A compound is found to contain 13.65 % carbon and 86.35 % fluorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 88.01 g/mol. The molecular formula for this compound is .

The molecular formula will be

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 C= 13.65 g

Mass of F = 86.35 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of F=

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

For C =

For F =

The ratio of C : F= 1: 4

Hence the empirical formula is

The empirical weight of = 1(12)+4(19)= 88g.

The molecular weight = 88.01 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=