Analytical chemistry
posted by Gloria on .
I did the next problem, i don't have the results for it and i just want to know if i calculated the mass for Fe2O3 the right?
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%

I 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 1.00fraction 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 Fe2O3mass FeOOH)/mass sample gives the same answer as fraction Fe2O3 = 1.00fraction FeOOH. 
Thank you very much!