Answer: FeO2
Explanation:
1.116 g Fe
0.480 g O
We want to compare the NUMBER of Fe and O atoms. We only have their masses, but since the molar mass of Fe is 5 times that of O, let's multiply the grams of O by 5 to have it "catch up" to the iron in ternms of number of atoms. If the molar masses were the same, the masses would be in direct proportion to the formula. Since they aren't, multiply by the molar mass ratio:
(1.116 grams Fe) x 1 = 1.116 g Fe
(0.480 g O) x 5 = 2.40 g O [Changes the O mass into a number that correlates with the number of atoms relative to iron],
If the molar masses were the same, this would be the ratio of O to Fe atoms in the compound. Normalize to the Fe number by dividing both by 1.116:
Fe = 1
O = 2.15
Not perfect, but an empiricle formula of FeO2 makes sense. I don't see enough information here to prove that this is also the molecular formula, but FeO2 is a valid compound. Fe(IV)O2,
1. The empirical formula of the compound is Fe₂O₃
2. The molecular formula of the compound is Fe₂O₃
1. Determination of the empirical formula of the compound.
Fe = 1.116 g
O = 0.480 g
Divide by their molar mass
Fe = 1.116 / 56 = 0.02
O = 0.480 / 16 = 0.03
Divide by the smallest
Fe = 0.02 / 0.02 = 1
O = 0.03 / 0.02 = 1.5
Multiply by 2 to express in whole number
Fe = 1 × 2 = 2
O = 1.5 × 2 = 3
Therefore, the empirical formula of the compound is Fe₂O₃
2. Determination of the molecular formula of the compound.
Empirical formula = Fe₂O₃
Molar mass = 5 × molar mass of O₂ = 5 × 32 = 160 g/mol
Molecular formula = empirical × n = molar mass
[Fe₂O₃]n = 160
[(2×56) + (3×16]n = 160
[112 + 48]n = 160
160n = 160
Divide both side by 160
n = 160 / 160
n = 1
Molecular formula = [Fe₂O₃]n
Molecular formula = [Fe₂O₃] × 1
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