ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: wildareal on December 24, 2010, 03:51:59 pm
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An impure sample of iron (II) sulfate, weighing 1.545g, was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate 0.315g, calculate the percentage of iron in the sample.
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Find the molar mass of Fe2O3, then find the percentage composition of iron in Fe2O3 according to molar mass.
Using that percentage, then, find that certain percentage of 0.315 grams
you will get a mass out of that.
that mass you just found / the impure mass sample x100 = percentage of iron in sample
If you don't get it still, I can work it out, but it's relatively simple.
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From memory, hope it's right (my brain has atrophied from disuse!)
n (Fe2CO3) = .315 / 171.8 = .0018335 mol
n (Fe) = .0018335 * 2 = .003667 mol
m (Fe) = .003667 * 55.9 = .205 grams
%m.sample (Fe) = .205/1.545 * 100/1 = 13.3%
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Or you could wait 34 seconds more for russ :D