Calculate the empirical formula of a laughing gas which is 63.62%nitrogen and 36.4% oxygen.

Lets take a 100g sample of Laughing Gas

Since there is 63.62% Nitrogen by mass, we have 63.62 grams of Nitrogen

Since there is 36.4% Oxygen by mass, we have 36.4 grams of Oxygen

Calculating the number of moles of Nitrogen and Oxygen

Moles of N = 63.62/14 = 4.5 moles

Moles of O2 = 36.4/16 = 2.25 moles

Empirical formula  = Moles of N/ Moles of O2

Empirical Formula = 4.5 / 2.25

N : O2 = 2 : 1

Therefore, the Empirical formula for Laughing Gas is N₂O

Kindly Mark , Thanks

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