Question 1.3
NCERT Class XI Chemistry
Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Concept
Empirical formula is the simplest integer ratio of the atoms of the constituents.Percentages are given, so take the mass of the oxide as 100 grams.
Then calculate the moles of Fe and O by using atomic masses as 56.00u and 16.00u.
The simplest ratio of the moles gives the empirical formula!
Video Explanation
Please watch this youtube video for a quick explanation of the solution:
Solution in Detail
$ \begin{aligned} &\text{Take 100 grams of the oxide}\\ &\text{(i) Iron (69.9\%) = 70 grams}\\ &\text{(ii) dioxygen } \text{(30.1\%) = 30 grams}\\\\ &\text{molar mass of oxygen } = 16.00\text{gm}\\ &\therefore \text{moles of O} = \frac{30}{16} = \frac{15}{8} \\\\ &\text{molar mass of Fe } = 56.00\text{gm}\\ &\therefore \text{moles of Fe} = \frac{70}{56} = \frac{5}{4}\\\\ &\text{simplest ratio Fe : O} = \frac 54: \frac {15}{8} = 2 : 3\\\\ &\text{Empirical formula is Fe}_2\text{O}_3\:\underline{Ans} \end{aligned} $
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