I have always had the notion of calculating weight percentage of Fe in FeO. It always seems that we are to calculate at times the weight percentage of FeO in Fe.Particularly when one weight percentage is involved.However "weight percentage of FeO in Fe" is something that seems extremely ridiculous on my part.I would be extremely grateful if someone can correct my wrong notion/give me an intuition to what I am asking for.

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    $\begingroup$ Please clarify the question and ideally post the exact question in the textbook. It makes little sense as written. $\endgroup$ – M. Farooq Aug 31 '19 at 15:37
  • $\begingroup$ @M.Farooq My question is how do I calculate weight of Fe from the weight of FeO if I have a mixture of Fe and FeO? $\endgroup$ – user586228 Aug 31 '19 at 15:44
  • $\begingroup$ I have read this three times and it does not make sense. Please do exactly what @M.Farooq suggested. $\endgroup$ – Ed V Aug 31 '19 at 15:55
  • $\begingroup$ Still makes no sense, so I will take a guess. You know that the mass % of Fe in FeO is 77.73%. Therefore, the reciprocal of 0.7773 is 1.2865. Hence, 1 unit (g, ton, mol, whatever) of Fe can be oxidixed to 1.2865 units (same as used for Fe). If this is not what you are attempting to ask about, then I have no clue. $\endgroup$ – Ed V Sep 1 '19 at 14:47
  • $\begingroup$ @EdV Your prediction of my question is absolutely correct.That is exactly what I am asking for.Like % mass of Fe in FeO is something that I understand but what does the reciprocal of that percentage give us?That is my question. $\endgroup$ – user586228 Sep 1 '19 at 14:49

Based on the comments and extended chat, this is the kind of question the OP actually wanted to ask: Consider a mixture of Fe and FeO. The total percentage (by weight) of iron is 86.638%. What is the percentage of iron metal in the mixture?

Answer: For simplicity, assume exactly 100 g of the mixture. Hence the total iron is 86.638g and the remaining 13.362g is oxygen. Therefore, the number of moles of oxygen atoms is 0.83516 moles, i.e., 13.362g divided by 15.9994g/mole. This is also the number of moles of Fe in FeO, so the mass of Fe in FeO is 46.639g, i.e., 0.83516 moles times 55.845 g/mole. Subtracting this from 86.638g yields 39.999g metallic iron. So the 100 g of mixture was 39.999g of Fe and 60.001g of FeO, i.e., 40% Fe and 60% FeO.


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