A compound contains only c, h, and n. it is 58.51% c and 7.37% h by mass. helium effuses through a porous frit 3.20 times as fast as the compound does. determine the empirical and molecular formulas of this compound. (type your answer using the format c2h2n for c2h2n.

Let us assume that there are 100 grams of substance present. The masses of the constituents will be:

C - 58.51 g

H - 7.37 g

N - 100 - 58.51 - 7.37 = 34.12 g

Next, we determine the moles of each substance by dividing by the atomic masses:

C - 58.51/12 = 4.88

H - 7.37/1 = 7.37

N - 34.12/14 = 2.44

Dividing all of the moles by the lowest number of moles, which is 2.44 of nitrogen, we get the ratio:

C - 2

H - 3

N - 1

The empircal formula becomes C₂H₃N

Next, we apply Graham's Law of diffusion, which mathematically is:

v₁/v₂ = √M₂ / √M₁

Let species 1 be helium, meaning v₁/v₂ = 3.2 and M₁ = 4

3.2 = √M₂ / √4

M₂ = 40.96

The mass of the empircal formula was:

12 * 2 + 1 * 3 + 14 = 41

The empircal formula is the molecular formula as their masses are approximately the same. This compound is C₂H₃N