Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

Answer:- Below-mentioned are the steps on how to determine the empirical formula for a compound that is 36.86% N and 63.14% O by mass:

1. Determine the moles of each element.
   • Convert the mass percents to grams.
   • Divide the mass of each element by its molar mass.

For example, the mass of nitrogen in 100 grams of the compound is 36.86 grams. The molar mass of nitrogen is 14.01 grams/mol, so the number of moles of nitrogen is 36.86 grams / 14.01 grams/mol = 2.631 moles.

The mass of oxygen in 100 grams of the compound is 63.14 grams. The molar mass of oxygen is 16.00 grams/mol, so the number of moles of oxygen is 63.14 grams / 16.00 grams/mol = 3.946 moles.

1. Divide the number of moles of each element by the smallest number of moles.
   • The smallest number of moles is 2.631 moles.
   • Dividing the number of moles of nitrogen by 2.631 moles gives 1.
   • Dividing the number of moles of oxygen by 2.631 moles gives 1.5.
2. Write the subscripts of the elements in the empirical formula.
   • The subscripts of the elements in the empirical formula are the smallest whole-number ratios of the number of moles of the elements.
   • In this case, the subscripts of nitrogen and oxygen are both 1.

Therefore, the empirical formula of the compound is N2O3.