Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively.

Asked by Abhisek | 1 year ago |  89

#### 1 Answer

##### Solution :-

Here,

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

No. of moles of Fe present in oxide

$$\dfrac{69.90}{55.85}$$

= 1.25

No. of moles of O present in oxide

=$$\dfrac{30.1}{16.0}$$

=1.88

Ratio of Fe to O in oxide,

= 1.25: 1.88

$$\dfrac{1.25}{1.25}:\dfrac{1.88}{1.25}$$

=$$1:1.5$$

= 2:3

Therefore, the empirical formula of oxide is $$Fe_{2}O_{3}$$

Empirical formula mass of $$Fe_{2}O_{3}$$

= [2(55.85) + 3(16.00)] gr

= 159. 7g

The molar mass of$$Fe_{2}O_{3}$$= 159.69g

Therefore n = $$\dfrac{Molar\;mass}{Empirical\;formula\;mass}$$

$$=\dfrac{159.69\;g}{159.7\;g}$$

= 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 $$Fe_{2}O_{3}$$ and n is 1.

Answered by Abhisek | 1 year ago

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