What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

A. NO

B. N2O3

C. NO2

D. N2O

D. N₂O

Explanation:

Let's assume we have 100 g of the compound. That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N * (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O * (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms. The empirical formula is therefore N₂O.