A compound is found to contain 26.94 % nitrogen and 73.06 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 104.02 g/mol. What is the molecular formula for this compound?

THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2

THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4

Explanation:

To calculate the empirical formula for the compound, we:

1. Write out the percentage weight of each elements

N = 26.94%

F = 73.06 %

2. Divide each by its atomic mass

( N= 14, F = 19)

N = 26.94 / 14 = 1.924

F = 73.06 / 19 = 3.845

3. Divide each by the smaller of the values

N = 1.924 / 1.924 = 1

F = 3.845 / 1.924 = 1,998

4. Round up to a whole number and write the empirical formula

N= 1

F = 2

So the empirical formula of the compound is N F2

To calculate the molecular formula, we:

(N F2 )n = molecular weight

( 14 + 19*2) n = 104.02

52 n = 104.02

n = 2.000

The molecular formula of the compound will be:

(N F2)2 = N2F4

In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4