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A -13.68 gram of sample of iron is heated in the presence of excess fluorine. A metal Fluoride is formed with a mass of 27.64g. Determine the empirical formula of the metal fluoride.

User Sherelle
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1 Answer

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Step-by-step 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

User Losee
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