Sure, let's break down the problem step-by-step for part (c) of the question, which involves determining the percentage of iron (Fe) in an iron ore sample following a gravimetric procedure.
### (c) (i) Writing the equation of the gravimetric factor for the analytical procedure
The gravimetric factor is used to relate the mass of the analyte (Fe) to the mass of the compound weighed (Fe2O3) in a gravimetric analysis.
Given:
- Molar mass of Fe (Fe) = 55.85 g/mol
- Molar mass of O (oxygen) = 16.00 g/mol
The gravimetric factor for converting Fe2O3 to Fe can be determined as follows:
1. Calculate the molar mass of Fe2O3:
- Fe2O3 consists of 2 atoms of iron (Fe) and 3 atoms of oxygen (O).
- Molar mass of Fe2O3 = [tex]\(2 \times 55.85 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol}\)[/tex]
- Molar mass of Fe2O3 = [tex]\(2 \times 55.85 + 3 \times 16 = 111.70 + 48.00 = 159.70\)[/tex] g/mol
2. Determine the gravimetric factor (GF) which represents the fraction of the mass of Fe in the mass of Fe2O3:
- The gravimetric factor = [tex]\(\frac{\text{mass of Fe in Fe2O3}}{\text{molar mass of Fe2O3}}\)[/tex]
- Mass of Fe in Fe2O3 = [tex]\(2 \times 55.85\)[/tex] g/mol = 111.70 g/mol
- Therefore, gravimetric factor = [tex]\(\frac{111.70 \, \text{g/mol}}{159.70 \, \text{g/mol}} = 0.6994\)[/tex]
So, the gravimetric factor equation is:
[tex]\[ \text{Gravimetric Factor} = \frac{2 \times \text{Molar Mass of Fe}}{\text{Molar Mass of Fe2O3}} \][/tex]
### (c) (ii) Determining the percentage of Fe in the ore
Given data:
- Mass of the iron ore sample = 2.2648 g
- Mass of pure Fe2O3 obtained = 1.0790 g
1. Calculate the mass of Fe in the obtained Fe2O3 using the gravimetric factor:
- Mass of Fe = Mass of Fe2O3 [tex]\(\times\)[/tex] Gravimetric factor
- Mass of Fe = 1.0790 g [tex]\(\times\)[/tex] 0.6994 = 0.7547 g
2. Determine the percentage of Fe in the ore sample:
- Percentage of Fe = [tex]\(\left(\frac{\text{Mass of Fe}}{\text{Mass of ore sample}}\right) \times 100\%\)[/tex]
- Percentage of Fe = [tex]\(\left(\frac{0.7547 \, \text{g}}{2.2648 \, \text{g}}\right) \times 100\%\)[/tex]
- Percentage of Fe = 33.32 %
Therefore, the percentage of Fe in the ore is approximately 33.32%.
