Question

Determine the empirical formula of a hydrate FexFy.ZH2O. It contains 0.676 g of iron combined with 0.460 g of fluorine and 0.870 g of water (H2O).

Answer 1

Hey there!:

Number of moles of iron:

Molar mass iron: 55.85 g/mol

0.676 / 55.85 => 0.012 moles of iron

Number of moles of fluorine:

Molar mass fluorine : 19 g/mol

0.460 / 19 => 0.024 moles of fluorine

Number of moles of water:

Molar mass water : 18.02 g/mol

0.870 / 18.02 => 0.048 moles of water

Simplest mol ratio is :

iron ( Fe ) = 0.012 / 0.012 => 1.0 mol

Florine ( F ) = 0.024 / 0.012 => 2.0 moles

Water ( H2O ) = 0.048 / 0.012 => 4 moles

Therefore:

Empirical formula is:

FeF2.4H2O

Hope that helps!