An analysis of a compound shows that it consists of 36.8% nitrogen and 63.2% oxygen

This question is incomplete, however, it is asking to calculate the empirical formula of the compound.

The empirical formula is N2O3

Explanation:

The following information about a compound was got from the question:

N = 36.8%

O = 63.2%

It can be said that the compound contains 36.8g of nitrogen (N) and 63.2g of oxygen (O).

Next, we convert each mass composition to moles by dividing by their respective molar masses.

N = 36.8g ÷ 14g/mol = 2.628mol

O = 63.2g ÷ 16g/mol = 3.95mol

Next, we divide each mole value by the smallest value (2.628mol).

N = 2.628mol ÷ 2.628 = 1

O = 3.95mol ÷ 2.628 = 1.5

We multiply each value by 2 to get a empirical ratio.

N = 2

O = 3

The ratio of N:O is 2:3, hence, the empirical formula is N2O3.

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