Question

3.08 x 10-3 mol fecl₃ was identified through titration from the 0.700 g impure sample. what is the percent fecl₃ in the original sample? ? % fecl₃

Explanation:

Step1: Calculate molar mass of FeCl₃

Fe: 55.85 g/mol, Cl: 35.45 g/mol. Molar mass $M = 55.85 + 3\times35.45 = 55.85 + 106.35 = 162.2$ g/mol.

Step2: Find mass of FeCl₃

Using $m = n\times M$, where $n = 3.08\times10^{-3}$ mol. So $m = 3.08\times10^{-3} \times 162.2$ g. Calculate: $3.08\times162.2\times10^{-3} \approx 0.500$ g.

Step3: Calculate percent composition

Percent = $\frac{\text{mass of FeCl}_3}{\text{mass of sample}} \times 100\% = \frac{0.500}{0.700} \times 100\% \approx 71.4\%$.

Answer:

$\approx 71.4\%$