Question

An oxide of nitrogen contains 25.9%nitrogen.what is it's emperical formula?

Answer 1

N2 =25.9 ÷ 14 = 1.85
O2 = 74.1 ÷ 16 = 4.63


N2 = 1.85 ÷ 1.85 = 1
Oxygen = 4.63 ÷ 1.85 = 2.5

N2o5 is he empirical farmUla hope it help you. .. Mark as brainlist plz PlzZ plz PlzZ plz PlzZ plz PlzZ plz

Answer 2

Hey there !

Solution :

Given Percentage = 25.9 %

So let there be 100 g of that oxide of nitrogen out of which 25.9 g is nitrogen.

So remaining amount of mass is 100 g - 25.9 g = 74.1 g 

Therefore this 74.9 g is amount of oxygen.

So let us calculate the number of moles.

Nitrogen :

Atomic Mass = 14 g

Given Mass = 25.9 ≈ 26 g

Number of Moles = Given Mass / Atomic Mass

=> Number of Moles = 26 / 14  = 1.85 Moles 

Oxygen :

Given Mass = 74.1 g ≈ 74 g

Atomic Mass = 16 g

Number of Moles = 74 / 16 = 4.625 Moles

Hence Number of moles must be divided with least number of moles

=> 1.85 is least, hence it must be divided by 1.85. 

So we get Nitrogen = 1.85 / 1.85 = 1 mole

Oxygen = 4.625 / 1.85 = 2.5

Since fraction cannot occur in Empirical formula it must be multiplied by a whole number to make the decimal a whole number. Hence it must be multiplied by 2.

Hence we get Nitrogen = 1 * 2 = 2 moles, Oxygen = 2.5 * 2 = 5 moles

Hence Empirical Formula = N₂O₅

Hope my answer helped :-)