) of Fe3+ taken = (1.050 g)/(55.845 g/mol) = 0.0188 mole.

Mole(s) of C5H8O2 taken = (1.365 g)/(100.1128 g/mol) = 0.0136 mole.

Mole(s) of C2H3O2- taken = (10.0 mL)*(1 L/1000 mL)*(0.500 M) = 0.005 mole.

0.0188 mole Fe3+ = (0.0188 mole Fe3+)*(3 moles C5H8O2/ 1 mole Fe3+) = 0.0564 mole C5H8O2.

0.0136 mole C5H8O2 = (0.0136 mole C5H8O2)*(1 mole Fe3+/3 mole C5H8O2) = 0.0045 mole Fe3+.

0.005 mole C2H3O2- = (0.005 mole C2H3O2-)*(1 mole Fe3+/3 mole C2H3O2-) = 0.0017 mole Fe3+.

Offcourse, C2H3O2- is the limiting reactant and the yield of the product is decided by the limiting reactant. The yield of the product is

0.005 mole C2H3O2- = (0.005 mole C2H3O2-)*(1 mole Fe(C5H7O2)3/3 mole C2H3O2-) = 0.0017 mole Fe(C5H7O2)3

Theoretical yield of Fe(C5H7O2)3 = (0.0017 mole)*(353.1594 g/mol) = 0.6004 g.

The actual yield of solid Fe(C5H7O2)3 is 0.350 g; therefore, the percent yield is (0.350 g)/(0.6004 g)*100 = 58.2945% 58.29% (ans).

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