What Is the Mass Percent of Iron in the Compound?

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Discussion Overview

The discussion revolves around calculating the mass percent of iron in an iron-containing compound based on the mass of Fe2O3 produced upon oxidation. Participants explore the stoichiometric relationships involved in the reaction and the calculations required to determine the mass percent of iron.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calculation attempting to find the mass percent of iron, but expresses uncertainty about the correctness of their approach and whether significant figures were considered.
  • Another participant notes that oxidation implies starting from a lower oxidation state, suggesting that iron could be in either the elemental form or as FeO, indicating a 2:1 ratio with Fe2O3.
  • A third participant reiterates the calculation steps for determining the number of moles of Fe2O3 and subsequently the moles of iron, but expresses confusion about the overall calculation.
  • A later reply identifies a mistake in the initial calculation, questioning the multiplication by the mass of the compound and clarifying the correct approach to find the mass of iron produced from the given mass of Fe2O3.
  • The corrected calculation results in a mass percent of approximately 15.8% for iron in the compound.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial calculation method, with some expressing confusion and others providing corrections. Multiple approaches to the problem are presented, indicating that the discussion remains unresolved regarding the best method to calculate the mass percent of iron.

Contextual Notes

There are unresolved issues regarding the assumptions made about the oxidation state of iron and the calculations involving significant figures. The discussion reflects varying interpretations of the stoichiometric relationships in the reaction.

Complexity
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Homework Statement



0.3106 g of an iron-containing compound yield 0.07017 g Fe2O3 upon oxidation. What is the mass percent of the iron in the compound??


Homework Equations



stoichiometry?


The Attempt at a Solution



(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)(0.3106g Fe)(1 mole Fe) = 2.729468003x10^-4 grams

Then I took 2.72x10^-4 grams / 0.07017 compound grams of Fe2O3 = .00388979 x 100 = .3889 percent .

What did I do wrong?? Was I off on sig figs?
 
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If an oxidation happens, you have to begin in a lower oxidation state, you only have two options:
Feº , or FeO

In both options it are a 2:1 proportion front a Fe2O3
 
Complexity said:
(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)

OK. Number of moles of Fe2O3.

Complexity said:
(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)

OK. Number of moles of Fe in Fe2O3.

Complexity said:
(0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)(0.3106g Fe)(1 mole Fe) = 2.729468003x10^-4 grams

No idea. It looks like ostrich, but has a head of giraffe.
 
Ok, I find the mistake; ¿Why you multiply by the mass of Fe 0.3106?

Compound with Fe 0.3106

Fe in the compound "x"

¿How much Fe produces Fe2O3?

0.07017 g Fe2O3 (1 mol Fe2O3 / 159.7 g Fe2O3)(2 mol Fe / 1 mol Fe2O3)---> mol of Fe

mol Fe x (55.845 g Fe / 1 mol Fe) = 0.04907 g Fe = x

Then

Rock = 0.3106 g
Fe in rock = 0.04907 g

Percentage (0.04907/0.3106) x 100 = 15.8%
 
Now it has a beak, as usual.
 
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