I have to determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ dioxygen by mass in it.

I have started as follows:

  1. For our convenience we take $\pu{100 g}$ of that iron oxide:
    We have $\pu{69.9 g}$ of iron in it and $\pu{30.1 g}$ dioxygen in it.

  2. We convert into amount of substance of each element:

    • $\text{amount of substance of iron} = 69.9/55.8 = 1.25$
    • $\text{amount of substance of dioxygen} = 30.1/32 = 0.94$

Is my method and calculations right?

  • $\begingroup$ Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms". $\endgroup$ – Martin - マーチン Feb 20 '18 at 10:49

First assume the mass of an oxide of iron to be $\pu{100 g}$ which implies mass of iron would be $\pu{69.9 g}$ and mass of oxygen would be $\pu{30.1 g}$. Now calculate amount of substance of both the elements by dividing the assumed mass by their atomic number.
Then calculate relative amount of substance of each element by dividing the obtained amount of substance of each element by the smallest of them. (E.g. suppose you've got $\pu{5 mol}$ of iron and $\pu{2 mol}$ of oxygen, divide 5 by 2 and 2 by 2 to get relative amounts of substance of both.)
Now just write relative amount of substances in the form of a ratio and bring the obtained ratio to least simplified form. (Remember the ratio should come in whole numbers.) After obtaining ratio, write the empirical formula as I did in this example. (eg. suppose I got this ratio $2:5$ from last example. So the empirical formula will be $\ce{Fe2O5}$ .)

Actually, you can skip the relative amount of substance step when you've got 2 elements in empirical formula, but you will need to perform all the steps when there are more than 2 elements as the answer comes to be different in some cases.


You've been doing everything correctly so far - what we have now is the relative amount of substance in moles of each element in this $\pu{100 g}$ compound. As you correctly calculated, we have $\pu{1.25 mol}$ of $\ce{Fe}$ and $\pu{0.94 mol}$ of $\ce{O2}$ - for the sake of simplicity we'll say it's $\pu{1.88 mol}$ of $\ce{O}$ (I multiplied by 2).

Now we just take this into a calculator! $1.88/1.25$ is approximately $1.5$, so the correct answer would be $\ce{Fe2O3}$. The reason we divide large by small is that it's easier to see whole number ratios from the values that are larger than one - $0.66$ being converted to $2/3$ may not be as obvious as $1.5$ being converted to $3/2$, for example.

Why did we divide? so we've calculated the relative ratio of the amounts of substances in this compound - this means for any compound of any mass, we'll always get the ratio of $3/2$ of $\ce{O/Fe}$, right? Which means the empirical formula is $\ce{Fe2O3}$.


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