anonymous
  • anonymous
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?
Chemistry
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Koikkara
  • Koikkara
\(Hint:\) \(\small\rm\color{green}{Check~out~this~similar~question~from~yahoo~!!}\) 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. \(\small\rm\color{blue}{Answer!}\) (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 \(\small\rm\color{blue}{This~is~how~you~solve~this~problem~!}\) \(But,\) \(\Large\rm\color{green}{Now~you~can~try~it~out~with~your~question!}\)
Koikkara
  • Koikkara
@YamadaTasnim \(Hope~It~helps~!!\)
Koikkara
  • Koikkara
\(Hint: Complete~Question~\) 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.