Answer :
Explanation:
A 13.68 g sample of iron is reacted in presence of excess fluorine and an iron fluoride with a mass of 27.64 g is formed.
We have to find the empirical formula of our compound. This compound has two elements: Fe and F. We can express it like:
metal fluoride = FeₐFₓ
Where a is the subscript of Fe and x is the subscript for F.
First we have to find the mass of F present in the compound. We know that our compound has a total mass of 27.64 g and the mass of Fe present is 13.68 g. Since it only has two elements we can find the mass of F.
mass of sample = mass of Fe + mass of F
mass of F = mass of sample - mass of Fe
mass of F = 27.64 g - 13.68 g
mass of F = 13.96 g
So we found that our sample has 13.96 g of F and 13.68 g of Fe. Now we can determine the number of moles of each element that are present in the sample by using their molar masses.
molar mass of Fe = 55.85 g/mol
molar mass of F = 19.00 g/mol
moles of Fe = 13.68 g * 1 mol/(55.85 g)
moles of Fe = 0.245 moles
moles of F = 13.96 g * 1 mol/(19 g/mol)
moles of F = 0.735 moles
By definition the empirical formula is "simplest whole number ratio of atoms present in a compound". So if we want to find the ratio between them we have to divide both of them by the smallest number.
a = subscript of Fe = 0.245/0.245 = 1
x = subscript of F = 0.735/0.245 = 3
So the empirical formula of our compound is:
empirical formula = FeₐFₓ
empirical formula = FeF₃
Answer: empirical formula = FeF₃
