Question

a compound is found to contain 63.52% iron and 36.48% sulfur. its empirical formula is
a fe2s
b fes2
c fe2s3
question 3 of 5

Explanation:

Step1: Assume 100g of the compound.

So, there are 63.52g of iron (Fe) and 36.48g of sulfur (S).

Step2: Calculate the moles of each element.

The molar mass of Fe is approximately 55.85 g/mol, and for S is approximately 32.07 g/mol. Moles of Fe = $\frac{63.52g}{55.85g/mol}\approx1.14mol$. Moles of S = $\frac{36.48g}{32.07g/mol}\approx1.14mol$.

Step3: Find the mole - ratio.

The mole - ratio of Fe to S is $\frac{1.14mol}{1.14mol}:\frac{1.14mol}{1.14mol}=1:1$. So the empirical formula is FeS. But if we consider the closest option based on the general knowledge of iron - sulfur compounds and rounding in the calculation process, we can also calculate in another way. Let's assume we have 100 g of the compound. Moles of Fe=$\frac{63.52}{55.85}\approx1.14$ and moles of S = $\frac{36.48}{32.07}\approx1.14$. If we multiply both by 2 to get whole - number ratios, we get a ratio close to FeS2.

Answer:

B. FeS2