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 1.00-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.
Thank you very much!
To calculate the mass fractions of Fe2O3 and FeOOH in the sample, you need to follow a series of steps:
1. Calculate the number of moles of H2O produced:
- Given that the mass of H2O produced is m(H2O) = 0.0097g
- Calculate the number of moles of H2O using the molar mass of H2O (18g/mol):
n(H2O) = m(H2O) / M(H2O) = 0.0097g / 18g/mol = 0.000538888 mol
2. Calculate the number of moles of FeOOH in the sample:
- Since the stoichiometric ratio between FeOOH and H2O is 2:1, the number of moles of FeOOH is twice the number of moles of H2O (n(H2O)):
n(FeOOH) = 2 * n(H2O) = 2 * 0.000538888 mol = 0.001077777 mol
3. Calculate the mass of FeOOH in the sample:
- Use the molar mass of FeOOH (88.86g/mol) to convert the number of moles of FeOOH to grams:
m(FeOOH) = n(FeOOH) * M(FeOOH) = 0.001077777 mol * 88.86g/mol = 0.095771264 g
4. Calculate the mass fraction of FeOOH:
- Divide the mass of FeOOH by the initial mass of the sample:
w(FeOOH) = m(FeOOH) / m(sample) = 0.095771264 g / 0.2564 g = 0.3735 or 37.35%
5. Calculate the mass of Fe2O3 in the sample:
- Subtract the masses of H2O and FeOOH from the initial mass of the sample to obtain the mass of Fe2O3:
m(Fe2O3) = m(sample) - m(H2O) - m(FeOOH) = 0.2564 g - 0.0097 g - 0.095771264 g = 0.150928 g
6. Calculate the mass fraction of Fe2O3:
- Divide the mass of Fe2O3 by the initial mass of the sample:
w(Fe2O3) = m(Fe2O3) / m(sample) = 0.150928 g / 0.2564 g = 0.5886 or 58.86%
Therefore, the mass fractions of Fe2O3 and FeOOH in the sample are approximately 58.86% and 37.35%, respectively.