Determine the emprical formula of an oxide of iron, which has 69.9% iron and 30.1% oxygen by mass.
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Explanation:
Atomic weight of iron is 56 amu
Atomic weight of oxygen is 16 amu.
= 69.9/56= 1.2.
= 30.1/16= 1.8
Simple ratio:
1.2/1.2=1. (1×2)
And, 1.8/1.2= 3/2. (3/2×2)
Fe = 2
and, O= 3
therefore, Fe2O3 is oxide obtained.
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