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please show work. When 1.5953g of an organic iron compound containing Fe, C H and O...

please show work. When 1.5953g of an organic iron compound containing Fe, C H and O was burning in O2, 2.9839g CO2 and 0.85395g of H2O were produced. In a seperate experiment to determine the mass percent of iron, 0.2836g of the compound yield 0.06407g of Fe2O3. What is the empirical formula of the compound?

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Answer #1


compound containing Fe, C H and O

mass of C present in sample = (2.9839/44)*12 = 0.814 grams

No of mol of C = 0.814/12 = 0.068 mol

No of mol of H present in sample = (0.85395/18)*2*1 = 0.095 mol

mass of Fe present aliquot = (0.06407/159.69)*(111.6900) = 0.045 grams

mass of Fe present in sample = 0.045*1.5953/0.2836 = 0.253 grams

No of mol of Fe = 0.253/55.845 = 0.0045 mol

mass of O = (1.5953 - (0.253+0.814+0.095)) = 0.433 grams

No of mol of O = 0.433/16 = 0.027

simplest ratio

C = 0.068/0.0045 = 15.11   = 15

H = 0.095/0.0045 = 21.11    = 21

Fe = 0.0045/ 0.0045 = 1     = 1

O = 0.027/0.0045 = 6        = 6


empirical formula

C15H21FeO6

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