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There is a series of nitrogen oxides with the general formula N?O?. What is the empirical formula of one that contains 30.45% nitrogen?

Question 13 options:

N2O


NO


NO2­


N2O3


N2O5

User Lilie
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1 Answer

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Answer: The empirical formula of one that contains 30.45% nitrogen is
NO_(2).

Step-by-step explanation:

Given: Mass of nitrogen = 30.45 g

Let us assume that the mass of given oxide is 100 grams.

As the atomic mass of nitrogen is 14.0067 g. So, moles of nitrogen will be calculated as follows.


Moles = (mass)/(molarmass)\\= (30.45 g)/(14.0067 g/mol)\\= 2.17 mol

Also, mass of oxygen = (100 - 30.45) g = 69.55 g

Atomic mass of oxygen is 15.9994 g/mol. So, moles of oxygen will be as follows.


Moles = (mass)/(molarmass)\\= (69.55 g)/(15.9994 g/mol)\\= 4.34 mol

The ratio of both the atoms is as follows.


(4.34)/(2.17) = 2

This means that gas has 2 moles of oxygen to 1 mole of nitrogen. Hence, the formula of oxide is
NO_(2).

Thus, we can conclude that the empirical formula of one that contains 30.45% nitrogen is
NO_(2).

User Evgeny  Bovykin
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