Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Asked by Abhisek | 1 year ago |  87

##### Solution :-

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:

$$\dfrac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}$$

$$\dfrac{69.9}{55.85}$$

= 1.25

Relative moles of O in iron oxide:

$$\dfrac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}$$

$$\dfrac{30.1}{16.00}$$

= 1.88

Simplest molar ratio of Fe to O:

= 1.25: 1.88

= 1: 1.5

$$\approx 2: 3$$

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

Answered by Abhisek | 1 year ago

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