Question

when 1.6713 g of organic iron compound containing Fe, C, H and O burned in O2, 3.1260g of CO2 and 0.89464g of H2O were produced. in a separate experiment to determine the mass percent of iron 0.2721g of the compound yield 0.06148g of Fe2O3. what is the empirical formula of the compound.

Answer 1

44gm of CO2 contains 12 gm of C. So, amount of C in 3.1260gm CO2 is 0.853 gm

18 gm H2O contains 2 gm of H2 . So, amount of H2 in 0.89464 gm H2O is 0.099 gm

Molar mass of Fe2O3 = 156gm/mol

156 gm fe2o3 contains 112 gm of Fe. Amount of 0.06148gm of Fe2O3 contains 0.06148 *112/156 =0.044 gm Fe

This amount of Fe is actually present in 0.2721 gm compound. So, amount of iron present in 1.6713 gm compound = 0.27 gm.

Sum up the data :

Amount of Fe in 1.6713gm of compound = 0.27 gm

Amount of C =0.853 gm

amount of H = 0.099 gm

amount of O2 = 1.6713 - (0.27 + 0.853 +0.099) = 0.45 gm

Moles of Fe =0.27/56 = 0.0048

Moles of C = 0.853/12 = 0.07

Moles of H =0.099/1 = 0.099

Moles of O2 = 0.45/16 =0.028

Mole ratio = Fe:C:H:O = 0.0048:0.07:0.099:0.028 = 1: 14.58 :20.63 :5.83 = 1:15:21:6

emperical formula = Fe C15 H20 O6