Question

when 1.6038 g of an organic iron compound Fe, C, H and O was burned in O2, 2.9998 g of CO2 and 0.85850 g of H2O were produced. In a separate experiment to determine the mass percent of iron, 0.2726 g of the compound yielded 0.06159 g of Fe2O3. What is the empirical formula of the compound?

Answer 1

compound + O₂ ---> CO₂ + H₂O

% of C in CO₂ is = [(molar mass of C)/(molar mass of CO₂)]x100

                        = (12.01g / 44.01 g) x 100 = 27.289 %

Hence, amount of carbon in 2.9998 g of CO2 is = (27.289/100) x 2.9998 = 0.8186 g C

Therefore, 1.6038 g of the organic iron compound contains 0.8186 g of C.

number of moles of carbon in the sample = 0.8186/12.01 = 0.0682 mol C

% of hydrogen in H₂O is = [(2 x molar mass of H)/(molar mass of H₂O)]x100

                                   = [(2 x 1.008g)/(18.015 g)]x100 = 11.19 %

Hence, amount of hydrogen in 0.8585 g of H₂O is = (11.19/100) x 0.8585 = 0.0961 g H

Therefore, 1.6038 g of the organic iron compound contains 0.0961 g H

number of moles of hydrogen in the sample = 0.0961/1.008 = 0.0953 mol H

% of Fe in Fe₂O₃ is = [(2 x molar mass of Fe)/(molar mass of Fe₂O₃)]x100

                                   = [(2 x 55.845g)/(159.69 g)]x100 = 69.94 %

Hence, amount of iron in 0.06159 g of Fe₂O₃ is = (69.94/100) x 0.06159 = 0.0431 g Fe

Then, 0.2726 g of the organic iron compound contains 0.0431 g Fe

Therefore, amount of Fe in 1.6038 g of the compound is =

number of moles of iron in the sample = 0.2534/55.845 = 0.0045 mol Fe

The given organic compound contains 0.8186 g of C(0.0682 mol C), 0.0961 g H(0.0953 mol H), 0.2534 g Fe(0.0045 mol Fe).

Hence, amount of oxygen in the compound = 1.6038 - (0.8186 + 0.0961 + 0.2534) = 0.4357 g

mol of oxygen in the compound = 0.4357/15.999 = 0.0272 mol O

The ratio of Fe : C : H : O is 0.0045 : 0.0682 : 0.0953 : 0.0272

Dividing them with the least number, 0.0045, the ratio becomes 1 : 15 : 21 : 6

Hence, Emperical formula of the compound is FeC₁₅H₂₁O₆