Acompound composed of only c and f contains 17.39 c by mass. what is its empirical formula

Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

C = 17.39 / 12 = 1.45

F = 82.61 / 19 = 4.35

Divide the two by the smaller number, so divide by 1.45

C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

So the empirical formula is:

CF3 

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.