# Analytical chemistry

posted by 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%

• Analytical chemistry - ,

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 1.00-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 - ,

Thank you very much!