Saturday

August 23, 2014

August 23, 2014

Posted by **Gloria** on Saturday, September 7, 2013 at 1:54pm.

A sample is a mixture of iron oxides (Fe2O3 and FeOOH). It's known that at 200°C this reaction occurs:

2FeOOH(s)-> Fe2O3(s) + H2O(g)

A 0.2564g sample is heated and we realize that the mass of the sample reduced for 0.0097g. Calculate the mass fractions of Fe2O3 and FeOOH in the sample.

m(H2O)=0.0097g

n(H2O)=0.0097g/18(g/mol)=0.000538888 mol

n(FeOOH)=2*n(H2O)=0.001077777 mol

m(FeOOH)=n*M(88.86 g/mol)=0.095771264 g

w(FeOOH)=m(FeOOH)/0.2564g= 37.35%

m(Fe2O3)=m(sample)-m(H2O)-m(FeOOH)=0.2564g - 0.0097g - 0.095771264g= 0.150928 g

w(Fe2O3)=0.150928g/0.2564g= 58.86%

- Analytical chemistry -
**DrBob222**, Saturday, September 7, 2013 at 6:17pmI think the answer for FeOOH is right (BUT note the problem asks for the fraction and not percent; therefore, 0.3735 is the faction). Also note that the 0.0097 has just two significant figures so you should adjust the fraction for just two s.f.

I don't agree with the last part. First, the problem states that the sample is composed of Fe2O3 and FeOOH (nothing about anything else there) so

fraction Fe2O3 must be 1.00-fraction FeOOH.

Second, I think the error is

mass Fe2O3 = mass sample - mass FeOOH. Subtracting the H2O means you are removing some of the mass of the FeOOH.

Note also that (mass Fe2O3-mass FeOOH)/mass sample gives the same answer as fraction Fe2O3 = 1.00-fraction FeOOH.

- Analytical chemistry -
**Gloria**, Saturday, September 7, 2013 at 6:35pmThank you very much!

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