Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1
Answers
Answer:
We have given,
Mass percent of iron (Fe) =69.9%
Mass percent of oxygen (0)=30.1%
Number of moles of iron present in the oxide =69.90/55.85 =1.25
Number of moles of oxygen present in the oxide%3D30.1/16.0
=1.88
Ratio of iron to oxygen in the oxide,
1.25:1.88
=1:1.5
=2:3
Therefore the empirical formula of the oxide is Fe203.
Empirical formula mass of Fe203= [2(55.85) +3(16)]g Molar
mass of Fe203=159.69g
Thus, n= Molar mass/Empirical Formula mass =159.69/159.7g
=0.999 = 1(approx.)
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe203 and n is 1.
Hence, the molecular formula of the oxide is Fe203.
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Answer:
Here,
Mass percent of Fe = 69.9%
Mass percent of O = 30.1%
No. of moles of Fe present in oxide
= \frac{69.90}{55.85}55.8569.90
= 1.25
No. of moles of O present in oxide
= \frac{30.1}{16.0}16.030.1
=1.88
Ratio of Fe to O in oxide,
= 1.25: 1.88
= \frac{1.25}{1.25}:\frac{1.88}{1.25}1.251.25:1.251.88
=1:1.51:1.5
= 2:32:3
Therefore, the empirical formula of oxide is Fe_{2}O_{3}Fe2O3
Empirical formula mass of Fe_{2}O_{3}Fe2O3
= [2(55.85) + 3(16.00)] gr
= 159.69 g
Therefore n = \frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}EmpiricalformulamassMolarmass=159.7g159.69g
= 0.999
= 1(approx)
The molecular formula of a compound can be obtained by multiplying n and the empirical formula.
Thus, the empirical of the given oxide is Fe_{2}O_{3}Fe2O3 and n is 1.
Hope it helps