Question

when 1.4613 g of an organic compound containing Fe, C, H , O was burned in 02(oxygen) . 2.7333 g of CO2 AND 0.78222 g of H20 were produced. in a separate experiment to determine the mass of iron 0.4676 g, the compound yielded 0.1056 g of Fe2O3. what is the empirical formula of the compuond?

Answer 1

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When 1.7358g of an organic iron compound containing Fe, C, H, and O was burned in O2, 3.2467g of CO2 and 0.92916g of H2O were produced.

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(0.06996 g Fe2O3) / (159.6887 g Fe2O3/mol) x
(2 mol Fe / 1 mol Fe2O3) x (55.8452 g Fe/mol) / (0.3096 g) =
0.158049 = 15.8049% Fe

(3.2467g CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) x
(12.01078 g C/mol) = 0.886065 g C

(0.92916 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) x
(1.007947 g H/mol) = 0.103972 g H

15.8049% of 1.7358 g = 0.274341 g Fe

Use the Law of Conservation of Mass:
(1.7358 g total) - (0.274341 g Fe) - (0.103972 g H) - (0.886065 g C) =
0.471422 g O

Convert to moles:
(0.274341 g Fe) / (55.8452 g Fe/mol) = 0.00491253 mol Fe
(0.886065 g C) / (12.01078 g C/mol) = 0.0737725 mol C
(0.103972 g H) / (1.007947 g H/mol) = 0.103152 mol H
(0.471422 g O) / (15.99943 g O/mol) = 0.0294649 mol O

Divide by the smallest number of moles:
(0.00491253 mol Fe) / 0.00491253 mol = 1.0000
(0.0737725 mol C) / 0.00491253 mol = 15.017
(0.103152 mol H) / 0.00491253 mol = 20.998
(0.0294649 mol O) / 0.00491253 mol = 5.9979

Round to the nearest whole numbers to find the empirical formula:
FeC15H21O6

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Answer 2

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Answer 3

\(Hint: Complete~Question~\)
When 1.7358g of an organic iron compound containing Fe, C, H, and O was burned in O2, 3.2467g of CO2 and 0.92916g of H2O were produced. In a separate experiment to determine the mass percent of iron , 0.3096g of the compound yielded 0.06996g of Fe2O3. What is the empirical formula of the compound.