A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by mass.
Question 1: The empirical formula for this compound is :

Question 2: The molar mass for this compound is 82.98 g/mol.
The Molecular formula for this compound is:

1. The empirical formula is  

2. The molecular formula is

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =

Moles of N =

Moles of F =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P =

For N =

For F =

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is

The empirical weight of = 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=