Anew compound contains nitrogen, hydrogen, boron, and fluorine. the assay values are: nitrogen, 13.360%; hydrogen, 3.8455%; boron, 10.312%. determine its empirical formula.

The empirical formula for the given compound is

Explanation:

We are given:

Percentage of N = 13.360 %

Percentage of H = 3.8455 %

Percentage of B = 10.312 %

Percentage of F = [100 - 13.360 - 3.8455 - 10.312] = 72.4825 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of N = 13.360 g

Mass of H = 3.8455 g

Mass of B = 10.312 g

Mass of H = 72.4825 g

To formulate the empirical formula, we need to follow some steps:

Moles of Nitrogen =

Moles of Hydrogen =

Moles of Boron =

Moles of Fluorine =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.937 moles.

For Nitrogen =

For Hydrogen =

For Boron =

For Fluorine =

The ratio of N : H : B : F = 1 : 4 : 1  : 4

Hence, the empirical formula for the given compound is

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.