Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.
Form the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+30.1= 100%
Step 1 calculation of simplest whole number ratios of the elements
Element |
Percentage |
Atomic mass |
Atomic ratio |
Simplest ratio |
Simplest whole no ratio |
Fe |
69.9 |
55.84 |
69.9/55.84=1.25 |
1.25= 1 |
2 |
O |
30.1 |
16 |
30.1/16 = 1.88 |
1.88=1.5 |
3 |
Step 2 writing the empirical formula of the compound
The empirical formula of the compound = Fe2 O3
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Why is mass dioxygen taken 16?
But the question said dioxygen
Excuse me actually the simplest whole number ratio should come Fe=1 & O=2 .why the answer is Fe=2 & O=3?
Correct ans sir
But it's a dioxygen
Why is mass of oxygen taken 16?
Why to double the simplest whole number ratio
Thank you
Its gud
I think they had double the simplest ratio than it become 2 and 3