Determine the empirical formula of an oxide of iron which has 77.7percent iron and 22.3 percent dioxygen by mass.
Answers
Explanation:
Answer : The empirical formula of the compound is,
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Fe = 69.9 g
Mass of O = 30.1 g
Molar mass of Fe = 55.85 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of Fe =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Fe =
For O =
The ratio of Fe : O = 1 : 1.5 = 2 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
Thus, the Empirical formula =
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Emperical Formula Of Iron Oxide is Fe2O3
Explanation:
Percent of Fe by mass = 69.9 % [As Given Above]
Percent of O2 by mass = 30.1 % [As Given Above]
Relative moles of Fe in Iron Oxide:
=
=
= 1.25
Relative moles of O in Iron Oxide:
=
=
= 1.88
Simplest molar ratio of Fe to O:
= 1.25: 1.88
= 1: 1.5
≈ 2:3