Question

When 1.6593 g of an organic iron compound containing Fe, C, H, and O was burned in O₂, 3.1036 g of CO₂ and 0.88821 g of H₂O were produced. In a separate experiment to determine the mass percent of iron, 0.2541 g of the compound yielded 0.05741 g of Fe₂O₃. What is the empirical formula of the compound?

Answer 1

first calculate the mass % of each of the compound.

44 g CO2 contain 12 g C

3.1036 g CO2 contain 3.1036 x 12 / 44 = 0.846 g C

% C = (0.846 / 1.6593 ) x 100

% C = 50.98

18 g H2O contain 2 g H

0.88821 g H2O contain 0.88821 x 2 / 18 = 0.09869 g H

% H = (0.09869 / 1.6593) x 100

% H = 5.948

159.69 g Fe2O3 contain 111.6 g Fe

0.05741 g Fe2O3 contain 0.05741 x 111.6 / 159.69 = 0.0401 g Fe

% Fe = (0.0401 / 0.2541) x 100

% Fe = 15.78

% O = 100 - 15.78 - 5.948 - 50.98 = 27.292