Question

You react 4.40 grams of iron filings (elemental iron) with excess sulfur in an experiment. You find that the mass of your iron sulfide product is 5.67 g. The empirical formula of that product is Fe3S2

Ir2S3

FeS

Fe2S

IrS

Answer 1

Sol.

As Mass of iron , Fe = 4.40 g

Molar Mass of Fe = 55.845 g/mol

So , Moles of Fe = 4.40 g / 55.845 g/mol = 0.0788 mol

Also , Mass of sulphur , S

= Mass of iron sulfide product - Mass of Iron

= 5.67 g - 4.40 g

= 1.27 g

Molar Mass of S = 32.065 g/mol

Moles of S = 1.27 g / 32.065 g/mol = 0.0396 mol  

Divide both the moles of Fe and S by the smallest number

Moles of Fe = 0.0788 / 0.0396 = 2

Moles of S = 0.0396 / 0.0396 = 1

Therefore ,

Emperical Formula of iron sufide is Fe2S