Answer:
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 Brainly, Thanks
