A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

A compound has 63.52% iron and 36.48% sulfur. I assume this is the percentage of the mass of the two elements as found in the compound of which we have to determine the empirical formula.

We know that the atomic mass of iron is 56 and the atomic mass of...

Unlock This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

A compound has 63.52% iron and 36.48% sulfur. I assume this is the percentage of the mass of the two elements as found in the compound of which we have to determine the empirical formula.

We know that the atomic mass of iron is 56 and the atomic mass of sulfur is 32.

Let the empirical formula of the compound be Fe(x)S(y)

As the mass of the sulfur in the compound and the mass of the iron in compound is in the ratio 36.48/63.52, we have: 32*y/ 56*x = 36.48/63.52

=> 32y/56x = 36.48/63.52

=> y/x = (36.48/63.52)*(56/32)

=> y/x = 1

This gives an equal number of sulfur and iron atoms in the compound.

Therefore the empirical formula of the compound is FeS.

Approved by eNotes Editorial Team