Question

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

Answer 1

N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

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

Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

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

Divide by the smallest to normalize:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

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