Empirical formula

Question

Please help, thank you :)

 

What is the empirical formula of a compound composed of 69.94% Fe and 30.06% O by mass?

Answer 1

Hi Beatrice,

The empirical formula gives us the ratio of atoms in a compound; in this case, we are looking for the ratio of Fe to O.

There are 3 steps to this calculation:

1) Covert mass% to moles

2) Divide both mole amounts by the lowest mole amount to get the mole ratio

3) adjust ratios to whole numbers

 

1) To convert from  mass % to moles, start by assuming you have 100 g of the compound.  (Any number of grams will work here but 100 g makes the calculations easier)

That means we have 69.94 g Fe and 30.06 g O.

 

Then we convert these to moles.

To convert our 69.94 g Fe to moles Fe, we divide by the atomic mass of Fe, 55.85g/mol.

69.94g / 55.85 g/mol = 1.25 moles Fe

To convert 30.06 g O to moles O, we divide by the atomic mass of O, 16.00g/mol

30.06g / 16.00g/mol = 1.88 moles O

 

So... In our theoretical 100 g sample, we have 1.25 moles Fe and 1.88 moles O

 

2) Divide both mole amounts by the lowest mole amount to get the mole ratio

 

1.25 moles is the lower of the two, so we divide both by 1.25.

Fe = 1.25/1.25 = 1

O = 1.88/1.25 = 1.5

So we know the ratio of Fe to O is 1:1.5

 

3) adjust ratios to whole numbers

We can't have 1.5 atoms of O in our formula, so we need to adjust the numbers while keeping the 1:1.5 ratio.

To do this, we multiply both by 2, so we now have 2:3 instead of 1:1.5.  The ratio is the same but now both sides are whole numbers.

 

Now all we have to do is plug these numbers into our formula;

Fe₂O₃